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sesenic [268]
3 years ago
15

A baseball of mass 0.143 kg is thrown from the roof of a building 23.1 m above the ground. Its initial velocity is 37.6 m/s at a

n angle of 45◦ above the horizontal. The acceleration of gravity is 9.81 m/s 2 . What is the maximum height above the ground that the ball reaches?
Physics
1 answer:
Olin [163]3 years ago
3 0

Answer:

A 0.17 kg baseball is launched from the roof of a building 14 m above the ground. Its initial velocity is 29 m/s at 40° above the horizontal. Assume any effects of air resistance are negligible.

(a) What is the maximum height above the ground that the ball reaches?

m

(b) What is the speed of the ball as it strikes the ground?

m/s

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An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
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To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

MV=mv

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

V = \frac{mv}{M}

Also we have that m/M = 1/7 times

On the other hand we have from law of conservation of energy that

W_f = KE

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

F_f*S = \frac{1}{2}MV^2

\mu M*g*S = \frac{1}{2}MV^2

\mu g*S = \frac{1}{2}( \frac{mv}{M})^2

\mu = \frac{1}{2} (\frac{1}{7}v)^2

\mu = \frac{1}{98}v^2

\mu = \frac{1}{g(98)(5.1)}v^2

Here we can apply the law of conservation of energy for light mass, then

\mu mgs = \frac{1}{2} mv^2

Replacing the value of \mu

\frac{1}{g(98)(5.1)}v^2  mgs = \frac{1}{2}mv^2

Deleting constants,

s= \frac{(98*5.1)}{2}

s = 249.9m

7 0
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Explain in detail what specific factors affect the momentum of a bicycle going down hill.
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Mass and velocity are the two terms which affect momentum of a bicycle going hill down.

Explanation:

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So, obviously with no doubt mass and velocity are the two terms which affect momentum.

Momentum(p) = Mass(m) * Velocity(v)

The momentum also depends upon the mass and speed of the object.

More the mass of the object more is the momentum.

Depending upon the gravity and bicycle's motion speed momentum varies.

Bicycle moves faster the down hill if it moves with some speed as it has lesser mass the momentum also will be less.

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Based on the information in the table, which elements are most likely in the same periods of the periodic table?
irinina [24]

Answer:

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6 0
3 years ago
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia
kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass M_{Io}=8.93\times10^{22}\ kg

Radius = 1821 km

Height h_{Io}=500\ km

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

g =\dfrac{GM_{Io}}{(R_{Io})^2}

Put the value into the formula

g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}

g=1.79\ m/s^2

Let  v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

H=\dfrac{v^2}{2g}

Using ratio of height of earth and height of Io

\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}

\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}

Put the value into the formula

\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}

\dfrac{H_{e}}{H_{Io}}=0.182

H_{e}=0.182\times H_{Io}

H_{e}=0.182\times500

H_{e}=91\ km

Hence, The height reached by the material on Earth is 91 km.

3 0
3 years ago
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