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makkiz [27]
3 years ago
15

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

Physics
1 answer:
eduard3 years ago
7 0

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

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A 10 kg ball is held above a building with a height of 30 m. What is the
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Answer: 2940 J

Explanation: solution attached:

PE= mgh

Substitute the values:

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6 0
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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
Phÿśïčš<br><br>write the relation between celsius and fahrenheit scale?​
Aleksandr-060686 [28]

Answer:

C/100 = (F-32) / 180

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5 0
3 years ago
80N force ;
lutik1710 [3]

The power of the engine is 320 W.

<u>Explanation:</u>

Power may be defined as the rate of doing work (or) work done per unit time. One unit of energy is used to do the one unit of work.

                          Power = Work done / Time taken

Given, Force = 80 N,    height = 5 m , final velocity = 4 m/s

To calculate the power, we must know the time taken.

To find the time, use the distance and speed formula which is given by

                              Time = Distance / speed

Here distance = 5 m and speed = 4 m/s

                              Time = 5 / 4 = 1.25 s.

Now,          Power = work done / time

                              = (F * d) / t  = (80 * 5) / 1.25

                  Power = 320 W.

The standard unit of power is watt (W) which is joule per second.

                                                               

4 0
3 years ago
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