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Nonamiya [84]
3 years ago
11

Steven carefully places a ????=1.95 kgm wooden block on a frictionless ramp so that the block begins to slide down the ramp from

rest. The ramp makes an angle of ????=54.3° up from the horizontal. Which forces do nonzero work on the block as it slides down the ramp?
Physics
1 answer:
NemiM [27]3 years ago
4 0

Answer:

The forces that do non-zero work on the block are gravity and normal reaction force

Explanation:

In this question we have given,

mass of the wooden block,m=1.95 kg

angle between ramp and horizontal surface, \alpha =54.3^o

in this case ramp is friction less so friction force will be zero and hence work done due to friction force will be zero.

As wooden block is sliding down due to gravity(gravity is pulling the block in downward direction).

so two forces will act on wooden block as it slides down the ramp that are gravity and normal reaction force.  

Therefore, the forces that do non-zero work on the block are gravity and normal reaction force.

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An object with a mass of 10 kg is accelerated upward at 2 m/sec2. What force is required?
telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

6 0
2 years ago
You and your friend are pushes hard against a stationary wall. If you push 3 times harder than your friend, then the amount of w
shtirl [24]

Answer:

Work = F * s    where s is the distance F moves

Since F is stationary, in this case, "no work" is done by either person

5 0
1 year ago
The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a
emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\  I_2= 0.125 \ foot-candles

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

To learn more about the distance refer to the link;

brainly.com/question/26711747

#SPJ1

6 0
2 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
A wheelbarrow is pushed with a force of 40 N. If 6,000 J of work is
stepladder [879]

Answer:

Distance = 150 meters

Explanation:

Given the following data;

Work done = 6,000 Joules

Force = 40 Newton

To find the total distance covered by the wheelbarrow;

Workdone = force * distance

Substituting into the formula, we have;

6000 = 40 * distance

Distance = 6000/40

Distance = 150 meters

Therefore, the total distance the wheelbarrow was pushed is 150 meters.

5 0
3 years ago
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