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Travka [436]
3 years ago
12

An old truck compresses a spring scale at the junkyard by 25 cm. The spring constant for the industrial scale at the junkyard is

39,200 N/m.
What is the force applied to the scale by the car?

_____N
Physics
2 answers:
vichka [17]3 years ago
8 0
Using F = Ke
F = 39200×(25÷100) = 39200×0.25
F= 9800N
MrMuchimi3 years ago
6 0

Answer:

Force, F = 9800 N

Explanation:

It is given that,

The spring constant for the industrial scale at the junkyard is, k = 39,200 N/m

The compression in the spring, x = 25 cm = 0.25 m

To find,

The force applied to the scale by the car.

Solution,

The Hooke's law gives the relation between the force and the compression in the spring. Its expression is given by :

F=-kx

F=39200\times 0.25

F = 9800 N

So, the force applied to the scale by the car is 9800 N.

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Answer:

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Explanation:

From the question we are told that

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And

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5 0
3 years ago
A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou
Sophie [7]

Answer:

442.36038 m or 1451.31362 ft

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P_1 = Initial pressure = 30.15 inHg

P_2 = Final pressure = 28.607 inHg

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1\ lb/ft^3=16.0185\ kg/m^3

1\ in=0.0254\ m

1\ m=3.28084\ ft

Density of mercury = 13560 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Difference in pressure is given by

P_1-P_2=\rho gh\\\Rightarrow h=\frac{P_1-P_2}{\rho g}\\\Rightarrow h=\frac{(30.15-28.607)\times 13560\times 0.0254\times 9.81}{0.075\times 16.0185\times 9.81}\\\Rightarrow h=442.36038\ m\\\Rightarrow h=442.36038\times 3.28084\\\Rightarrow h=1451.31362\ ft

The height of the building is 442.36038 m or 1451.31362 ft

4 0
3 years ago
a car slows down from 22 m/s to 3 m/s at a constant rate of 3.5 m/s squared. how many seconds are required before the car is tra
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Answer:

9.05 s

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6 0
3 years ago
The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and
Salsk061 [2.6K]
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

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c = length of one side = 6 m
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We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
 
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3 years ago
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3 years ago
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