A heavy rock and a light rock are dropped from the same height and experience no significant air resistance as they fall. Which of the following statements about these rocks are correct?O Just as they are about to reach the ground, the heavier rock has more kinetic energy than the lighter rockO The moment they were released, both rocks had the same amount of gravitational potential energy O Both rocks have the same kinetic energy when they reach the ground O The heavier rock reaches the ground before the lighter rock. .O Both rocks different acceleration as they are about to reach the ground

Answer:

the correct answer is C     +, - , +

Explanation:

The electric field for positive charges is outgoing and for negative charges it is directed towards the charge.

Let's apply this to our case:

On plates A and B the field goes to the right, therefore plate A must be positive

In plates B and C the way it goes to the left, so the field is selected from it, this implies that plate C is positive

therefore plate B must be negative for both cases

when checking the correct answer is C     +, - , +

Answer:

Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.

The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.

Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.

A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.

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Answer:

1/Re= 1/R1 + 1/R2

Explanation:

Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by  [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the rate of R changes when R₁ = 117 Ω and

R₂ = 112 Ω is 0.25 Ω/min

For a given resistor connected in parallel;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]

Making R from the left-hand side the subject of the formula, then:

[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]

Given that:

Now, replacing the values in the above previous equation, we have:

[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]

However, the differentiation of R with respect to time t will give us the rate at which R is changing when R1=117Ω and R2=112Ω.

So, by differentiating the given equation of the resistor in parallel with respect to time t;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex],   we have:

[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]

[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]

Therefore, we can conclude that the rate at which R is changing R1=117Ω and R2=112Ω is 0.25 Ω/min

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Answer:

  a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]  ,       t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Explanation:

To solve this exercise we must use Newton's second law

For the block

let's set a reference system with the x axis parallel to the plane

X axis

         Wₓ - T = m a

Y axis  

         N- W_y = 0

         N = W_y

for pulley

          ∑τ = I α

           T R = (½ m_p R²) α

let's use trigonometry for the weight components

         sin β = Wₓ / W

         cos β = W_y / W

         Wx = W sin β

angular and linear variables are related

          a = α R

          α = a / R

we substitute and group our equations

         W sin β - T = m a

         T R = ½ m_p R² (a / R)

         W sin β - T = m a

                        T = ½ m_p a

we solve the system of equations

         W sin β = (m + ½ m_p) a

          a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]

let's find the time to travel the distance (d) through the block

          x = v₀ t + ½ a t²

          d = 0 + ½ a t²

          t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

Answer:

h=0.142m=14.2cm

Explanation:

ρ=8.96 g/cm3=8960 kg/m3

d=0.1m

ρ=m/V - - >ρ=m/[π*((d/2)^2) *h] - - >

8960=10/[π*((0.1)^2) *h] - - >

h=0.142m

Answer:

FORCE as for my answer....

Answer:

Universe

Explanation:

I took the quiz.

Answer:

[tex]21.6\ \text{kg m}^2[/tex]

[tex]3.672\ \text{Nm}[/tex]

[tex]54.66\ \text{revolutions}[/tex]

Explanation:

[tex]\tau[/tex] = Torque = 36.5 Nm

[tex]\omega_i[/tex] = Initial angular velocity = 0

[tex]\omega_f[/tex] = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

[tex]\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2[/tex]

Torque is given by

[tex]\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2[/tex]

The wheel's moment of inertia is [tex]21.6\ \text{kg m}^2[/tex]

t = 60.6 s

[tex]\omega_i[/tex] = 10.3 rad/s

[tex]\omega_f[/tex] = 0

[tex]\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2[/tex]

Frictional torque is given by

[tex]\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}[/tex]

The magnitude of the torque caused by friction is [tex]3.672\ \text{Nm}[/tex]

Speeding up

[tex]\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}[/tex]

Slowing down

[tex]\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}[/tex]

Total number of revolutions

[tex]\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}[/tex]

[tex]\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}[/tex]

The total number of revolutions the wheel goes through is [tex]54.66\ \text{revolutions}[/tex].

