The purple arrow on the graph shows speed and direction.
What is this arrow called?

A. an acceleration arrow
B. a vector
C. a force arrow
D. a speed arrow

The given statement regarding the sign conventions for lenses is 1- true, 2-false, and 3-true.

The magnification, m, is negative for inverted images. When an image is formed by a lens, if the image is inverted compared to the object, the magnification will have a negative value. The first statement is true.

Virtual images appear on the opposite side of the lens from the object. Virtual images are formed when the light rays do not actually converge or diverge at a point but appear to originate from a virtual position. They are always formed on the same side of the lens as the object. The image distance for virtual images is positive. The second statement is false.

Real images appear on the opposite side of the lens from the object. Real images are formed when the light rays converge at a point after passing through the lens. They are formed on the opposite side of the lens from the object. The image distance for real images is negative. The third statement is true.

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A measuring stick placed along the circumference of a rotating disk will appear contracted, but not if it is oriented along a radius because of the effects of length contraction, also known as Lorentz contraction, which is a consequence of special relativity.

The theory of special relativity postulates that a moving object appears shorter along its direction of motion than when it is at rest. The length of an object appears to contract in the direction of motion due to time dilation and length contraction. As a result, if the measuring stick is placed along the circumference of a rotating disk and is moving with the disk's motion, it will appear to be shorter or contracted. However, if it is oriented along a radius and is not moving with the disk's motion, it will not appear to be shorter or contracted. Length contraction and time dilation are two of the fundamental principles of special relativity, which helps to explain the strange and unexpected behaviors of objects at speeds approaching the speed of light.

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To show that the electric field inside the insulating sphere is given by E = ρ(r - b)/(3ε₀), where ρ is the charge density, r is the distance from the centre of the sphere, b is the displacement of the centre from the origin, and ε₀ is the permittivity of free space, we can use Gauss's law.

Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. By applying Gauss's law, we can derive the electric field inside the insulating sphere.

Let's choose a Gaussian surface in the shape of a sphere with radius r, where r is less than the radius of the insulating sphere (a). Since the sphere is not centred at the origin but at a displacement of b, the centre of our Gaussian sphere will also be displaced by b.

According to Gauss's law, the electric flux through this Gaussian surface is given by:

Φ = E * A

where Φ is the electric flux, E is the electric field, and A is the area of the Gaussian surface.

Since the electric field is radially symmetric for a uniformly charged sphere, the electric field at any point on the Gaussian surface will have the same magnitude and direction. Therefore, the electric field can be taken out of the dot product with the area vector, and we have:

Φ = E * A = E * 4πr²

Now, we need to determine the charge enclosed by this Gaussian surface. Since the sphere has a uniform charge density (ρ), the charge enclosed within a sphere of radius r is given by:

Q = (4/3)πr³ρ

Now, applying Gauss's law, we have:

Φ = Q / ε₀

Substituting the expressions for Φ and Q, we get:

E * 4πr² = (4/3)πr³ρ / ε₀

E = (1/3) * r * ρ / ε₀

Since r is the distance from the origin, and the sphere is displaced by b, we can rewrite r as (r - b). Therefore:

E = ρ(r - b) / (3ε₀)

Therefore, we have shown that the electric field inside the insulating sphere is given by E = ρ(r - b) / (3ε₀), where r is the distance from the origin, b is the displacement of the sphere from the origin.

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Answer:

Polaroid fliter

Explanation:

light can be polarized by using Polaroid filters

Polaroid fliter are made of special material that is capable of blocking one of the two planes of vibration of an electromagnetic wave

hope this is useful--(have a good day)

According to the equations of motion, the time taken to reach the ground is 5 seconds.

