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2 years ago

How do I solve this problem

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

5 0

Answer:

it is light

Explanation:

the arrow that says light is on the glass it must be near from tungsten

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A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children

trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

Explanation:

The centripetal acceleration is given by:

$a_{c} = \frac{v^{2}}{r}$

Where:

$v^{2}$ : is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

$a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}$

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0

2 years ago

Read 2 more answers

WARRIOR [948]

Answer:

Explanation:distance-time =speed

a,V =s/t

V=2m/2s

V=1m/s

b,v=s/t

V=80m/40s

V= 2m/s

The average speed is 2m/s

8 0

1 year ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

3 years ago

If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final

Alisiya [41]

5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

Force on the ball = 5N

time taken = 0.1s

Unknown:

final speed of the ball = ?

Solution:

According to newton's second law "the net force on a body is the product of its mass and acceleration".

Force = mass x acceleration equation 1

Acceleration =

V is the final velocity

U is the initial velocity

T is the time taken

U = O since it is a stationary body;

a = $\frac{V}{T}$

Input "a" into equation 1

F = m x $\frac{V}{T}$

5 = 0.1 x $\frac{V}{0.1}$

V = 5m/s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0

3 years ago

Help pleasee i give brainliest

MA_775_DIABLO [31]

Answer:

A is a solid. C is a gas. In solid an liquid the particals are touching. In C, the particals have less affect on each other because they are so far apart.

5 0

3 years ago