Magnet A has twice the magnetic field strength of Magnet B and at a certain distance pulls on magnet B with a force of 100 N. The amount of force that magnet A exerts on magnet B is A. at or about 50 N. B. exactly 100 N. C. more information is needed.

Answer:

0.7 hours

Explanation:

From the way back, we can calculate the distance between Irina's work and Irina's home. In fact, we know that the car takes 0.4 hourse traveling at 27 mph, so the distance covered should be

When Irina rides to work with her bike, she travels at a speed of 16 mph. So we can find the time she takes by dividing the total distance (10.8 miles) by her speed:

Acceleration due to gravity on this planet will be 3.802 m / s^2

What are equations of motion?

Equation of motion are defined as equations that describe the behavior of a physical system in terms of its motion as a function of time

Using equation of motion

u=0

s= 2.3 m

t = 1.1 sec

to find = g (acceleration due to gravity on this planet)

s = u t + 1/2 (a ) (t ^(2))

s = 1/2 (g) (t^2)

2.3 = 1/2 (g) (1.1^2)

g = 2 * 2.3 /(1.1)^2

g = 4.6 /1.21= 3.802

correct answer is b) g = 3.802 m / s^2

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Part A: The enmeshed cars were moving at a velocity of approximately 8.66 m/s just after the collision.

Part B: Car A was traveling at a velocity of approximately 8.55 m/s just before the collision.

How to compute the above velocities

To find the speed of car A just before the collision in Part B, you can use the principle of conservation of momentum.

The total momentum of the system before the collision should equal the total momentum after the collision. You already know the total momentum after the collision from Part A, and now you want to find the velocity of car A just before the collision.

Let's denote:

- v_A as the initial velocity of car A before the collision.

- v_B as the initial velocity of car B before the collision.

In Part A, you found that the enmeshed cars were moving at a velocity of 8.66 m/s at an angle of 60 degrees south of east. You can split this velocity into its eastward and southward components. The eastward component of this velocity is:

v_east = 8.66 m/s * cos(60 degrees)

Now, you can use the conservation of momentum to set up an equation:

Total initial momentum = Total final momentum

(mass_A * v_A) + (mass_B * v_B) = (mass_A + mass_B) * 8.66 m/s (the final velocity you found in Part A)

Plug in the known values:

(1900 kg * v_A) + (1500 kg * v_B) = (1900 kg + 1500 kg) * 8.66 m/s

Now, you can solve for v_A:

(1900 kg * v_A) + (1500 kg * v_B) = 3400 kg * 8.66 m/s

1900 kg * v_A = 3400 kg * 8.66 m/s - 1500 kg * v_B

v_A = (3400 kg * 8.66 m/s - 1500 kg * v_B) / 1900 kg

Now, plug in the values from Part A to find v_A:

v_A = (3400 kg * 8.66 m/s - 1500 kg * 8.66 m/s) / 1900 kg

v_A = (29244 kg*m/s - 12990 kg*m/s) / 1900 kg

v_A = 16254 kg*m/s / 1900 kg

v_A ≈ 8.55 m/s

So, car A was going at approximately 8.55 m/s just before the collision in Part B.

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Answer:

1.1 x 10⁵m/s²

Explanation:

Given parameters:

Velocity  = 452m/s

distance  = 0.93m

Unknown:

Acceleration of the bullet  = ?

Solution:

To solve this problem, we use one of the kinematics equation which is given below:

           V² = U²  + 2aS

V is the final velocity

U is the initial velocity  = 0m/s

a is the unknown acceleration

S is the distance traveled  

 So;

          452²   = 0²  + (2 x a x 0.93)

          204304  = 1.86a

               a  = 1.1 x 10⁵m/s²

Final answer:

The acceleration of the bullet in the gun barrel can be calculated using the kinematic equation for motion. By substituying the given values into the equation, we find the acceleration to be approximately 1.095 x 10^5 m/s^2.

Explanation:

The subject of this question is Physics, specifically a topic under mechanics known as kinematics. The problem given can be solved using kinematic equations which are used to describe the motion of an object without considering the forces that cause it to move. In this case, the final velocity (vf) of the bullet is given as 452 m/s, the initial velocity (vi) is assumed to be 0 as it starts from rest, and the distance (d) is given as 0.93 m. We are asked to determine the value of acceleration (a).

Using the kinematic equation vf2 = vi2 + 2ad and substituting the given values, we get (452 m/s)2 = 0 + 2*a*0.93 m. We can rearrange to solve for acceleration to get: a = (452 m/s)2 / (2*0.93 m) = 109523.66 m/s2.

So, the acceleration of the bullet in the gun barrel is approximately 1.095 x 105 m/s2.

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