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Ivanshal [37]
3 years ago
8

A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of work during this pull.

Physics
1 answer:
Wittaler [7]3 years ago
7 0

Answer:

<h2>602.08 N</h2>

Explanation:

The force supplied by the train can be found by using the formula

f =   \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{578000}{960}  \\  = 602.083333...

We have the final answer as

<h3>602.08 N</h3>

Hope this helps you

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A 1500 kg car moving with a speed of 20 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of
storchak [24]

Answer:

-100000 N.

Explanation:

Force: This can be defined as the product of the mass of a body and it's acceleration. The S.I unit of Force is Newton(N). The Formula of force is given as,

F = ma ........................... Equation 1

Where F = Average force exerted on the car, m = mass of the car, acceleration of the car, a = acceleration of the car.

a = (v-u)/t..................... Equation 2

Where v = Final velocity, u = Initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............. Equation 3

Given: m = 1500 kg, u = 20 m/s, v = 0 m/s (brought to rest), t = 0.3 s.

Substitute into equation 3

F = 1500(0-20)/0.3

F = 1500(-20)/0.3

F = -100000 N.

Note: The negative sign is due to the fact that the force exerted on the car by the pole is equal and opposite the force of the car.

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How would the body be affected if red blood cells and low levels of homoglobinHow would the
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What is the connection between electrons and protons
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Electrons and protons are connected with one another in term of charge and size.

Explanation:

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4 years ago
Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190
julia-pushkina [17]

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

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\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

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8 0
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