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2 years ago

15

The voyager 1 spacecraft is currently 20.7 billion km from earth and heading out of our solar system. how long does it take radi

o messages from voyager 1 to reach us

1 answer:

sveta [45]2 years ago

8 0

Answer:

20.7 billion = 20.7E9 converting to units of 10

20.7E9 km = 20.7E12 m

t = S /v = 20.7E12 m / 3.0E8 m/s = 6.9E4 s = 69000 s

69000 s = 1150 min = 19.2 hrs

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Tomas wants to paint his bedroom. He goes to the store and buys two cans of paint.

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C an inclined plate
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3 years ago

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how much time would it take for an airplane to reach its destination if it traveled at an average speed of 790 km/hr for a dista

NeTakaya

I think your question is incomplete because the distance between destination and departure point isn't given in the question

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3 years ago

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

$F-f=ma$

Put the value into the formula

$340-f=60\times2.0$

$f=340-60\times2.0$

$f=220\ N$

We need to calculate the coefficient of friction

Using formula of friction force

$f= \mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{220}{60\times9.8}$

$\mu =0.37$

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

$F = ma$

$a=\dfrac{F}{m}$

$a=\dfrac{220}{100}$

$a=2.2\ m/s^2$

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0

3 years ago

What is your speed (in m/s) if you walk 1000m in 20 minutes? (Hint: how many seconds are in a minute)

labwork [276]

Answer:

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Explanation:

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3 years ago

A heat engine operates between two reservoirs at 300K. During each cycle, it absorbs 200 calories from the high temperature rese

yanalaym [24]

(a) Zero

The maximum efficiency (Carnot efficiency) of a heat engine is given by

$\eta=1-\frac{T_C}{T_H}$

where

$T_C$ is the low-temperature reservoir

$T_H$ is the high-temperature reservoir

For the heat engine in the problem, we have:

$T_C = 300K$

$T_H = 300K$

Therefore, the maximum efficiency is

$\eta=1-\frac{300}{300}=0$

(b) Zero

The efficiency of a heat engine can also be rewritten as

$\eta = \frac{W}{Q_H}$

where

W is the work performed by the engine

$Q_H$ is the heat absorbed from the high-temperature reservoir

In this problem, we know

$\eta=0$

Therefore, since the term $Q_H$ cannot be equal to infinity, the numerator of the fraction must be zero as well, which means

W = 0

So the engine cannot perform any work.

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3 years ago