Answer:

Current is calculated as the amount of charge that pass  a point on a circuit per time.

Explanation:

Electric current is defined as the amount of charge that passes a point on a circuit per unit time. The mathematical definition of electric current is given by :

[tex]I=\dfrac{q}{t}[/tex]

Where

q is a charge (Coulomb)

t is time (in seconds)

The SI unit of electric current is A. It is equivalent to C/s. So, the correct option is (A).

Answer:

a) < 3 , -4 >

b) < 3 , -4 >

Explanation:

a) If you can imagine this, adding vectors is like putting them "tip to tail", where you put the beginning point of vector B to the end point of vector A (or vice versa).  Your new vector (A+B) would be from the "tip" of vector A to the "tail" of vector B.

Mathematically, this is the same as adding the x-components of each vector together as well as the y-components.

Vector A: 3 units along the positive x-axis: < 3 , 0 >

Vector B: 4 units along the negative y-axis: < 0 , -4 >

A+B = < 3 , 0 > + < 0 , -4 > = < (3+0) , (0+(-4)) > = < 3 , -4 >

b) Subtracting is like adding a negative, so you could use the same "tip to tail" visual by adding the negative of vector B instead (which is B in the opposite direction).

Vector A: < 3 , 0 >

Vector B: < 0 , -4 >

Vector -B: < 0 , 4 >

A-B = A+(-B) = < 3 , 0 > + < 0 , 4 > = < (3+0) , (0+4) > = < 3 , 4 >

Answer:

iEvaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)Evaluate  for \(x=2.\)

Explanation:

Distance travelled in one round will be equal to the circumference of the disc i.e [tex] 2 \pi r[/tex]

Radius=50cm

Circumference= [tex] 2 \times \frac{22}{7} \times 50cm=> \frac{2200}{7} cm[/tex]

If the disc rotates at speed of 100rpm that means it completes 100 rotation in a minute(60 second)

So, in 30 seconds it will complete 50 rotation.

1 rotation = [tex] \frac{2200}{7} cm [/tex]

[tex] 50 rotation=\frac{2200}{7} \times 50cm \\\\ \frac{110000}{7} cm[/tex]

Break it in decimal.

Your answer will be 15714.2 cm

Answer:

The correct answer is - 93.24×10^4 kg/m^3.

Explanation:

Given:

height of cylinder: 50.5 mm

diameter = 52.0

then radius will be diameter/2 = 52/2 = 26

Formula:

Density = mass/ volume

Volume = πr^2h

solution:

Now the volume of a cylinder is v = (22/7)×r^2×h

= 22/7×26×26×50.5

= 107261.59 mm^3  

Now volume in cubic meter V =10.7261 ×10^(-5) m^3

So density d = m/V = 1/(10.7261 ×10^(-5))

Or d = 93.24×10^4 kg/m^3

Answer:

BI/BII = 1

Explanation:

The magnetic field due to a solenoid is given by the following formula:

[tex]B = \mu nI\\[/tex]

where,

B = Magnetic Field due to solenoid

μ = permeability of free space

n = No. of turns per unit length

I = current passing through the solenoid

Now for the first solenoid:

[tex]B_1 = \mu n_1I_1 \\[/tex]

For the second solenoid:

[tex]B_2 = \mu n_2I_2\\[/tex]

Dividing both equations:

[tex]\frac{B_1}{B_2} = \frac{\mu n_1I_1}{\mu n_2I_2}\\[/tex]

here, no. of turns and the current passing through each solenoid is same:

n₁ = n₂ and I₁ = I₂

Therefore,

[tex]\frac{B_1}{B_2} = \frac{\mu nI}{\mu nI}\\[/tex]

BI/BII = 1

This question is incomplete, the complete question is;

A rifle fires a 2.10 × 10⁻² kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 9.10 × 10⁻² m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Answer:

the spring constant is 303.5 N/m

Explanation:

Given the data in the question;

