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A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strike

vivado [14]

Answer:

Explanation:

Ball is thrown downward:

initial velocity, u = - 20 m/s (downward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(a) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2}=(-20)^{2}+2\times 9.8 \times 60$

v = 39.69 m/s

(b) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = - 20 - 9.8 t

t = 2 second

Now the ball is thrown upwards:

initial velocity, u = 20 m/s (upward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(c) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2}=(-20)^{2}+2\times 9.8 \times 60$

v = 39.69 m/s

(d) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = + 20 - 9.8 t

t = 6.09 second

3 0

3 years ago

After creating a prototype, which of the following would not be and appropriate next step in the design process of a new technol

Dafna11 [192]

The answer would be to research the need. This should have been done before the project began.

6 0

4 years ago

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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a

zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

$\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s$

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

4 0

3 years ago

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A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must

mezya [45]

We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)

4 0

3 years ago

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during a space shuttle launch about 830,000 kg of fuel is burned in 8 min. the fuel provides the shuttle with a constant thrust,

kirill115 [55]

The shuttles acceleration in the creases as the fuel is burned because the acceleration of the obect as produced by net force is directly proportional to the magnitude of the net force.

8 0

3 years ago

Plsase help meeeeeee​

Plsase help meeeeeee