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2 years ago

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When Stone B collides with Stone A during Test 2, Stone B stops and Stone A begins to move. Which statement BEST describes the m

omentum in Test 2?
A
Stone A moves away from Stone B at a velocity of 4 m/s, and the momentum is conserved.

B
Stone A moves away from Stone B at a velocity of 2 m/s, and the momentum is not conserved.

C
Stone A moves away from Stone B at a velocity of 4 m/s, and the momentum is not conserved.

D
Stone A moves away from Stone B at a velocity of 2 m/s, and the momentum is conserved.

1 answer:

Anna11 [10]2 years ago

5 0

Answer:

A

Explanation:

According to law of conservation of momentum, the total momentual in the system will be conserved

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A proton is placed in an electric field of intensity 700 N/C. What are the magnitude and direction of the acceleration of this p

olya-2409 [2.1K]

Answer:

Acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

Explanation:

We have given a proton is placed in an electric field of intensity of 700 N/C

So electric field E = 700 N/C

Mass of proton $m=1.67\times 10^{-27}kg$

Charge on proton $e=1.6\times 10^{-19}C$

So electric force on the proton $F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N$

This force will be equal to force due to acceleration of the proton

According to newton's law force is given by F = ma

So $1.67\times 10^{-27}\times a=1.120\times 10^{-16}$

$a=0.67\times 10^{11}m/sec^2$

So acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

6 0

3 years ago

The drawing shows 6 point charges arranged in a rectangle. The value of q is 2.83 uC and the distance d is 0.123 m. Find the tot

vova2212 [387]

the total electric potential at location P, which is at the center of the rectangle is 0V.

The charges placed at the corner of the rectangle are same in magnitude but different in charge. hence the total electric potential will be same in magnitude but different in charge and will be cancelled. Similarly, all the total electric potential will be cancelled and resultant will be zero.

<h3>What is total electric potential?</h3>

The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).
More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion. In order to prevent the test charge from gaining kinetic energy or radiating, the travel across the field is also meant to occur with very little acceleration.
The electric potential at the reference location is, by definition, zero units. Any point may be used as the reference point, but typically it is earth or a point at infinity.

To learn more about total electric potential with the given link

brainly.com/question/14776328

#SPJ4

3 0

2 years ago

Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L

snow_lady [41]

Answer:

$Q=81.56\ W/m^2$

Explanation:

Given that

$T_1= 210 K$

$T_2= 150 K$

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

$Q=\sigma (T_1^4-T_2^4)\ W/m^2$

So now by putting the values

$Q=\sigma (T_1^4-T_2^4)\ W/m^2$

$Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2$

$Q=81.56\ W/m^2$

So rate of heat transfer per unit area

$Q=81.56\ W/m^2$

8 0

3 years ago

A runner of mass 53.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cen

const2013 [10]

Answer:

0.336 rad/s

Explanation:

$\omega_1$ = Angular speed of the turntable = -0.2 rad/s

R = Radius of turntable = 2.9 m

I = Moment of inertia of turntable = $76\ kgm^2$

M = Mass of turn table = 53 kg

$v_1$ = Magnitude of the runner's velocity relative to the earth = 3.6 m/s

As the momentum in the system is conserved we have

$Mv_1R+I\omega_1=(I + MR^2)\omega_2\\\Rightarrow \omega_2=\dfrac{Mv_1R+I\omega_1}{I + MR^2}\\\Rightarrow \omega_2=\dfrac{53\times 3.6-76\times 0.2}{76+53\times 2.9^2}\\\Rightarrow \omega_2=0.336\ rad/s$

The angular velocity of the system if the runner comes to rest relative to the turntable which is the required answer is 0.336 rad/s

4 0

3 years ago

1. The head of a rattlesnake can accelerate 50.0 m/s in striking a victim. If a car could do

Maslowich

Answer: 0.492 s

Explanation:

We can use the following equation to solve this problem:

$V=V_{o}+a.t$

Where:

$V=24.6 m/s$ is the car’s final velocity

$V_{o}=0 m/s$ is the car’s initial velocity (it starts from rest)

$a=50 m/s^{2}$ is the car's acceleration

$t$ is the time we have to find

Isolating $t$ :

$t=\frac{V-V_{o}}{a}$

$t=\frac{24.6 m/s-0 m/s}{50 m/s^{2}}$

Finally:

$t=0.492 s$

6 0

3 years ago