All categories

4 years ago

A little girl kicks a soccer ball. It goes 10 feet and comes back to her. How is this possible?

1 answer:

8 0

A classic puzzle...

She either kicked it at a wall exactly 10 foot in front of her, where the ball rebounded off the wall.

Or, she kicked the ball straight up, vertically, at a 90 degree angle, where due to the law of gravity, which states that anything that goes up must come down, when the soccer ball reaches exactly 10 feet, it falls back down.
(Note: This is nearly impossible to achieve -- exactly 10 feet.)

You might be interested in

What is the acceleration of the object? m/s²

Orlov [11]

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

7 0

2 years ago

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$

$kg\dfrac{m^2}{s^2}=Joule$

$P=\dfrac{kgm^2}{s^3}$

So, the answer is D. the proper replacement unit for one joule per second

8 0

3 years ago

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 563 K, and r

Schach [20]

Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= $\frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}$

$= \frac{-1760}{563} + \frac{1760}{354}$

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

8 0

4 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago

A pendulum with a length of 1.00 m is released from an initial angle of 12.5°. After 1 000 s, its amplitude has been reduced by

Viefleur [7K]

Answer:

A student is conducting a pendulum experiment. Which of the following pieces of safety equipment would be the most vital to conduct this test?

Explanation:nduebidndo eyn h ehj jd

jknsjk Jjkfnjkjnuifvr

7 0

3 years ago