Using;

s = ut + 1/2gt^2

s = distance

u = initial velocity

t = time taken

g = acceleration due to gravity

Note that u = 0 m/s since the object was dropped from a height

Substituting values;

125 = 1/2 × 10  × t^2

125 = 5t^2

t^2 = 125/5

t^2 = 25

t = 5 secs

Velocity on reaching the ground is obtained from

v = u + gt

Where u = 0 m/s

v = gt

v = 10 × 5

v = 50 m/s

At half the time it takes to reach the ground;

s =  ut + 1/2gt^2

Where u = 0 m/s

s =  1/2gt^2

s = 1/2  × 10  × (2.5)^2

s = 31.25 m

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Answer:

(i) time taken by the ball to reach the

ground is 5 sec.

(ii) velocity of the ball on reaching the

ground is 50 m/s.

(iii) the height of the ball at half the time it

takes to reach the ground is 31.25 m.

Step-by-step explanation:

(i) time taken by the ball to reach the

ground

[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]

[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]

[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]

Hence, the ball taken 5 sec to reach the ground.

[tex]\begin{gathered}\end{gathered}[/tex]

(ii) velocity of the ball on reaching the

ground

[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]

[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]

Hence, the velocity of ball is 50 m/s.

[tex]\begin{gathered}\end{gathered}[/tex]

(iii) the height of the ball at half the time it

takes to reach the ground.

[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]

Hence, the height of the ball to reach the ground is 31.25 m.

[tex]\underline{\rule{220pt}{3.5pt}}[/tex]

The magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is 5.68 × 10⁴ N/C.

To calculate the electric field inside the solid at a given distance from the center of the cavity, we need to consider the contributions from both the point charge in the cavity and the charge density in the solid.

Let's break down the calculation step by step:

1. Electric field due to the point charge in the cavity:

The electric field at a point inside the solid due to the point charge in the cavity can be calculated using the formula:

E_point = k * |Q_point| / r²

where

E_point is the electric field due to the point charge,

k is the Coulomb's constant (8.99 × 10^9 N m²/C²),

|Q_point| is the magnitude of the point charge (-3.00 μC = -3.00 × 10⁻⁶C),

and r is the distance from the center of the cavity to the point inside the solid (9.50 cm = 0.095 m).

Substituting the values into the formula, we get:

E_point = (8.99 × 10⁹ N m²/C) * |-3.00 × 10⁶ C| / (0.095 m)²

E_point = 2.85 × 10⁷N/C

2. Electric field due to the charge density in the solid:

The electric field at a point inside the solid due to the charge density can be calculated using the formula:

E_density = (k * ρ * r) / (3ε0)

where

E_density is the electric field due to the charge density,

ρ is the charge density (7.35 × 10^(-4) C/m³),

r is the distance from the center of the cavity to the point inside the solid (9.50 cm = 0.095 m),

and ε0 is the permittivity of free space (8.85 × 10⁻¹² C²/N m²).

Substituting the values into the formula, we get:

E_density = [(8.99 × 10⁹N m²/C²) * (7.35 × 10⁻⁴C/m³) * (0.095 m)] / (3 * 8.85 × 10⁻¹² C²/N m²)

E_density = 1.06 × 10^8 N/C

3. Total electric field inside the solid:

To find the total electric field at the given point inside the solid, we need to sum the contributions from the point charge and the charge density. Since the charges have opposite signs, we subtract the magnitudes:

E_total = |E_point| - |E_density|

E_total = 2.85 × 10⁷N/C - 1.06 × 10⁸N/C

E_total = -7.78 × 10⁷N/C

However, the electric field is a vector quantity, and its direction is radial, pointing from the center of the cavity towards the point inside the solid.

Therefore, the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is:

|E_total| = 7.78 × 10⁷N/C

The magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity is 5.68 × 10

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The distance - time graph of the humming bird is missing, so i have attached it.

Answer:

Instantaneous velocity = 0.5 m/s

Explanation:

From the attached graph, at time t = 1 s, the corresponding distance is 0.5 m.

Instantaneous velocity is the velocity at that point.

