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Ugo [173]
3 years ago
12

A 3500 kg truck is parked on a 7.0∘ slope. How big is the friction force on the truck?

Physics
2 answers:
Usimov [2.4K]3 years ago
7 0

Answer:

Frictional force, f = 4180.11 N

Explanation:

It is given that,

Mass of the truck, m = 3500 kg

It is parked on a 7 degrees slope. Let f is the frictional force acting on the truck. The force acting on it parallel to the slope is equal to the friction force required to prevent the truck from moving. It is given by :

f=mg\ sin\theta

f=3500\times 9.8 \ sin(7)

f = 4180.11 N

So, the friction force on the truck is 4180.11 N. Hence, this is the required solution.                          

Arada [10]3 years ago
6 0

Answer:

4180 N

Explanation:

In order to keep the truck balanced and stationary, the force of friction on the truck must be equal to the component of the weight of the truck acting parallel to the slope.

The component of the weight of the truck acting parallel to the slope is:

mg sin \theta

where

m = 3500 kg is the mass of the truck

g = 9.8 m/s^2 is the acceleration of gravity

\theta=7.0^{\circ} is the angle of the slope

Therefore, the force of friction must be equal to

F=(3500)(9.8)(sin 7.0^{\circ})=4180 N

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Anna71 [15]

Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

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        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          Em_{f} = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m

6 0
3 years ago
What is the function of the commutator in a DC motor? A. to spin the magnet B. to change direct current into alternating current
sergejj [24]

Answer

Correct Answer is B(to change direct current into alternating current)

Explanation

commutation

The main aim of commutation is to maintain the torque acting on armature of DC motros in the same direction.

commutator  a switch which is used to convert the direction of flow of electric current  it is connected to the armature of DC motor.

In DC motors, voltage across the armature is alternating in nature. The direction of  this current is reverses by Commutator when armature of DC motors  passes through magnetic neutral axis to maintain the unidirectional. Hence, commutator is used to change direct current into alternating current.


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3 years ago
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AlladinOne [14]
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3 years ago
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Answer:

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5 0
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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
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