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3 years ago

Hell please thanks!!!!!!’

Physics

1 answer:

Slav-nsk [51]3 years ago

5 0

Answer:

liquid phase

Explanation:

it is liquid phase because molecules are not that tightly packed as solid and not that far away from each other as in gas phase.

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A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of

garik1379 [7]

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m$

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s$

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s$

c) Her velocity when her feet hit the water is 7.3942875 m/s

3 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Changing the speed of a synchronous generator changes A) the frequency and amplitude of the output voltage. B) only the frequenc

Dmitrij [34]

Answer:

A) the frequency and amplitude of the output voltag

Explanation:

Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.

Changing the speed regulator will change the engine throttle setting to maintain the speed.

While the power, torque, current, fuel flow rate and torque angle will have decreased.

8 0

3 years ago

What is it about copper that makes it a good conductor

natta225 [31]

It’s because conductors have nearly zero resistance to the flow of electrons that go through them. This leaves the electrons free to move and current can travel with full strength.

6 0

3 years ago

Read 2 more answers

Relativistic velocity is of the order of _____ of velocity of light A- 1/15th of the velocity of light B-1/20th of the velocity

ratelena [41]

Answer:

Relativistic velocity is of the order of 1/10th of the velocity of light

Explanation:

We define relativistic speed (or velocity) as a speed that is a significant fraction of the speed of light: c = 3*10^8 m/s

Such that for these speeds, the special relativity theory starts to apply (the relativity effects starts to apply).

Usually, we define relativistic speeds as those that are of the order (or larger) of c/10, which is one-tenth of the speed of light.

Then the correct option is C:

Relativistic velocity is of the order of 1/10th of the velocity of light

4 0

3 years ago