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2 years ago

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What net force would be needed to accelerate a 2.95kg bowling ball at 6.25m/s^2?

1 answer:

aksik [14]2 years ago

3 0

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A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

4 years ago

There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

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3 years ago

How does increasing the tension of a spring affect a wave on the spring?

Liono4ka [1.6K]

More energy is being used in the movement

8 0

3 years ago

Read 2 more answers

Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?

Oksanka [162]

Answer:

1.58 Hz

Explanation:

The frequency of the simple pendulum is given by

f = 1/T

= 1/2π√g/l

In this problem, I = 10.0 cm = 0.1 m

f = 1/2π√9.8/0.1

= 1.58 Hz

7 0

4 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull

DiKsa [7]

Answer:

$F_{net} = 232.8 N$

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

$F_x = F_1 + F_2cos45 + F_3cos45$

here we know that

$F_1 = 77.3 N$

$F_2 = 61.7 N$

$F_3 = 147 N$

$F_x = 77.3 + 61.7 cos45 + 147 cos45$

$F_x = 224.9 N$

Similarly in Y direction we will have

$F_y = F_3 sin45 - F_2 sin45$

$F_y = (147 - 61.7)sin45$

$F_y = 60.3 N$

Now the net force on the donkey is given as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$F_{net} = \sqrt{224.9^2 + 60.3^2}$

$F_{net} = 232.8 N$

Now direction of force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{60.3}{224.9}$

$\theta = 15^o$ towards right so it is -15 degree

3 0

3 years ago