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3 years ago

A man throws a football straight into the air. As it rises, it slows down. Which

Physics

1 answer:

mario62 [17]3 years ago

8 0

Answer:

if it is slowing down than it is gaining potential energy

Explanation:

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A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m

harina [27]

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

6 0

3 years ago

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

3 years ago

PLEASE HELP ASAP!!!!!!!1

ArbitrLikvidat [17]

Age of dino: Mesozoic era

End of earth a dessert: End of the Mesozoic

a layer: Paleozoic

7 0

2 years ago

The options are stationary, running, and walking

Sunny_sXe [5.5K]

Answer:

A is walking

B is stationary

C is running

D is walking

E is stationary

4 0

2 years ago

An object moves in one dimension according to the function x(t)=13at3, where a is a positive constant with units of ms3. During

9966 [12]

Answer:

B) 1/5 ba^2 T^5

Explanation:

The dissipated energy is given by the work done over the object by the force F=-bv. The work is given by the following formula:

$dW=Fdx$

you derivative the function f(x) and replace v by the derivative dx/dt you obtain:

$v=\frac{dx}{dt}=at^2\\\\dx=at^2dt\\\\W=\int_0^{T} Fdx=-\int_0^Tvbdx=-\int_0^Tb(at^2)(at^2dt)\\\\W=-ba^2\frac{T^5}{5}=-\frac{1}{5}ba^2T^5$

hence, the dissipated energy is 1/5 ba^2 T^5

7 0

3 years ago