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3 years ago

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A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

he price is now Traveling at 5.000 feet per second. What acceleration did the socket deliver?

1 answer:

aleksandr82 [10.1K]3 years ago

6 0

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

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When and where did the first nuclear reactor generate electricity

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<h2>Answer: On December 20th, 1951 in Idaho, United States. </h2>

The world's first experimental nuclear power plant was the Experimental Breeder Reactor Number One (EBR-I), which was built in a desert in Idaho, United States.

This reactor made history when, on December 20th, 1951, four 200-watt light bulbs were illuminated by means of atomic energy, specifically by nuclear fission reaction.

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3 years ago

two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second

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Answer:

Approximately $0.88\; {\rm m \cdot s^{-1}}$ to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Since momentum of this system (of the astronauts) conserved:

$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}$ .

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$\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

The final momentum of the first astronaut ( $m = 64\; {\rm kg}$ , $v = 0.8\; {\rm m\cdot s^{-1}}$ to the left) would be $p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}}$ to the left.

Let $p_{2}$ denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be $(p_{1} + p_{2})$ .

$\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}$ .

Hence, $p_{2} = (-p_{1})$ . In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude ( $51.2\; {\rm kg \cdot m \cdot s^{-1}}$ ) but opposite in direction (to the right versus to the left.)

Rearrange the equation $p = m\, v$ to obtain an expression for velocity in terms of momentum and mass: $v = (p / m)$ .

$\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}$ .

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2 years ago

In fast-pitch softball, a pitcher swings her arm from straight overhead in a circle, releasing the 0.196 kg ball at the bottom o

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Answer:

560.06714 Nm

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = $\pi$ (Half rotation)

v = Velocity of bat = 29.8 m/s

M = Mass of bat = 11.3 kg

m = Mass of ball = 0.196 kg

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$I=\dfrac{1}{3}MR^2+mR^2\\\Rightarrow I=\dfrac{1}{3}11.3\times 0.984^2+0.196\times 0.984^2\\\Rightarrow I=3.83687577\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=3.836875776\times 145.96958\\\Rightarrow \tau=560.06714\ Nm$

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