Ask question

Ask question

All categories

3 years ago

13

Avi, a gymnast, weighs 40 kg. She is jumping on a trampoline that has a spring constant value of 176,400 StartFraction N over m

EndFraction. If she compresses the trampoline 20 cm, how high should she reach? meters

2 answers:

Lina20 [59]3 years ago

7 0

Answer:

9

Explanation:

Because its right!

Fiesta28 [93]3 years ago

6 0

Answer:

9

Explanation:

You might be interested in

A truck is moving with a certain uniform velocity. It is accelerated uniformly by 0.75 m/s^2. After 20 seconds , the velocity be

scoundrel [369]

Answer:

Vi = 5 m/s

Explanation:

let (a) acceleration = 0.75 m/s²

(t) time = 20 seconds

Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)

find Initial velocity (Vi)

Vf - Vi

a = -----------

t

Vi = Vf - (a * t) = 20 - (0.75 * 20)

Vi = 5 m/s

5 0

3 years ago

Which location on the map above is a source of North Atlantic deep water?

bija089 [108]

The answer might be C ? hope it's right

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Chuge%5Cmathfrak%5Cpurple%7Bhello...%7D%20%5C%5C%20%20%5C%5C%20%5Chuge%5Cmathfrak%5Cgreen%7B

andre [41]

Energy is "the ability to do work". Energy is how things change and move. It takes energy to cook food, to drive to school, and to jump in the air. Different forms of Energy. Energy can take a number of different forms.

6 0

3 years ago

Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri

Bess [88]

Answer: $0.75\ cm$

Explanation:

Given

Wavelength of light $\lambda=500\ nm$

Screen is $D=1\ m$ away

Distance between two adjacent bright fringe is $\Delta y=\dfrac{\lambda D}{d}$

When same experiment done in water, wavelength reduce to $\dfrac{\lambda }{\mu}$

So, the distance between the two adjacent bright fringe is $\Delta y'=\dfrac{\lambda D}{\mu d}$

Keeping other factor same, distance becomes

$\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm$

3 0

3 years ago

A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig

Gre4nikov [31]

Answer:

The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

Explanation:

From the question we are told that

The distance between earth and Retah is $d = 20 \ light \ hours = 20 * 3600 * c = 72000c \ m$

Here c is the peed of light with value $c = 3.0*10^8 m/s$

The time taken to reach Retah from earth is $t = 25 \ hours = 25 * 3600 =90000 \ sec$

The velocity of the spacecraft is mathematically evaluated as

$v_s = \frac{d }{t}$

substituting values

$v_s = \frac{72000 * 3.0*10^{8} }{90000}$

$v_s = 2.40*10^{8} \ m/s$

The time elapsed in the spacecraft’s frame is mathematically evaluated as

$T = t * \sqrt{ 1 - \frac{v^2}{c^2} }$

substituting value

$T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }$

$T = 54000 \ s$

=> $T = 15 \ hours$

So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame

7 0

3 years ago