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EleoNora [17]
2 years ago
10

A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble. Is the change of the bo

ulder’s momentum in one second greater than, less than, or equal to the change of the pebble’s momentum in the same time period?
Physics
1 answer:
Mandarinka [93]2 years ago
8 0

Answer:

equal

Explanation:

According to the Newton's second law of motion, the force acting on the body is directly proportional to the rate of change of momentum.

If the force is same, then the change in momentum is also same for a given time.

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GenaCL600 [577]
I think it’s true!!!
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2 years ago
A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is t
Korolek [52]

Answer:

static friction acting opposite to the direction of travel

Explanation:

Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.

Explanation:

8 0
2 years ago
Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marbl
Lostsunrise [7]
Conservation of linear momentum:

m*v inital = m*v final

0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v

(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)

0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s

Answer 1.8 m/s (positive, to the right).

 

4 0
2 years ago
A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height
Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

v_x = 23.2 m/s

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

v_y(t)= v_{y0} -gt (1)

where

v_{y0}=0 is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s

Substituting t into (1) we find the final vertical velocity

v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}

6 0
2 years ago
A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a
Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

y = 0.049 \cos(7t) .......(1)

The general equation of a wave is given by :

y=A\cos(\omega t) .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0
2 years ago
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