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stich3 [128]
3 years ago
14

Which of these reactions is responsible for energy radiating from stars, including the Sun?

Physics
1 answer:
snow_tiger [21]3 years ago
3 0

Answer: nuclear fusion

Explanation: give me a brainliest

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Consider an electrical transformer has 10 loops on its primary coil and 20 loops on its secondary coil. What is the voltage in t
Arlecino [84]

Answer:

240 V

Explanation:

number of turns in primary coil, Np = 10

Number of loops in secondary coil, Ns = 20

Voltage in primary coil, Vp = 120 V

Let the voltage in secondary coil is Vs.

So, Vs / Vp = Ns / Np

Vs / 120 = 20 / 10

Vs / 120  = 2

Vs = 240 V

Thus, the voltage in secondary coil is 240 Volt.

4 0
3 years ago
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A
vazorg [7]

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

8 0
3 years ago
(c)
grigory [225]

Answer:

Both AC and DC describe types of current flow in a circuit. In direct current (DC), the electric charge (current) only flows in one direction. Electric charge in alternating current (AC), on the other hand, changes direction periodically.

7 0
2 years ago
A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts
jeka57 [31]
The answer is: 0 Newtons
4 0
2 years ago
Read 2 more answers
The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm
DIA [1.3K]

Answer:

T_{max} = 4.735\,kN\cdot m

Explanation:

The shear stress due to torque can be calculed by using the following model:

\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}

The maximum torque on the section is:

T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}

The Torsion Constant for the circular tube is:

J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})

J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]

J_{tube} = 4.560\times 10^{-6}\,m^{4}

Now, the require output is computed:

T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}

T_{max} = 4.735\,kN\cdot m

7 0
3 years ago
Read 2 more answers
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