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Drupady [299]
3 years ago
15

A soldier throws a grenade horizontally from the top of a cliff. Which of the following curves best describes the path taken by

the grenade?
1. Circle
2. Ellipse
3. Hyperbola
4. Parabola
Physics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

4. Parabola

Explanation:

The motion of the flight of the grenade is a projectile motion, it's shape is best illustrated using a parabolic diagram.

It's not a circle as the path is not totally round.

It's not an ellipse as the path of motion is not a a completely bounded shape.

Its not a hyperbola as a hyperbola is an open curve with two branches.

Its a parabolic shape because a parabola has just one branch

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Explanation:

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What is the net force experienced by the rope? Include both the magnitude and direction
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All the physical world objects that comers in the contact to exert the force to each other. The contact forces are different from their names and what type of force they exert.

Explanation:

The cables and the ropes are the useful objects that exert the forces that can efficiently transfer the force from a significant distance.

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Which of the following adaptations best protects birds from bird catchers?
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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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