!!! i'll give brainliest to whoever can answer this for me !!!

A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0

3 years ago

A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction

taurus [48]

Answer:

31m/s

23.17°

Explanation:

Given the following :

From the diagram attached :

AB = 15m/s, BC = 18m/s, AC = a = resultant

Resolving Velocity into both vertical and horizontal component.

Kindly see attached picture for detailed explanation.

8 0

3 years ago

Read 2 more answers

Decribe an experiment to show that pressure increase with decrease in the area of surface

topjm [15]

Answer:

for example the studs are made in football player boot because to increase pressure with descrease in area of surface

6 0

2 years ago

A cannonball is fired on flat ground

algol [13]

hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.

This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.

V₀ = 420m/s and θ₀ = 53.0°

So, when the cannonball is fired it has horizontal and vertical components:

V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s

V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s

When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:

Vy = V₀y - g tₐ = 0

tₐ = V₀y/g

tₐ = (335.43m/s)/(9.8m/s²) = 34.23s

Then, the maximum height is reached in the instant tₐ = 34.23s:

h = V₀y tₐ - 1/2g tₐ²

hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²

hmax = 11481.77m - 5741.29m

hmax = 5740.48m

3 0

3 years ago

Prove law of conservation of momentum ,??

ziro4ka [17]

Explanation:

$\law \: of \: conservation \: of \: momentum)$

$\frac{initial}{m1u1 + m2u2} = \frac{fi}{m 2u2}$

<h3>BEFORE COLLISION)  INITIAL MOM. OF BODY A </h3>

PiA = m1u1

After collision) .................. B I. e piB = m2u2

<h2>FINÅL MOMENTUM OF BØDY A I. e PfA= m1v1</h2>

FINAL .......... B I. e PfB = m2v2

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7 0

2 years ago