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4 years ago

What is the transfer of heat by the flow of a heated material?

1 answer:

7 0

Heat flux is a quantitative, vectorial representation of heat-flow through a surface. ...Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes.

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A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?

STatiana [176]

Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
P = 833.33 Watts

So, your final answer is 833.33 Watts

5 0

4 years ago

Jake drops a book out of a window from a height of 10 meters. At what velocity does the book hit the ground?

user100 [1]

G = 9.81 m/sec^2) g = 9.81 $\frac{meters}{ second^{2} }$

Solving for velocity : 

$velocity^{2}$ = 2gh 
v = $2gh^{ \frac{1}{2} }$
v = (2 x 9.81 x 10)^1/2 
v = 196.2 m/sec (answer)

8 0

3 years ago

The diffraction limit is a limit on: The diffraction limit is a limit on: A telescope's size. A telescope's angular resolution.

Mnenie [13.5K]

Answer:

A telescope's angular resolution.

Explanation:

Diffraction limit is a minimum angular separation of two sources and it can be distinguished by the telescope. This angle is known as the diffraction limit. It is proportional to the wavelength of light and it has an inverse relation with the diameter of the telescope. Mathematically it is defined as

θ = 1.22λ/d

where θ is the angle, λ wavelength and d is the diameter of the objective mirror (lenz).

7 0

3 years ago

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

$c=\frac {F}{y}$

Since F is gm but taking the given mass to be F

$c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930$

By substitution now considering F to be 3.3 kg

$y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm$

8 0

3 years ago

The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T · m2 when it is placed in a region of

valkas [14]

Answer:

The strength of the magnetic field is 3.5 x 10⁻³ T

Explanation:

Given;

magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²

angle of inclination of the field, θ = 42.0°

radius of the circular plate, r = 8.50 cm = 0.085 m

Generally magnetic flux in a uniform magnetic field is given as;

Φ = BACosθ

where;

B is the strength of the magnetic field

A is the area of the circular plate

Area of the circular plate:

A = πr²

A = π (0.085)² = 0.0227 m²

The strength of the magnetic field:

B = Φ / ACosθ

B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

B = 3.5 x 10⁻³ T

Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T

3 0

3 years ago