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SpyIntel [72]
2 years ago
11

Use the formula from Newtons second law, namely Fnet=∆t=∆P that the Net force exerted on the object is equal to the product of t

he mass and acceleration of the object​
Physics
1 answer:
Aleksandr-060686 [28]2 years ago
6 0

Answer:

Explanation:

Fnet = Δt = ΔP

Fnet = ΔP/Δt

= mv - mu / t

= m(v - u) / t

But we know v = u + at

=> a = v - u / t

Therefore,

Fnet = ma

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f biofuel production and use reduces our consumption of imported fossil fuels, we may become less vulnerable to the adverse impacts of supply

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Why is a specimen smaller than 200 nm not visible with a light microscope?
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Because the specimen is very small with a light microscope
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3 years ago
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

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4 0
3 years ago
If we increase the force applied to an object, and all other factors remain the same, the amount of work will
soldier1979 [14.2K]
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.

-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.

-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.

-- But if you push on a battleship, even with a much bigger force ...
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3 years ago
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An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t
Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

At one point

Wide= 16.0 m

Deep = 3.8 m

Water flow = 2.8 cm/s

At a second point downstream

Width of canal = 16.5 m

Water flow = 11.0 cm/s

We need to calculate the depth

Using Bernoulli theorem

A_{1}V_{1}=A_{2}V_{2}

Put the value into the formula

16.0\times3.8\times2.8=16.5\times x\times 11.0

x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}

x=0.938\ m

Hence,  The depth of the water at this point is 0.938 m.

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3 years ago
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