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3 years ago

You are falling off the edge.. What should you do to avoid falling..

Physics

2 answers:

FrozenT [24]3 years ago

4 0

You could try not to panic and try ur best to get up. If not, call for help.

Ugo [173]3 years ago

3 0

If it is a matter of which way you are going you could lean forward. It would help to put all the weight opposite of where you are falling.

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COULD SOMEONE PLS DO THESE FOR ME <br> I’LL LOVE U FOREVER AND EVER

sladkih [1.3K]

I can help you do them for yourself :)

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4 years ago

n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat

musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0

3 years ago

Two astronauts of mass 100 kg are 2 m apart in outer space. What is the

fredd [130]

The force of gravity between the astronauts is $1.67\cdot 10^{-7}N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

$m_1 = 100 kg\\m_2 = 100 kg$

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

$F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

3 years ago

An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False

Umnica [9.8K]

Hi There! :)

An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P

7 0

3 years ago

Read 2 more answers