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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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(b) 74.56338m.

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We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

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                                  x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2

                                 F =ma

                                 V(t) = V_{o} +at

                                 x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2

(a) Calculating velocity right before going up the ramp.

 Wooden block is going on a straightaway and has net for on it.

         F_{n} =F-F_{s} = F-uF_{n}  = 100N-0.4*9.8m/s^2*5kg =81N

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      a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .

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          x(t) = 12m = 16.2m/s^2*t^2

         t = 1.217s

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(b) How far up the ramp will the wooden block go before stopping.

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