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blsea [12.9K]
3 years ago
6

The weight of an astronout is 60kg on earth, find the wight of the same object on the planet where the gravitational attraction

has been reduced to 1/10 of the earths pull
Physics
1 answer:
Marianna [84]3 years ago
3 0

The astronaut's weight is not 60 kg anywhere, because kg is a unit of mass, not weight.

If the astronaut's mass is 60 kg, then his weight is (60 kg)x(acceleration of gravity).

That's 588 Newtons on Earth, and 58.8 Newtons on a planet with 1/10 Earth's gravity.

The astronaut's mass of 60 kg goes with her, and doesn't depend on where she is.

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A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60
mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

z(t) = \sqrt{x^{2} + y(t)^{2}}

z(t) = \sqrt{70^{2} + 60t^{2}}

Now, differentiate the above eqn w.r.t 't':

\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600

\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t

For t = 4 s:

\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s

4 0
3 years ago
Question 1 of 10
blsea [12.9K]

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5 0
3 years ago
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A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight
LuckyWell [14K]

Answer:

At an angle of 30^{\circ}

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}

6 0
2 years ago
12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?
ziro4ka [17]
  • initial velocity=u=0m/s
  • Final velocity=v=25m/s
  • Time=t=30s

\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}

\\ \tt\longmapsto Acceleration=\dfrac{25}{30}

\\ \tt\longmapsto Acceleration=0.8m/s^2

6 0
3 years ago
PLEASE HELP ME I JUST CAN'T FIGURE THIS ONE OUT
kupik [55]
Indeed the answer is c!!
4 0
3 years ago
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