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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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Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
3 years ago
A circular wire loop of radius 15.0 cm carries a current of 2.60
Korolek [52]
Part (a): Magnetic dipole moment

Magnetic dipole moment = IA, I = Current, A = Area of the loop
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Indicate whether a change in the value of each of the following determinants of demand leads to a movement along the demand curv
Inessa05 [86]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Change in market price is m<span>ovement along the demand curve. </span>

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Margaret [11]

Answer:

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Explanation:

The air in the tube can be considered an ideal gas,

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In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

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Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

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If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

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6 0
2 years ago
In general, how did the water pressure in the tank change when mass was added to the fluid?
MissTica

Answer:

As the height increases the pressure must increase.

Explanation:

When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression

           P = ρ g h

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As the height increases the pressure must increase.

3 0
2 years ago
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