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2 years ago

15

A 3 kg drone starts at 5 meter above the ground and reached 25 meter above the

1 answer:

docker41 [41]2 years ago

4 0

Recall that work is the amount of energy transferred to an object when it experiences a displacement and is acted upon by an external force. It is given a symbol of W and is measured in joules (J).

W=\vec{F}\cdot \Delta \vec{d}

We can use this formula to determine the work done by very specific forces, generating specific types of energy. We will examine three types of energy in this activity: gravitational potential, kinetic, and thermal. Before we start deriving equations for gravitational potential energy and kinetic energy, we should note that since work is the transfer and/or transformation of energy, we can also write its symbol as \Delta E.

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The force a magnet exerts on another magnet, on iron or a similar metal, or on moving charges is

In-s [12.5K]

The answer is a magnetic force.

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3 years ago

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

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3 years ago

A centrifuge was used to separate into it’s components. What conclusions out the sample can be formed based on the technique use

Tatiana [17]

-- The sample was a fluid.

-- It was a mixture or a suspension ... NOT a solution.

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3 years ago

Which type of erosion and deposition is most common in coastal areas around the Gulf of Mexico

Alja [10]

<span> most common in coastal areas around the gulf of mexico is mostly by river

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3 years ago

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QUICK WILL MARK BRAINLIEST

Tju [1.3M]

Answer:

37.1°

Explanation:

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3 years ago