8. Imagine a clock whose hands move in distinct "clunks."

A 48.0-kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like t

marusya05 [52]

Answer:

The astronaut is moving at a speed of 0.36m/s

Explanation:

Speed here corresponds to velocity

The astronaut's mass = 48kg

velocity of astronaut = ?

mass of socket = 0.72kg

velocity of socket = 5m/s

mass of the spanner = 0.8kg

velocity of spanner = 8m/s

change in time = 0.05 -0 = 0.05sec

mass of the mallet = 1.2kg

velocity of mallet = 6m/s

change in time = 9.9 -0 = 9.9sec

To find the astronaut velocity, we would calculate the total momentum which is the astronaut.

∑momentum (M) = ∑astronaut momentum

∑M = ∑astronaut M

∑astronaut M = M of socket + M of spanner + M of mallet

momentum = mass × velocity

(mass × velocity)of astronaut = (0.72×5) + (0.8×8) + (1.2×6)

48 × velocity of astronaut= 3.6 + 6.4 + 7.2

48 × velocity of astronaut= 17.2

velocity of astronaut = 17.2/48

velocity of astronaut = 0.36m/s

The astronaut is moving at a speed of 0.36m/s

5 0

3 years ago

What will happen to force?

telo118 [61]

Answer:

it will double because im right

8 0

2 years ago

Imagine a raindrop starting from rest in a cloud 2 km in the air. If it fell with no air friction at all, it would accelerate to

LenKa [72]

Answer:

2) 433 mph

Explanation:

The final velocity of the raindrop as it reaches the ground can be found by using the equation for a uniformly accelerated motion:

$v^2 = u^2 + 2ad$

where

v is the final velocity

u = 0 is the initial velocity (the raindrop starts from rest)

a = g = 9.8 m/s^2 is the acceleration due to gravity

d = 2 km = 2000 m is the distance covered

Solving for v,

$v=\sqrt{u^2 +2gd}=\sqrt{0^2+2(9.8 m/s^2)(2000)}=198 m/s$

And keeping in mind that

1 mile = 1609 metres

1 hour = 3600 s

The speed converted into miles per hour is

$v=198 \frac{m}{s}\cdot \frac{3600 s/h}{1609 m/mi}=433 mph$

5 0

3 years ago

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

$k= \frac{F}{x}$

$k= \frac{mg}{x}$

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s

4 0

2 years ago

How do you find the velocity after a collision

Evgen [1.6K]

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

Explanation:

7 0

3 years ago