Ask question

Ask question

All categories

3 years ago

13

The table shows information about four students who are running around a track. Which statement is supported by the information

in the chart?
Chart

vvvvvvvvvvvvvvvvvvvvvv

Autumn 50 kg 4 m/s

Mohammed 57 kg 3 m/s

Lexy 53 kg 3 m/s

Chiang 64 kg 5 m/s

Which statement is supported by the information in the chart?

Autumn has more kinetic energy than Chiang.

Mohammed has less kinetic energy than Autumn. This is the answer. I got tiered of waiting

Lexy has more kinetic energy than Mohammed.

Chiang has less kinetic energy than Lexy.

2 answers:

murzikaleks [220]3 years ago

7 0

Answer:

Mohammed has less kinetic energy than Autumn

Explanation:

The kinetic energy of each student is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the student

v is the speed of the student

Let's use the formula above to calculate the kinetic energy of each student:

- Autumn: $K=\frac{1}{2}(50 kg)(4 m/s)^2=400 J$

- Mohammed: $K=\frac{1}{2}(57 kg)(3 m/s)^2=256.5 J$

- Lexy: $K=\frac{1}{2}(53 kg)(3 m/s)^2=238.5 J$

- Chiang: $K=\frac{1}{2}(64 kg)(5 m/s)^2=800 J$

Therefore, by looking at the numbers, we see that the correct answer is

Mohammed has less kinetic energy than Autumn

mafiozo [28]3 years ago

5 0

Answer:

The answer is B. Mohammed has less kinetic energy than Autumn.

Explanation:

You might be interested in

In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text

Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second , change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

= 4.859 x 10⁻² oscillations .

5 0

3 years ago

How does light and heat energy from the sun reach earth?

defon

Explanation:

uuubbv. very ecrcvtyfyhc g you f gg and you are the one to give me the chance for a little bit of time and effort into this is 6.0 and if

6 0

3 years ago

How can you justify that force is transferred to lift and throw soil using shovel?<br>

Natali [406]

Forces are needed to lift, turn, move, open, close, push, pull, and so on. When you throw a ball, you are using force to make the ball move through the air. More than one force can act on an object at the same time.

4 0

2 years ago

Kepler's First Law states that the shape of planetary orbits is a/an ________________ ?

harkovskaia [24]

Answer:

elliptical orbit

Explanation:

There are three laws of planetary motion, which are called Kepler's law of planetary motion.

First Law : It states that all the planets revolve around the sun in an elliptical path and the sun is at one focus of that elliptical path.

4 0

3 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago