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2 years ago

8

The mass of a basketball is three times greater than the mass of a softball . Compare the momentum’s of a softball and a basketb

all if they both are moving at the same velocity

1 answer:

Lemur [1.5K]2 years ago

4 0

They are moving the softball of basketball

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If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?

yaroslaw [1]

Answer:

735watts

Explanation:

Power = Force*distance/time

Given

Force = mg = 50 * 9.8'

Force = 490N

distance = 3m

time = 2s

Substitute

Power = 490*3/2

Power = 1470/2

Power = 735Watts

Hence the minimum power output is 735watts

7 0

2 years ago

URGENT!!!!!!!<br><br> PLEASE HELP WITH THIS PHYSICS PROBLEM

Levart [38]

Explanation:

Let

$x_1$ = distance traveled while accelerating

$x_2$ = distance traveled while decelerating

The distance traveled while accelerating is given by

$x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2$

$\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2$

$\:\:\:\:\:= 1125\:\text{m}$

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

$v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}$

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

$v^2 = v_0^2 + 2ax_2$

$0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2$

Solving for $x_2,$ we get

$x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}$

$\:\:\:\:\:= 4327\:\text{m}$

Therefore, the total distance x is

$x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}$

$\:\:\:\:= 5452\:\text{m}$

3 0

2 years ago

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the waterÂs surface to knoc

sergij07 [2.7K]

Answer:

.012

Explanation:

Take the mass of the fish and divide it by the mass of the water:

65/.30=216.667

Divide the given speed by the value we found above:

2.5/216.667=.0115

Answer can be rounded up to .012

3 0

2 years ago

Read 2 more answers

A particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.

mrs_skeptik [129]

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

V1 = 2.35 m/s

V2 = - 8.72 m/s

T1 = 3.42s

T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

#SPJ1

4 0

1 year ago

The magnitude of a component of a vector must be

den301095 [7]

Less than or equal to the magnitude of the vector

6 0

2 years ago