Thus;

Instantaneous velocity = 0.5/1

Instantaneous velocity = 0.5 m/s

Solution :

Given :

Force, F = 500 N

Let [tex]$ \vec F = F_x\ \hat i + F_y\ \hat j$[/tex]

      [tex]$|\vec F|=\sqrt{F_x^2+F_y^2}$[/tex]

∴ [tex]$F_x=F \cos 60^\circ = 500 \ \cos 60^\circ = 250 \ N$[/tex]

  [tex]$F_y=-F \cos 30^\circ = -500 \ \cos 30^\circ = -433.01 \ N$[/tex]   (since [tex]$F_y$[/tex] direction is in negative y-axis)

[tex]$F=250 \ \hat i - 433.01 \ \hat j$[/tex]

So scalar components are : 250 N and 433.01 N

     vector components are : [tex]$250 \ \hat i$[/tex] and [tex]$-433.01\ \hat j$[/tex]

1.  Scalar components along :

 x' axis = 500 N, since the force is in this direction.

 [tex]$F_{x'}= F \ \cos \theta = 500\ \cos \theta$[/tex]

  Here, θ = 0° , since force and axis in the same direction.

So, cos θ = cos 0° = 1

∴ [tex]$F_{x'}=500 \times 1=500\ N$[/tex]

[tex]$F_{y'}= F \ \sin \theta = 500\ \sin 0^\circ=500 \times 0=0$[/tex]

[tex]$F_{y'}=F\ cos \theta$[/tex] but here θ is 90°. So the force ad axis are perpendicular to each other.

[tex]$F_{y'}=F\ \cos 90^\circ= 500 \ \cos 90^\circ = 500 \times 0=0$[/tex]

∴ [tex]$F_{x'}= 500\ N \text{ and}\ F_{y'}=0\ N$[/tex]

2. Scalar components of F along:

x-axis :

[tex]$F_x=F\ \cos \theta$[/tex],   here θ is the angle between x-axis and F = 60°.

[tex]$F_x=500 \times \cos60^\circ=250\ N$[/tex]

y'-axis :

[tex]$F_{y'}=F\ \cos \theta$[/tex],   here θ is the angle between y'-axis and F = 90°.

[tex]$F_{y'}=500 \times \cos90^\circ=500\times 0=0\ N$[/tex]

∴ [tex]$F_{x}= 250\ N \text{ and}\ F_{y'}=0\ N$[/tex]

The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.

The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In this case, the mountain goat is standing on a slope.

When an object is on an inclined surface, the normal force can be split into two components: one perpendicular to the slope (normal to the surface) and one parallel to the slope (tangential to the surface).

The component parallel to the slope is responsible for counteracting the gravitational force pulling the object down the slope.

In this scenario, the mountain goat is standing on the slope, and we can assume it is in equilibrium, meaning it is not sliding down the slope. Therefore, the parallel component of the normal force is equal in magnitude and opposite in direction to the gravitational force acting down the slope.

The weight of the mountain goat is given as 650 N. This is the magnitude of the gravitational force acting on the goat.

The normal force acting on the goat is equal in magnitude but opposite in direction to the gravitational force. So the normal force is also 650 N.

The normal force acting on the mountain goat is 650 N. It supports the weight of the goat and keeps it in equilibrium on the slope.

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Answer:

Geothermal Energy.

Explanation:

Answer:

C. The lever applies three times more force than you hand can apply.

Explanation:

Since it's advantage is 3, that means you'll have to multiply the input of it by 3, making this apply 3x more force than your hand.

Hope this helped! <3

Source(s): Me and a bit of g*ogle for clarification

Answer & Explanation:

This can be explained with Newton's first law, Inertia - if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.  

When you walk, your entire body is in a forwarding motion. If your toe hits an object (in this case, a chair), only this unfortunate toe will stop while the rest of your body continues its forwarding motion, resulting in you falling forward.

The correct answer is iii) hydrogen. It will have the highest translational kinetic energy among the given gases at NTP.

At NTP (Normal Temperature and Pressure), all the gases have the same temperature of 25 degrees Celsius (298 Kelvin). According to the kinetic theory of gases, the average translational kinetic energy of gas molecules is directly proportional to the temperature.

