1) 1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the magnitude of the centripetal force acting on the car? *

1 point

1 N

100 N

1000 N

10000 N


2) 1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the magnitude of the acceleration of the car? *

1 point

0.1 m/s^2

1 m/s^2

10 m/s^2

100 m/s^2

Answer:

Do you have the answers for the unit 6 test?

Explanation:

PLZ

Answer:

The constant force to be exerted on the rope is 221.55 N

Explanation:

Given;

mass of the merry, m = 235 kg

radius, r = 1.5 m

number of revolution per second, = 0.4 rev/s

time of motion, t= 2.00 s

The angular acceleration is given by;

[tex]\alpha = \frac{0.4 \ rev}{s} *\frac{2\pi \ rad}{rev} *\frac{1}{2.0\ s} = 1.257 \ rad/s^2[/tex]

Torque is given by;

τ = F.r

Also torque in uniform solid disk is given by;

τ = ¹/₂mr²α

Thus, equating the two;

F.r = ¹/₂mr²α

F =  ¹/₂mrα

F = ¹/₂(235)(1.5)(1.257)

F = 221.55 N

Therefore, the constant force to be exerted on the rope is 221.55 N

Answer:

answer is C on edge 2021

Explanation:

Answer:

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.

Explanation:

spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.

Let any rolling ball of mass (m ) is traveling with velocity v ,

common effect on ball (N) = mg

because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.

hence,

fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.

Answer:

e. 55.0 m

Explanation:

Given;

diameter of the aluminum wire, d = 0.865 mm

radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m

resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m

resistance of the wire, R = 2.48 Ω

The resistance of a wire is given by;

[tex]R = \frac{\rho \ L}{A} \\\\[/tex]

where;

L is length of the wire

A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²

Substitute the givens and solve for L,

[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]

Therefore, the length of the wire is 55.0 m

Answer:100m west

Explanation:

Answer:

distance travelled = 3000m

Explanation:

distance travelled = average speed x time

                               =30m/s*100s

                                =3000m

Answer:

Explanation:

Step one:

given data

we are told that 1 large pizza can be cut into 8 even slices

Required

we want to find how many slices are there in 4 large pizzas

Step two:

so if 1 pizza has 8 slices

       4 pizza will have x

cross multiply we have

x= 8*4

x=32 slices

Answer:

false ..........false

Answer:

FALSE                                            

Explanation:

Answer:

The correct reaction force in response to Heidi's action force is:

c. The friction is equal to 660 N since the beam is not accelerating.

Explanation:

Heidi's action force does not affect the beam.  Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect.  The reaction force is what makes the dog to move because it acts on it.  According to Newton's Third Law of Motion, forces always come in action-reaction pairs.  This Third Law states that for every action force, there is an equal and opposite reaction force.  This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."

Answer:

a

Explanation:

all the other ones take years and hard work but learning japanese is pretty easy and you dont need to work to hard

Answer:

-20°C

Explanation:

The specific heat capacity of ice using the cgs system is 0.5cal/g°C

The enthalpy change is calculated as follows

ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.

10cal = 2g×0.5cal/g°C×∅

∅=10cal/(2g×0.5cal/g°C)

∅=10°C

Final temperature= -30°C+ 10°C= -20°C

Answer:

Explanation:

Heat generated by the coil is expressed as;

Heat generated = Power ×Time

Heat generated = IVt

I is the current = 4A

V is the voltage= 5V

t is the time = 4hrs

Convert to secs

t = 4(3600) = 14400secs

Substitute into the formula

Heat generated = 4×5×14400

Heat generated = 20×14400

Heat generated = 288000Joules

Heat generated =288KJ

Answer:

depends on how talll the building is but lets say its 100 ft tall 12MPH

Explanation:

Answer:

around 9.81m/s i think

Explanation:

Answer:

If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

Explanation:

Answer:

5,000

Explanation:

Vf = Vi + a * t

Answer:

The final temperature of the bullets is 327.3 ºC.

Explanation:

Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:

[tex]K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0[/tex] (1)

Where:

[tex]K_{A,o}[/tex], [tex]K_{A}[/tex] - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.

[tex]K_{B,o}[/tex], [tex]K_{B}[/tex] - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.

[tex]U_{A,o}[/tex], [tex]U_{A}[/tex] - Initial and final internal energies of the 15-g bullet, measured in joules.

[tex]U_{B,o}[/tex], [tex]U_{B}[/tex] - Initial and final internal energies of the 7.75-g bullet, measured in joules.

By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:

[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0[/tex] (2)

Where:

[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the 15-g and 7.75-g bullets, measured in kilograms.

[tex]v_{A,o}[/tex], [tex]v_{A}[/tex] - Initial and final speeds of the 15-g bullet, measured in meters per second.

[tex]v_{B,o}[/tex], [tex]v_{B}[/tex] - Initial and final speeds of the 7.75-g bullet, measured in meters per second.

[tex]c[/tex] - Specific heat of lead, measured in joules per kilogram-Celsius degree.

[tex]T_{o}[/tex], [tex]T[/tex] - Initial and final temperatures of the bullets, measured in Celsius degree.

