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3 years ago

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Physics

1 answer:

timama [110]3 years ago

3 0

Answer:

20.7N

Explanation:

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What would the weight of an astronaut be on Saturn if his mass is 68 kg and acceleration of gravity on Saturn is 10.44 m/s2? Ple

alex41 [277]

Here's the part you need to know:

   (Weight of anything) =

   (the thing's mass)
times
       (acceleration of gravity in the place where the thing is) .

   Weight = (mass ) x (gravity) .

That's always true everywhere.
You should memorize it.

For the astronaut on Saturn . . .

   Weight = (mass ) x (gravity) .

   Weight = (68 kg) x (10.44 m/s²)

   = 709.92 newtons .
__________________________________

On Earth, gravity is only 9.8 m/s².
So as long as the astronaut is on Earth, his weight is only

   (68 kg) x (9.8 m/s²)

   = 666.4 newtons .

Notice that his mass is his mass ... it doesn't change
no matter where he goes.

But his weight changes in different places, because
it depends on the gravity in each place.

4 0

2 years ago

What is the mechanical advantage of a single pulley

Viktor [21]

the answer is 1 because a single pulley has 1 rope

5 0

3 years ago

Read 2 more answers

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$