The formula for translational kinetic energy is given by:

K.E. = (3/2) k T

Where:

K.E. is the translational kinetic energy

k is the Boltzmann constant (1.38 × 10^-23 J/K)

T is the temperature in Kelvin

Since the temperature is the same for all the gases at NTP, the gas with the highest translational kinetic energy will be the one with the lightest molecules. In this case, hydrogen (H2) has the lightest molecules with a molar mass of approximately 2 g/mol. Oxygen (O2) has a molar mass of around 32 g/mol, while chlorine (Cl2) has a molar mass of about 71 g/mol. Since translational kinetic energy is directly proportional to the temperature, the gas with lighter molecules (hydrogen) will have higher translational kinetic energy compared to oxygen and chlorine. option(iii)

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The tension in the string is approximately 46.7 N. To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.

To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.

At the top of the circle:

The downward gravitational force (mg) and the tension in the string (T) act downward.

The net force in the upward direction is provided by the centripetal force (Fc).

At the bottom of the circle:

The downward gravitational force (mg) and the tension in the string (T) act downward.

The net force in the upward direction is the sum of the centripetal force (Fc) and the car's weight (mg).

We can set up the following equations of motion at the top and bottom of the circle:

At the top:

T - mg = Fc ...(1)

At the bottom:

T + mg = Fc + mg ...(2)

We can substitute the expression for the centripetal force (Fc = mv^2 / r) into the equations:

At the top:

T - mg = mv^2 / r ...(3)

At the bottom:

T + mg = mv^2 / r + mg ...(4)

Now we can solve these equations to find the tension in the string.

At the top:

T - mg = mv^2 / r

T = mv^2 / r + mg ...(5)

At the bottom:

T + mg = mv^2 / r + mg

From equation (5), we can substitute the expression for T:

mv^2 / r + mg + mg = mv^2 / r + mg

2mg = mv^2 / r

Now we can solve for the tension (T):

T = mv^2 / r - mg

T = (1.61 kg)(12.0 m/s)^2 / 5.00 m - (1.61 kg)(9.8 m/s^2)

T ≈ 46.7 N

Therefore, the tension in the string is approximately 46.7 N.

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A. The magnitude of the electric field in the region r ≤ a is zero.

B. The magnitude of the electric field in the region a < r < b is zero.

C. The magnitude of the electric field in the region r ≥ b is Q/ε₀, where ε₀ is the permittivity of free space.

D. The charge on the inside surface of the hollow sphere is +Q.

E. The charge on the exterior surface of the hollow sphere is +2Q.

A. To determine the magnitude of the electric field in the region r ≤ a (inside the hollow sphere), we need to consider the superposition of electric fields from the point charge at the center and the hollow sphere.

The electric field inside a conducting hollow sphere is zero. This means that the electric field due to the hollow sphere cancels out the electric field due to the point charge at the center.

Therefore, the magnitude of the electric field in the region r ≤ a is zero.

B. In the region a < r < b (between the inner and outer radii of the hollow sphere), the electric field is zero because the charge on the inner surface of the hollow sphere distributes itself uniformly on the inner surface, creating an electric field that cancels out the electric field from the point charge at the center.

Therefore, the magnitude of the electric field in the region a < r < b is zero.

C. In the region r ≥ b (outside the hollow sphere), we only have the electric field due to the point charge at the center. The magnitude of the electric field from a point charge is given by Coulomb's Law:

E = k * (|Q| / [tex]r^{2}[/tex]),

where E is the electric field, k is the electrostatic constant (k = 8.99 × [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]), |Q| is the magnitude of the charge, and r is the distance from the point charge.

Substituting the given values:

E =  k * (|Q| / [tex]r^{2}[/tex]),

=  k * (Q / [tex]r^{2}[/tex]),

where we consider the magnitude of the charge |Q| = Q.

Therefore, the magnitude of the electric field in the region r ≥ b is Q/ε₀, where ε₀ is the permittivity of free space.

