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AURORKA [14]
3 years ago
6

A man is walking away from a lamppost with a light source h = 6 m above the ground. the man is m = 2 m tall. how long is the man

's shadow when he is d = 8 m from the lamppost?
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer;

= 4 m is the length of the man's shadow.

Explanation;

2/x=6/(8+x) cross multiply.

6x=2(8+x)

6x=16+2x

6x-2x=16

4x=16

x=2=16/4

x=4 m. is the length of the man's shadow.

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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9  N

4 0
3 years ago
Two people are each
docker41 [41]

Answer:

the rope should break

Explanation:

she with equal amounts of pulling are on each side then the rope should slowly start to tare apart and snap/break.

hope this helps you

3 0
2 years ago
The table and graph below show the distances traveled by two different objects. (3 points)
Aleks04 [339]

Answer:c

Explanation: the speed of object a changes but b travels at constant speed

5 0
3 years ago
A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup
Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

15.8 min*\frac{60s}{1min}=948s

The equation of motion you need is:

v_f = a*t+v_0

where v_f is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

v_f = a*t = 20.2*948 = 19,149.6 m/s

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

x=\frac{v_0+v_f}{2}t

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km

Then the ship traveled 384,000-9,076.9104*2 = 361,846.1792km at constant speed, which means that it traveled:

f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min

That added to the other interval times:

t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min

5 0
3 years ago
How many newtons of force are represented by the following amount: 3 kg·m/sec^2?
HACTEHA [7]

1 newton is the force that accelerates
1 kilogram of mass at the rate of 1 m/s² .

So in SI base units,  1 newton  =  1 kg-m/s² .

                       and    3 kg-m/s²  =  3 newtons .

4 0
3 years ago
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