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evablogger [386]
3 years ago
7

7. A pool cue ball moving at 2.1 m/s east collides in a straight line with the eight ball at rest. Find the velocity of the eigh

t ball after the collision if momentum is conserved and
A. the cue ball stops. (5 points)
B. the cue ball continues on in the same direction at 0.10 m/s. (5 points)
C. the cue ball continues on in the same direction at 0.70 m/s. (5 points)
Physics
2 answers:
Hitman42 [59]3 years ago
8 0

the answer is letter B


MArishka [77]3 years ago
6 0
I think the answer is C the cue ball con....
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As an tempature decreases what happens to the rate of radiation?
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As the temperature decreases, the rate of radiation goes down, but the radiation exists as long as the temperature is above the absolute zero, which is actually 0 Kelvin. 0 Kelvin equals -273°C or -460°F. All objects in the world radiate if above that temperature.
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3 years ago
A piece of transparent meterial which has one or two spherical surfaces is called​
Iteru [2.4K]

Answer:

Lens

Explanation:

Lenses fit this description.

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3 years ago
A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo
Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be a m/s².

Given:

Mass of the block is, m=5\ kg

Force of pull is, F=40\ N

Frictional force on the block is, f=8\ N

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

F_{net}=F-f=40-8=32\ N

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0
3 years ago
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
3 years ago
If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m
algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

H=ut+\dfrac{1}{2}at^2

Here, a = g

H=\dfrac{1}{2}gt^2            

H=\dfrac{1}{2}\times 9.8\times (2.261)^2

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

7 0
2 years ago
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