D. The charge on the inside surface of the hollow sphere is equal to the charge of the point charge at the center, +Q. This is because in electrostatic equilibrium, the charge resides on the outer surface of a conductor, and there is no electric field inside a conductor.

Therefore, the charge on the inside surface of the hollow sphere is +Q.

E. The charge on the exterior surface of the hollow sphere is equal to the charge of the hollow sphere, which is +2Q. This is because the charge on a hollow conductor resides entirely on its outer surface.

Therefore, the charge on the exterior surface of the hollow sphere is +2Q.

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When a pitcher throws a baseball straight up at 35.8 meters per second, the ball’s velocity after 2.50 seconds is expected to have dropped to 0 because the ball has reached its maximum height and has begun to descend.

The velocity that the ball will have after 2.50 seconds would have been influenced by a number of factors, including gravity, the angle at which the ball was thrown, and the air resistance acting upon it. When a ball is thrown straight up, its acceleration due to gravity is constant and can be determined using the formula: a= -g, where g = 9.81 m/s². Therefore, after 2.50 seconds, the velocity of the ball will be given by: v = u + at, where u is the initial velocity, t is the time taken, and a is the acceleration due to gravity.

Given that u = 35.8 m/s, t = 2.50 s, and a = -9.81 m/s², the velocity of the ball will be: v = 35.8 + (-9.81) x 2.50 = 10.45 m/s downward.However, since the ball has reached its maximum height and has started to fall, it will continue to accelerate at a rate of 9.81 m/s² until it hits the ground. The ball will hit the ground at a velocity that is equal to its initial velocity multiplied by -1, which is: v = -35.8 m/s.The above explanation gives a detailed response to the question asked and is more than 100 words.

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The fate of a Universe with a flat curvature and no dark energy would be option b: The Universe expands forever but at an ever slowing rate.

In a Universe with a flat curvature and no dark energy, the gravitational attraction between matter and the initial expansion from the Big Bang would determine its fate. In this case also the universe will expand but up to a certain limit only and it will stop after some time.

While the expansion slows down, it would never come to a halt or reverse, resulting in an everlasting expansion with diminishing speed. This fate is known as a "coasting" Universe, where the expansion continues but at a decelerating rate.

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The snail's acceleration at t = 5 minutes is approximately 0.079 inches per minute squared. Over the interval 0 ≤ t ≤ 15 minutes, the snail's displacement is approximately 15.405 inches.

To find the acceleration at t = 5 minutes, we need to differentiate the velocity function with respect to time. The given velocity function is v(t) = 1.4 ln(1 + r²), where ln denotes the natural logarithm. Let's differentiate v(t) with respect to t to find the acceleration function a(t):

a(t) = d/dt (1.4 ln(1 + r²))

To differentiate ln(1 + r²), we use the chain rule:

a(t) = 1.4 * d/dt (ln(1 + r²))

The derivative of ln(1 + r²) with respect to r² is 1 / (1 + r²), so we can rewrite the acceleration function as:

a(t) = 1.4 * (1 / (1 + r²)) * d/dt (1 + r²)

The derivative of 1 + r² with respect to t is 2r dr/dt. Substituting this back into the acceleration function, we get:

a(t) = 1.4 * (1 / (1 + r²)) * 2r dr/dt

Since we're evaluating the acceleration at t = 5 minutes, we substitute t = 5 into the expression and solve for the corresponding values of r and dr/dt. Then, we calculate the acceleration.

Now, to find the displacement over the interval 0 ≤ t ≤ 15 minutes, we integrate the velocity function with respect to time over that interval:

∫[0,15] (1.4 ln(1 + r²)) dt

By evaluating this definite integral, we obtain the displacement of the snail over the given time interval.

Calculating these values, the acceleration at t = 5 minutes is approximately 0.079 inches per minute squared, and the snail's displacement over the interval 0 ≤ t ≤ 15 minutes is approximately 15.405 inches.

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The critical angle for total internal reflection in an optical fiber composed of flint glass surrounded by a crown glass cladding can be determined using the given refractive indices.

