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3 years ago

6

A 4kg object is at rest. How much force is required to get the object to a velocity of 20m/s in 2 seconds? (Show work and includ

e units)

1 answer:

Ugo [173]3 years ago

8 0

Explanation:

<h2>Answer:-</h2>

The Object was at rest. So, Initial Velocity is Zero.

[Initial Velocity]u = 0
[Final Velocity]v = 20 m/s
[Time]t = 2 seconds
Mass = 4kg
Force = ?

We know that:-

$\sf{Force = Mass \times \dfrac{(v-u)}{t}}$

Applying it, we get:-

$\sf{Force = 4 \times \dfrac{(20-0)}{2}}$

$\sf{Force = 2 \times \dfrac{(20)}{2}}$

$\sf{Force = 2 \times 10}$

$\sf{Force = 20 \ N \ (Newton)}$

Hope it helps :)

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In as small room fan of rating 50 watt is used for 10hrs 2bulb of rating 10v are used for 8hrs daily.calculate monthly. electric

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Answer:

The electric bill for June is Rs198000

Explanation:

Convert volt to watt, but in order to do so I need to know the amps and since it is not provided I converted if the amps was 1.

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3 years ago

You find a micrometer (a tool used to

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Explanation:

A micrometer is a measuring device or an instrument which is used to measure very minute measurements very accurately and precisely. It is mathematical tool used to provide accurate measurement for any mechanical components.

Now, if the micrometer that I have found is badly bent, it would provide faulty or wrong measurements both in terms of precision and accuracy when compared to a high quality meter stick.

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3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

The current through a 10-ohm resistor connected to a 120-V power supply is The current through a 10-ohm resistor connected to a

Mazyrski [523]

Answer:

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Explanation:

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2 years ago

If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?

storchak [24]

Answer:

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Explanation:

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so the correct answer is

d. 100 meters

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3 years ago