There is a potential energy associated with the spring;

we know that potential energy = kinetic energy

mgh = [tex]\frac{1}{2}[/tex]kx²

where k is the spring constant and x is the compression

as the pallet is fired, the spring gives kinetic energy which is converted into gravitational potential energy

so

m = 2.10 × 10⁻²

g = 9.81 m/s²

h = 6.10 m

x = 9.10 × 10⁻² m

we substitute

mgh = [tex]\frac{1}{2}[/tex]kx²

2.10 × 10⁻² × 9.81 × 6.10 =  [tex]\frac{1}{2}[/tex] × K × ( 9.10 × 10⁻² )²

1.256661 = 0.0041405 × K

K = 1.256661 / 0.0041405

K = 303.5 N/m

Therefore, the spring constant is 303.5 N/m

Answer:

okay

Explanation:

please I don't know

Answer:

a)   v = 0.9167 m / s,  b)  A = 0.350 m,  c)  v = 0.9167 m / s, d)  A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

          v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

           λ = x

           λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

          T / 2 = t

           T = 2t

period and frequency are related

           f = 1 / T

           f = 1 / 2t

we substitute

           v = x / 2t

           v = 5.50 / 2 3

           v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

          2A = y

           A = y / 2

           A = 0.700 / 2

           A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

      v = 0.9167 m / s

d) the amplitude is

           A = 0.500 / 2

           A = 0.250 m

Answer:

yes

it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.

Answer:

I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.

Answer: Mercury

Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.

Answer:

Explanation:

I jus did it on usatestprep

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

Answer:

r = 1,224 10⁻² m

Explanation:

For this exercise let's use Newton's second law

          F = m a

the force is magnetic

         F = q v x B

The bold letters indicate vectors, the module of this expesion is

          F = q v B

The direction of the force is found by the right hand rule

thumb points in the direction of the velicad + x

fingers extended in the direction of B -z

the palm is in the direction of the force + and

the acceleration of the proton is cenripetal

          a = v² / r

we substitute

          q v B = m v² / r

          r = [tex]\frac{m \ v}{q \ B}[/tex]

let's calculate

         r = [tex]\frac{1.7 \ 10^{-27} \ 5.760063 \ 10^7 }{1.6 \ 10^{-19} \ 0.5 }[/tex]

         r = 1,224 10⁻² m

Answer:

Option B. 0.016 Hz

Explanation:

From the question given above, the following data were obtained.

Period (T) = 60 s

Frequency (f) =?

Frequency and period are related according to the following equation:

Frequency = 1 / period

f = 1/T

With the above formula, we can obtain the frequency of the ride as follow:

Period (T) = 60 s

Frequency (f) =?

f = 1/T

f = 1/60

f = 0.016 Hz

Thus, the rider's frequency is 0.016 Hz

Answer: 51

Explanation:

Given

Output is 23 V

The square armature side is [tex]a=5.1\ cm[/tex]

Magnetic field [tex]B=0.5\T[/tex]

Rate of revolution [tex]n=55\ rev/s[/tex]

Angular speed

[tex]\omega =2\pi n\\\omega=2\pi \times 55=110\pi\ rad/s[/tex]

Peak voltage is given by

[tex]E_{peak}=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=\dfrac{E_{peak}}{B\omega A}\\\\N=\dfrac{23}{0.5\times 110\pi\times (0.051)^2}\approx 51[/tex]

So, there are approximate 51 loops

Answer:

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

[tex]2.898\times 10^{-10}\ \text{m}[/tex] x-ray region

Explanation:

T = Temperature

b = Constant of proportionality = [tex]2.898\times 10^{-3}\ \text{m K}[/tex]

[tex]\lambda[/tex] = Wavelength

[tex]T=10^4\ \text{K}[/tex]

From Wein's law we have

[tex]\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}[/tex]

The wavelength of the radiation will be [tex]2.898\times 10^{-7}\ \text{m}[/tex] and it is in the ultraviolet region.

[tex]T=10^7\ \text{K}[/tex]

[tex]\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}[/tex]

The wavelength of the radiation will be [tex]2.898\times 10^{-10}\ \text{m}[/tex] and it is in the x-ray region.

Answer:

hi I thinks its number 3

Explanation:

hope you have a nice day