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elena-14-01-66 [18.8K]
3 years ago
15

NHT0015

Physics
1 answer:
poizon [28]3 years ago
3 0

Answer:

(1) 1×10⁻⁴

Explanation:

From the question,

α = (ΔL/L)/(ΔT)............. Equation 1

Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.

Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C

Substitute these values into equation 1

α  = 1×10⁻⁴/10

α = 1×10⁻⁵ °C⁻¹ .

β = (ΔA/A)/ΔT................... Equation 2

Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.

make ΔA/A the subject of the equation

ΔA/A = β×ΔT.......................... Equation 3

But,

β = 2α.......................... Equation 4

Substitute equation 4 into equation 3

ΔA/A = 2α×ΔT................ Equation 5

Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹

Substitute into equation 5

ΔA/A = ( 2)×(1×10⁻⁵)×(5)

ΔA/A  = 10×10⁻⁵

ΔA/A  = 1×10⁻⁴

Hence the right option is (1) 1×10⁻⁴

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a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
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Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

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You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
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Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
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  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

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  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

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