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Murljashka [212]
3 years ago
14

If earth squeezed to the size of moon.Calculate the weight of 100kg mass in the squeezed earth.​

Physics
1 answer:
Ugo [173]3 years ago
5 0

Answer:

833.9kg that's the answer

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Why is the unit of work is a derived unit<br>​
masha68 [24]

Answer:

Since energy can be measured as work, we can write energy = force x distance. Thus SI derived unit of energy has the units of newtons x meter or kg m2/s2.

Explanation:

5 0
2 years ago
Read 2 more answers
If the engine receives 6.45 kJ of heat energy from the reservoir at 520 K in each cycle, how many joules per cycle does it rejec
Oxana [17]

Answer:

3.72 kJ

Explanation:

QH = 6.45 kJ

TH = 520 K

Tc = 300 K

Qc = ?

By use of Carnot's theorem

Qc / QH = Tc / TH

Qc / 6.45 = 300 / 520

Qc = 3.72 kJ

6 0
3 years ago
A 0.35 m2 coil with 50 turns rotates at 5 radians per sec2 in a magnetic field of 0.6 Tesla. What is the value of the rms curren
soldier1979 [14.2K]

Answer:

11.25 amps

Explanation:

For transformers, the magnetic flux

\Phi _{max} = \beta  \times A

Therefore;

\Phi _{max} = 0.6  \times 0.35 = 0.21 \ Weber

Ф = Фmax (cosωt) = 0.21·(cos(5·t))

From Faraday's law of induction, we have;

ε = -N × dΦ/dt

Which gives;

dΦ/dt = -1.05(sin (5t) )

ε = -N × dΦ/dt = -50× -1.05(sin (5t) )

ε = 52.5(sin (5t) )

I = ε/R = 52.5(sin (5t) )/3.3 = 15.9091(sin (5t) ) amps

The peak current is therefore = 15.9091 amps

The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.

5 0
3 years ago
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

4 0
3 years ago
According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt
scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength(\lambda _1)=450 nm

Second wavelength(\lambda _2)=700 nm

According wien's diplacement law

\lambda T=constant

where \lambda =wavelength

T=Temperature

Let T_1 and T_2 be the temperatures corresponding to \lambda _1 & \lambda _2 respectively.

\lambda _1\times T_1=\lambda _2\times T_2

\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}

\frac{T_1}{T_2}=\frac{700}{450}=1.55

Thus object with \lambda 450 nm is 1.55 times hotter than object with wavelength \lambda =700 nm

8 0
3 years ago
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