All categories

3 years ago

-Select ALL that apply-

1 answer:

8 0

The correct options are:
A) 0 K
<span>F) the temperature at which motion is at a minimum

In fact, absolute zero corresponds to zero degrees in the Kelvin scale: 0 K. At this temperature, the motion of the molecules of any substance is minimum (it is zero, to be precise).

(note: there are some weird signs in the answers. if there is a negative sign in front of </span> $273^{\circ}C$ in option D), then option D is correct as well, because $0 K =-273^{\circ} C$ ).

You might be interested in

What is the relationship between amplitude and frequency of a wave?.

kondaur [170]

Answer:

The relationship between the wave's amplitude and frequency is such that it is inversely proportional to the frequency. The amplitude decreases as the frequency increases. The amplitude increases as the frequency decreases. The higher the energy of a wave, the higher the amplitude. The lower the energy, the lower the amplitude. Energy has no effect on wavelength, speed, or frequency, only the amplitude.

Explanation:

5 0

1 year ago

Please help solving this physics question

sp2606 [1]

True
False
False.
True
True
False

4 0

2 years ago

A truck accelerates from a stop to 27m/sec in 9 minutes. What was the trucks acceleration?

olchik [2.2K]

The acceleration is 3 m/s per minute, or 0.05 m/s per second.

5 0

3 years ago

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$

The sound intensity of train is 1000 times greater than that of the library.

3 0

2 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago