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kaheart [24]
3 years ago
8

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, A)compression B)w

ork C)radiation is being done.
Physics
2 answers:
Gnesinka [82]3 years ago
7 0
Salutations!

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy,  _______________ is being done.

<span>If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done. Energy being transferred and the object begins to move is called work.

Thus, your answer is option B.

Hope I helped (:

Have a great day!</span>
Mademuasel [1]3 years ago
5 0

B) work is being done

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A sealed box contains a monatomic ideal gas. The number of gas atoms per unit volume is 5.00 * 1020 atoms&gt;cm3, and the averag
Veronika [31]

Answer:

Pressure,P=6×10^3Pa

Explanation:

The gas has an ideal gas behaviour and ideal gas equation

PV=NKT

T= V/N p/K ...eq1

Average transitional kinetic energy Ktr=1.8×10-23J

Ktr=3/2KT

T=2/3Ktr/K....eq2

Equating eq1 and 2

V/N p/K = 2/3Ktr/K

Cancelling K on both sides

P= 2/3N/V( Ktr)

Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3

P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23

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6 0
3 years ago
X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a
FrozenT [24]

Answer:

5.1645\times 10^{-11}\ m

Explanation:

n = Order = 1

c = Speed of light = 3\times 10^8\ m/s

f = Frequency = 3\times 10^{18}\ Hz

\theta = Angle = 75.5^{\circ}

Lattice spacing is given by

d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m

The lattice spacing of the crystal is 5.1645\times 10^{-11}\ m

6 0
3 years ago
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

v_{rms,o2}^2 *M_{o2} =3RT_{o2}

solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

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7 0
4 years ago
People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T
Dafna11 [192]

Answer:

The travel would take 6.7 years.

Explanation:

The equation for an object moving in a straight line with acceleration is:

x = x0 + v0 t + 1/2a*t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

In a movement with constant speed, a = 0 and the equation for the position will be:

x = x0 + v t

where v = velocity

Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74  m/s²:

Seconds in half a year:

1/2 year = 1.58 x 10⁷ s

x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)²  = 1.59 x 10¹⁵ m

Now let´s see how much time it takes the travel to the nearest star after this half year.

The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²

v = v0 + a t

Since the spacecraft starts from rest, v0 = 0

v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s

Using the equation for position:

x = x0 + v t

4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t

(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t

t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.

The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.

7 0
3 years ago
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