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luda_lava [24]
2 years ago
5

During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.

Physics
1 answer:
Allushta [10]2 years ago
6 0
Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

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There is a reflection at the boundary between two materials if there is a change of __________ at the boundary.
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Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g
djyliett [7]

Answer:

a) m = 69.0 kg

b) release some gas in the opposite direction to the astronaut's movement

Explanation:

a) Let's use Newton's second law

         F = m a

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         m = 60.0 / 0.870

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b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved

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One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.

3 0
2 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

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v=\sqrt{19.6}

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We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

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Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

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