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sertanlavr [38]
3 years ago
9

You toss a racquetball directly upward and then catch it at the same height you released it 1.82 s later. assume air resistance

is negligible. (a) what is the acceleration of the ball while it is moving upward?
Physics
1 answer:
Anton [14]3 years ago
5 0
Neglecting air resistance, the acceleration of the ball is
the acceleration of gravity ... 9.8 m/s² downward.

It doesn't matter what you toss, what it's mass is, what it weighs,
what color it is, how much it cost, what its shape or size is, how
fast you toss it, in what direction, or how long it's in the air.

Its horizontal acceleration is zero and its vertical acceleration
is 9.8 m/s² downward, from the moment it leaves your hand
until the moment somebody catches it or it hits the ground.
 
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You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find
wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

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3 years ago
A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the c
UkoKoshka [18]

Answer:0.00125 watts

Explanation:

resistance=50 ohms

Current=5 milliamps

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8 0
3 years ago
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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

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Answer:

<h3><em>2</em><em>4</em><em>7</em><em>9</em><em> </em><em>Newton</em></h3>

<em>Sol</em><em>ution</em><em>,</em>

<em>Mass</em><em>=</em><em>1</em><em>0</em><em>0</em><em> </em><em>kg</em>

<em>Accele</em><em>ration</em><em> </em><em>due</em><em> </em><em>to</em><em> </em><em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>2</em><em>4</em><em>.</em><em>7</em><em>9</em><em> </em><em>m</em><em>/</em><em>s^</em><em>2</em>

<em>Now</em><em>,</em><em>.</em>

<em>weight = m \times g \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 100  \times 24.79 \\  \:  \:  \:  \:  \:  \:  = 2479 \: newton</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>

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3 years ago
What was the result of the Mexican army's victory over the Texan garrison at the Alamo?
sasho [114]

The result of the Mexican victory was that fallen defenders became heroes to the cause of Texan independence.<span> The Battle of the Alamo took place between February 23 and March 6, 1836 and became the central episode of the Texas Revolution . After this thirteen-day battle, the Mexican troops of General President Antonio Lopez de Santa Anna began an attack on San Antonio de Bexar, the current San Antonio in Texas. The Battle of the Alamo fought the army of Mexico against a group of Texan rebels, mostly American settlers. More than four thousand men from Santa Ana stood in front of the Alamo Fort , the last stronghold of the rebels, which barely reached 187. The Alamo was not a fortress prepared to withstand a siege. It is believed that all the rebels of the Alamo died in the siege, but Santa Anna came to lose up to about 900 men during the days that lasted the fight. However, the worst result for Santa Ana was precisely the resistance that the Texan rebels had in the Alamo, which fostered the fighting spirit of the Texans. A few days later, on March 14, 1836, Texas became independent from Mexico and a month later, Santa Ana was imprisoned.</span>

3 0
3 years ago
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