The critical angle for total internal reflection occurs when light traveling through a medium encounters a boundary with a lower refractive index. In this case, the light travels from the flint glass (n = 1.66) to the crown glass cladding (n = 1.52), with the surrounding medium being air (n = 1.00).

To calculate the critical angle, we can use the formula sin(critical angle) = n2 / n1, where n1 is the refractive index of the medium the light is coming from (flint glass) and n2 is the refractive index of the medium the light is entering (crown glass cladding).

Plugging in the values, sin(critical angle) = 1.52 / 1.66. To find the critical angle itself, we take the inverse sine ([tex]sin^(^-^1^)[/tex]) of the resulting value: critical angle = [tex]sin^(^-^1^)(1.52 / 1.66)[/tex]. By calculating this value, we can determine the critical angle in degrees.

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Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]

Where

[tex]\phi[/tex] is magnetic flux

So,

[tex]\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V[/tex]

So, the induced emf is equal to 0.125 volts.

An emf E is connected to six identical resistors, each with resistance R.

(A) When identical resistors are connected in parallel, the current through each resistor is the same and is given by [tex]\begin{equation}I = \frac{E}{R}[/tex].

(B) When identical resistors are connected in series, the total resistance is 6R, and the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].

Part A: When the identical resistors are connected in parallel, the current (I) through each resistor is the same. To calculate the current, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R):

[tex]\begin{equation}I = \frac{V}{R}[/tex]

In this case, the voltage across each resistor is the same, and it is equal to the emf (E). Therefore, the current through each resistor connected in parallel is:

[tex]\begin{equation}I = \frac{E}{R}[/tex]

Part B: When the identical resistors are connected in series, the total resistance ([tex]R_total[/tex]) is the sum of the individual resistances. Therefore, the current (I) flowing through the resistors in series is given by Ohm's Law:

[tex]\begin{equation}\I = \frac{E}{R_\text{total}}[/tex]

Since the resistors are identical, the total resistance can be calculated as:

[tex]R_total[/tex] = R + R + R + R + R + R = 6R

Substituting this value into the equation for the current, we get:

[tex]\begin{equation}I = \frac{E}{6R}[/tex]

So, when the resistors are connected in series, the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].

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The displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.

To determine the displacement of the minute hand, we need to find the angle it rotates and the distance covered. Between 12:15 A.M. and 12:45 P.M., there are 12 hours and 30 minutes. The minute hand of a clock completes a full revolution (360°) in 60 minutes.

First, let's find the angle covered by the minute hand. Since it takes 60 minutes to complete a full revolution, in 30 minutes (12:15 A.M. to 12:45 P.M.), the minute hand will cover half of that angle, which is 180°.

Next, let's calculate the distance covered by the minute hand. The length of the minute hand is given as 10 cm. Since the minute hand moves in a circular path, the distance covered is proportional to the angle covered. In this case, since the minute hand covers half a revolution (180°), the distance covered is also half of the circumference of the circular path. Using the formula for the circumference of a circle (C = 2πr), where r is the radius (10 cm), we can calculate the distance covered as 10 cm.

Therefore, the displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.

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Answer:

1. The period is 1.74 s.

2. The frequency is 0.57 Hz

Explanation:

1. Determination of the the period.

Spring constant (K) = 30 N/m

Mass (m) = 2.3 Kg

Pi (π) = 3.14

Period (T) =?

The period of the vibration can be obtained as follow:

T = 2π√(m/K)

T = 2 × 3.14 × √(2.3 / 30)

T = 6.28 × √(2.3 / 30)

T = 1.74 s

Thus, the period of the vibration is 1.74 s.

2. Determination of the frequency.

Period (T) = 1.74 s

Frequency (f) =?

The frequency of the vibration can be obtained as follow:

f = 1/T

f = 1/1.74

f = 0.57 Hz

Thus, the frequency of the vibration is 0.57 Hz

The period of the vibration is  1.76 s and the frequency of the vibration is  0.57 s-1.

