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Leto [7]
2 years ago
12

You apply a very small force, say 0.001 newtons, to a very large truck, with a mass of 2000 kilograms. What can you say for sure

about what will happen to the truck?
The truck will accelerate (move from rest to motion) as long as that tiny force is larger than any force of friction that opposes it.

The truck will not move because small forces cannot move large objects.

The truck will accelerate (move from rest to motion) as long as the small applied force is large enough to overcome the inertia of the truck.

An extremely small force can never accelerate such a large truck. The truck will not move under any circumstance.
Physics
1 answer:
Fofino [41]2 years ago
5 0

Answer:

it will stay still. unless its in space.

Explanation:

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A dragster runs the quarter mile in 8.96 s. What is the car's velocity (in ft/s) at the finish line?
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The first thing you should know to answer this question is the following conversion:
 1mi = 5280feet
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 Answer:
 the car's velocity (in ft / s) at the finish line is 147.32 feet / s
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The two main states of mechanical energy are ___________ and potential energy.
shutvik [7]
<span>Potential energy and Kinetic energy</span>
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Read 2 more answers
The speed of sound is 346 m/s. If a sound wave travels at a frequency of 55 Hz, what would its wavelength be?​
Korvikt [17]

Answer:

6.29 meters.

Explanation:

, where v is the speed of wave and f is the frequency of wave.

We are given that ,

The speed of sound is 346 m/s.

i..e v=346 m/s

A sound wave travels at a frequency of 55 H.

i..e f=55

the wavelength would be 6.29 meters.

This is based on another brainly answer

Link: brainly.com/question/12538018

3 0
2 years ago
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

80 a = 32. 4 × 10 m/s ²  (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0
3 years ago
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