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3 years ago

You are riding a bicycle. You apply a forward force of 150 N, and you and the

Physics

2 answers:

yawa3891 [41]3 years ago

7 0

Answer:

1.67 m/s2

Explanation:

other guys is a f moron

ella [17]3 years ago

3 0

Answer:

B. 0.60 m/s2

- Hope this helped!

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A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to

Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

3 0

3 years ago

A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55

Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

$\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s$

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0

3 years ago

Pls help for brainliest:)

Rasek [7]

Answer:

Explanation:

<h3>#CARRY ON LEARNING</h3><h3>#MARK ON LEARNING</h3><h3>#HELPING HAND</h3>

6 0

3 years ago

You kick a soccer ball off the ground with a velocity of 15 m/s at an angle of 40° above the horizontal

jeyben [28]

At its maximum height, the ball will have zero vertical velocity, so the ball's velocity at this point is exactly equal to its horizontal velocity.

At any time <em>t</em>, the horizontal component of its velocity is

<em>v</em> = (15 m/s) cos(40°) ≈ 11.49 m/s

so at the highest point of its trajectory, the ball has a velocity of about 11.49 m/s pointed in the positive horizontal direction.

8 0

3 years ago

You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the s

olga_2 [115]

To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.

Centrifugal acceleration is determined by

$a_c = \frac{v^2}{r}$

Where,

v = Velocity

r = Radius

From the given data of the moon we know that gravity there is equivalent to

$a = 1.62m/s$

While the radius of the moon is given by

$r = 1.74*10^6m$

If we rearrange the function to find the speed we will have to

$v = \sqrt{ar}$

$v = \sqrt{1.6(1.74*10^6)}$

$v = 1.7km/s$

The speed for this to happen is 1.7km/s

3 0

3 years ago