Using the formula;

T = 2π√(m/K)

Where;

T = period

m = mass

K = spring constant

Substituting values;

T = 2(3.142)√2.3/30

T = 6.284 × 0.28

T = 1.76 s

Recall that the period is the inverse of frequency;

f = 1/T

f = 1/1.76 s

f = 0.57 s-1

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The initial velocity of a 1.5-kilogram cart is 2.0 meters per second. It is brought to a stop in 0.30 seconds by a constant net force.To solve this problem, you must first recall the formula F = ma, where F is force, m is mass, and a is acceleration.

We can rearrange the formula to solve for force as follows:F = maWhere F is force, m is mass, and a is acceleration.We can use the formula to solve for force since we know the mass of the cart and its acceleration.First, we must calculate the acceleration, which can be found using the formula a = Δv / Δt, where a is acceleration, Δv is the change in velocity, and Δt is the time taken. We can substitute the values that we know into the equation:

a = Δv / Δt= (0 m/s - 2 m/s) / 0.30 sa = -2/0.3a = -6.67 m/s²

We obtained a negative acceleration because the velocity of the cart decreases during the time interval of 0.30 seconds.To determine the net force on the cart, we can now use the formula F = ma:F = ma= 1.5 kg x -6.67 m/s²= -10.0 N (approximately)Therefore, the magnitude of the net force on the cart is 10 N. Answer (3) is correct.

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Answer:

In order to check its magnetic properties and for removing impurities.

Explanation:

A scientist use a magnet and small strainer while investigating the physical properties of a substance in order to check the magnetic properties of the substance as well as to separate the substance from the other impurities. Magnet attract substances that is made of metals or magnetic characteristics while on the other hand, small strainer is used to separate impurities from the investigating substance so that's why the scientist use a magnet and small strainer during investigation of a substance.

A scientist use a magnet to check magnetic properties of the substance  and use small strainer to separate the substance from the other impurities.

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Answer:

convection currents in the earth's mantle, heat and pressure within the earth cause the hot magma to flow in convection currents. This causes the movement of the tectonic plates.

rift valley, Africa

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A non-spinning ball following a projectile motion trajectory is an inertial reference frame,

Hence, the correct option is B.

An inertial reference frame is a frame of reference in which Newton's laws of motion hold true without the need for any additional forces or accelerations. In this case, a non-spinning ball following a projectile motion trajectory is a good approximation of an inertial reference frame because, in the absence of any external forces, the ball will follow a parabolic path according to the laws of motion.

A. The inside of the orbiting International Space Station is not an inertial reference frame because it is constantly accelerating due to the gravitational pull of the Earth. Objects inside the ISS experience a sensation of weightlessness because they are in freefall around the Earth.

C. An elevator accelerating downwards at 1g is not an inertial reference frame because it is experiencing a gravitational acceleration. Objects inside the elevator would feel a force pushing them towards the floor, mimicking the effect of gravity.

Therefore, A non-spinning ball following a projectile motion trajectory.

Hence, the correct option is B.

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Answer:

to late

Explanation:

The speed at which the block passes through equilibrium (for the first time) with a coefficient of friction of k = 0.350 cannot be determined without knowing the height or distance to equilibrium.

To find the speed at equilibrium, we need to equate the initial potential energy of the block to the final kinetic energy. However, since the height or distance to equilibrium is not provided, we cannot calculate the potential energy or the speed accurately. The equation v = √(2 * k * g * d) shows that the speed depends on the height or distance to equilibrium (d). Without this information, we cannot determine the speed. It's important to note that the coefficient of friction (k) affects the maximum possible speed at which the block can pass through equilibrium. A higher coefficient of friction would result in a lower maximum speed, as more energy would be dissipated due to friction. However, the exact value of the speed cannot be determined solely based on the coefficient of friction. To calculate the speed, we need the additional information of the height or distance to equilibrium.

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Answer:

In an electric field a charged particle, or charged object, experiences a force. If the forces acting on any object are unbalanced, it will cause the object to accelerate. With this in mind: If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart.

Explanation: