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3 years ago

A projectile is defined as:

1 answer:

8 0

A projectile is defined as an object that has an initial velocity and follows a path that dependent on the gravitational acceleration.
The gravitational force initially will act on the oposite direction until it reach the maximum height, and right after that, the gravitational force will accelerate the speed of the projectile.

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Which is larger, the Sun's pull on Earth or Earth's pull on the Sun?

Juliette [100K]

B it’s the most logical answer

3 0

4 years ago

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a

s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

6 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

solmaris [256]

Answer:

114.075 N

798.525 Nm

Explanation:

C = Drag coefficient = 0.5

ρ = Density of air = 1.2 kg/m³

A = Surface area = 9 m²

v = Velocity of wind = 6.5 m/s

r = Height of the tree = 7 m

Drag equation

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 0.5\times 9\times 6.5^2\\\Rightarrow F=114.075\ N$

Magnitude of the drag force of the wind on the canopy is 114.075 N

Toque is given by the product of force and radius

$\tau=F\times r\\\Rightarrow \tau=114.075\times 7\\\Rightarrow \tau=798.525\ Nm$

Torque exerted on the tree, measured about the point where the trunk meets the ground is 798.525 Nm

5 0

3 years ago

A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th

Elis [28]

Answer:

The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

$y=y_0+V_0t-gt^2/2$

We are only interested in the vertical motion, so that equation is all what you need.

1. Water ballon is thrown horizontally at sped 2.00 m/s.

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

$y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s$

2. Water ballon thrown straight down at 2.00 m/s

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

$0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s$

To solve the equation you can use the quadratic formula.

$t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92$

You get two times. One of the times is negative, thus it does not have physical meaning.

3. Conclusion:

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

5 0

3 years ago

Explain two ways in which water’s properties help sustain life. Earth Science

hjlf

Answer:

Water is essential for all living things. Water's unique density, high specific heat, cohesion, adhesion, and solvent abilities allow it to support life.

6 0

3 years ago

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