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GaryK [48]
4 years ago
15

a shot putter accelerates a 7.3kg shot from rest to 14m/s in 1.5r seconds. what average power was developed?

Physics
1 answer:
Sedaia [141]4 years ago
3 0

Explanation:

power =  \frac{energy \: expended}{time}

m = 7.3kg

u = 0

v = 14m/s

t = 1.5sec

P = (0.5×7.3×14²) ÷ 1.5

P = 476.93

P = 477 watt

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The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant =  8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
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Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe
Ede4ka [16]

The final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.

The given parameters;

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  • Mass of the second astronaut, = m₂
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  • Final velocity of the second astronaut, = v_f_2

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f

if v₂ > v₁, then v_1_f > v_2_f, to conserve the linear momentum.

Thus, the final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

5 0
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Radda [10]

Answer:

\theta=34 \textdegree

Explanation:

From the question we are told that:

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Angle \theta =28.0

Coefficient of static friction \alpha =0.680

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For

\sum_y=0

N=mgcos \theta

for

\sum_x=0

F_{s}=mgsin\theta

Where

F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta

F_{s}=0.68*55*9.8*cos 28

F_{s}=323.62N

Therefore

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mote1985 [20]

Explanation:

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