Now we clear the final temperature of the bullets:

[tex](m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})][/tex]

[tex]T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex]

[tex]T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}[/tex] (3)

If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the final temperature of the collided bullets is:

[tex]T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]

[tex]T = 852.534\,^{\circ}C[/tex]

Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:

[tex]\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0[/tex] (4)

[tex]U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)[/tex]

If we know that [tex]T_{o} = 30\,^{\circ}C[/tex], [tex]T = 327.3\,^{\circ}C[/tex], [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 295\,\frac{m}{s}[/tex], [tex]v_{B,o} = 375\,\frac{m}{s}[/tex], [tex]v_{A} = 0\,\frac{m}{s}[/tex], [tex]v_{B} = 0\,\frac{m}{s}[/tex] and [tex]c = 128\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then latent heat received by the bullets during impact is:

[tex]U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)[/tex][tex]U = 331.872\,J[/tex]

The maximum possible latent heat ([tex]U_{max}[/tex]), measured in joules, that both bullets can receive during collision is:

[tex]U_{max} = (m_{A}+m_{B})\cdot L_{f}[/tex] (5)

Where [tex]L_{f}[/tex] is the latent heat of fusion of lead, measured in joules per kilogram.

If we know that  [tex]m_{A} = 15\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.75\times 10^{-3}\,kg[/tex] and [tex]L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}[/tex], then the maximum possible latent heat is:

[tex]U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)[/tex]

[tex]U_{max} = 557.375\,J[/tex]

Given that [tex]U < U_{max}[/tex], the final temperature of the bullets is 327.3 ºC.

Answer: Half

Explanation:

Given

The object is in uniform circular motion with constant speed.

Radial acceleration of the object is given by

[tex]\Rightarrow a_r=\dfrac{v^2}{r}[/tex]

Where, [tex]a_r=\text{Radial acceleration}[/tex]

[tex]v=\text{speed}\\r=\text{distance from the axis of rotation}[/tex]

If the radial distance [tex]r[/tex] is doubled, i.e. [tex]2r[/tex]

Radial acceleration is

[tex]\Rightarrow a'_r=\dfrac{v^2}{2r}\\\\\Rightarrow a'_r=\dfrac{a_r}{2}[/tex]

Radial acceleration becomes half of the initial value.

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Answer:

a = 5.5 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.

ΣF =m*a

where:

F = Force [N] (units of Newtons)

m = mass = 20 [kg]

a = acceleration [m/s²]

Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.

[tex]230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ][/tex]

[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

➡ 150hrs.

[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

Answer:half-life (T1/2) of this isotope =

Explanation:

The number of nuclei of any radioactive substance at a given time is expressed by

Nt = N0e⁻kt

Nt=decay of material  at a time t, =3110 decays per minute

N=decays at t=0, 8255 decays per minute

k=constant

Nt=N0e−kt

3110= 8255 e⁻k(4.50)

3110/ 8255=e−k(4.50)

0.3767 =  e−k(4.50)

In 0.3767  =   -k (4.50)

0.976=-4.5k

k=0.976/4.5

=0.2159

Also we know that t 1/2= time that it takes half the original material to decay.it is  related to the rate constant by

T₁/₂=ln  2 / k

Therefore half-life (T1/2) of this isotope

T₁/₂=ln  2/0.2159

T₁/₂=3.12 days

Answer:

The value is [tex]I = 0.48 \ kg m^2[/tex]

Explanation:

From the question we are told that

   The coordinate of the knee is  [tex]knee = (1.2,2.7)[/tex]

    The coordinate of the ankle is [tex]ankle = (1.5,2.1)[/tex]

    The mass of the shank is  [tex]m = 3.2 \ kg[/tex]

 Generally the length between the knee and the ankle is mathematically represented as

      [tex]D = \sqrt{( 1.5 - 1.2)^2+ [ 2.1- 2.7]^2 }[/tex]

=>  [tex]D =0.67 \ m[/tex]

Generally the  moment of inertia of the shank about the proximal end point is mathematically represented as

         [tex]I= m * \frac{D^2}{3}[/tex]

=>       [tex]I = 3.2 * \frac{0.67^2}{3}[/tex]

=>       [tex]I = 0.48 \ kg m^2[/tex]

Answer:

B: lodestone

Explanation:

Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.

Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

Answer:

80kg = 133 Newtons I'm pretty sure this is right.

The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Weight of the object is force applied due to gravity, So force = mass.gravity .

on the  moon gravity is 1/6 of that on the earth.

so weight or force = 80*(9.8/6)

                               = 130 Newtons

The weight or gravitational force of an object with mass 80 kg on the moon is 130 Newtons.

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#SPJ6

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

K.E₂ = mg(h - 2R)

The kinetic energy of the car is expressed as K.E = mg(h - 2R). The energy of the object by the integrity of its movement is understood as kinetic energy.

The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of velocity.

The following data is observed from the figure;

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop =?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

According to the law of conservation of energy, total energy is defined as the sum of kinetic energy and potential energy.

Total energy = kinetic energy+potential energy

On applying the law  of conservation of energy for top and bottom of the loop;

K.E₁ + P.E₁ = K.E₂ + P.E₂

0 + mgh = K.E₂ + mg(2R)

K.E₂ = mg(h - 2R)

Hence the kinetic energy of the car is expressed as K.E = mg(h - 2R).

To learn more about kinetic energy refer to the link;

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