All categories

2 years ago

Csggggresddgdfgthxefecdtvh

Physics

2 answers:

Inessa05 [86]2 years ago

8 0

Answer:

csggggresdd

...........

SOVA2 [1]2 years ago

3 0

Answer:

yes :)

Explanation:

You might be interested in

If a stereo has a dial that changes the volume of the sounds the stereo makes, what is the dial doing? a. changing the number of

Over [174]

If the stereo has a dial that changes the volume of the sounds the stereo makes, the one that being done by the dial is : D. Changing the amplitude of the sound waves

Hope this helps

8 0

3 years ago

Read 2 more answers

HELP ASAP TIMED TEST

balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

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

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

$U_1 = mgH$

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

$\displaystyle U_2=\frac{1}{2}mv^2$

Since the energy is conserved, U1=U2

$\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]$

For the speed to be double, we need to drop the snowball from a height H', and:

$\displaystyle mgH'=\frac{1}{2}m(2v)^2$

Operating:

$\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]$

Dividing [2] by [1]

$\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}$

Simplifying:

$\displaystyle \frac{H'}{H}=4$

Thus:

H' = 4H

Correct choice: b 4H

4 0

2 years ago

Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecule

Colt1911 [192]

Answer:

8563732.58906 Pa

3992793.23326 Pa

5708.00923 J

Explanation:

V = Volume

N = Number of molecules = $3\times 6.023\times 10^{23}$

T = Temperature = 300 K

b = $7\times 10^{-29}\ m^3$

$k_$ = Boltzmann constant = $1.38\times 10^{-23}\ J/K$

P = Pressure

We have the equation

$P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa$

The pressure is 8563732.58906 Pa

For isothermal expansion

$P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa$

The pressure is 3992793.23326 Pa

Work done is given by

$dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J$

The work done is 5708.00923 J

7 0

2 years ago

When energy is transferred to air, what happens to the particles of air? (1 point) They move slower. They move faster. They cool

zubka84 [21]

This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster

5 0

3 years ago

Read 2 more answers

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$