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2 years ago

The mass of a large car is 1000 kg. How much force would be required to accelerate the car at a rate of 3msec2

Physics

1 answer:

pogonyaev2 years ago

8 0

Answer:

3000 N

Explanation:

We have,

• Mass, m = 1000 kg

• Acceleration, a = 3 m/s²

We have to find force required, F.

$\longrightarrow$ F = ma

$\longrightarrow$ F = 1000 × 3 N

$\longrightarrow$ F = 3000 N (Answer)

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A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.

sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0

3 years ago

What does weight require?

kherson [118]

Answer:

Explanation:

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7 0

2 years ago

Which of two factors influence the weight of an object due to gravitational pull?

Maslowich

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5 0

3 years ago

A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang

Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0

2 years ago

REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part

sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

$W_{fr}$ = ΔEm

$W_{fr}$ = $Em_{f}$ -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

$Em_{f}$ = $W_{fr}$ + Em₀

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

½ m v² = -fr d + mgh

v = √(-fr d 2/m + 2 gh)

v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

5 0

2 years ago