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3 years ago

6

a ball is projected horizontally from the top of a hill with a velocity of 30 m per second if it reaches the ground 5 seconds la

ter the height of the hill is

2 answers:

alina1380 [7]3 years ago

7 0

The height of the hill is 125m.

Reptile [31]3 years ago

7 0

Answer:

The height of the hill is 125m.

Explanation:

Since the ball is projected horizontally from the top of a hill, there is no vertical component of the velocity of projection. Therefore, so far as motion in the vertical direction is concerned, the ball is just dropped, ie its initial velocity u is merely 0 m/s. It reaches the ground after 5 seconds from the moment of projection, under the action of accerelation due to gravity.

Using,

s = u t + ½ a t². In this expression u= 0 m/s, a = 10m/s², t = 5 s. Substituting in the equation we get,

s = 0× t + ½ ×10 m/s²× 5²s²= 5× 25 m = 125m.

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an astronaut on an eva has wandered dangerously far away from the shuttle. she has also exhausted all the fuel in her jet pack.

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The conservation of the momentum allows to find the result of how the astronaut can return to the spacecraft is:

Throwing the thruster away from the ship.

The momentum is defined as the product of the mass and the velocity of the body, for isolated systems the momentum is conserved. If we define the system as consisting of the astronaut and the evo propellant, this system is isolated and the internal forces become zero. Let's find the moment in two moments.

Initial instant. Astronaut and thrust together.

p₀ = 0

Final moment. The astronaut now the thruster in the opposite direction of the ship.

$m_f$ = m v + M v '

where m is propellant mass and M the astronaut mass.

As the moment is preserved.

0 = m v + M v ’

v ’= $- \frac{m}{M} \ v$

We can see that the astronaut's speed is in the opposite direction to the propeller, that is, in the direction of the ship.

The magnitude of the velocity is given by the relationship between the masses.

In conclusion, using the conservation of the momentun we can find the result of how the astronaut can return to the ship is:

Throwing the thruster away from the ship.

Learn more here: brainly.com/question/14798485

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2 years ago

you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll

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Technically, it should roll forever.

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3 years ago

You whirl a stone in a horizontal circle in such a way that the stone is in uniform circular motion. Which of the following is t

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a. The direction of the stone's velocity changes as it moves around the circle.

b. The magnitude of the stone's velocity does not change.

d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.

Above given are true for the given situation.

<u>Answer:</u> Option A, B and D

<u>Explanation:</u>

Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.

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3 years ago

Where is the energy in a glucose molecule stored?

Semmy [17]

Answer:

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3 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago