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mel-nik [20]
4 years ago
13

An object slides down a very smooth ramp, with negligible friction. It slides with constant acceleration a under the action of t

wo forces: the supporting force n exerted by the ramp, and the force m g due to gravity exerted by the earth. What can you say about how the magnitude of n compares to the magnitude of m g?
1. n is less than m g.
2. Their magnitudes are equal, as is almost always true on the earth’s surface.
3. None of these
4. n is greater than m g.

Physics
1 answer:
sveta [45]4 years ago
3 0

Answer:

In statics condition and if is a body is over plane surface N = mg

In all cases which a body is on a ramp force N will be smaller than mg, and depending on the level of the ramp (bigger the ramp smaller force N)

Explanation:

See annex (Figure 1)

We can see with the help of force diagram the relation between  N force  and mg

Lets look first the body on the ramp assuming there is not movement

The  sum of forces alog y axis = 0

mg* cos∠ BAC = N (normal force)

So N is always a fuction of mg and of the cosine of the inclination angle, and the cosine of angle varies from value 1 when the angle is 0 (look the case where the block is on the surface without inclination N = mg ).

And as the inclination increase the value of cosine will become smaller and so will N force which is directly proportional of cosine of the inclination angle

Therefore we can say that force N is always minor than mg unless in extreme case where the block is over the surface in which case N takes its maximum value N = mg

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Grace drives her car 168 km in 2 hours. What is her average speed in kilometers per hour?
wlad13 [49]

Answer:

84kliometers

Explanation:

divide one hundred and sixty eight kilo meters by two hours

5 0
3 years ago
A weightlifter must exert 25 Newtons of force to
kakasveta [241]

Work=Force \: × \: displacement \\ => W=Fs \\  \\ Given, \\ Force=25 \: N \\ Displacement=2m \\  \\ so \: work = 25 \: N \:  \times 2m \\  =  > work = 25kgm {s}^{ - 2} \times 2m \\  =  > work = 50kg {m}^{2} {s}^{ - 2}  \\  =  > work = 50J

This is the answer.

Hope it helps!!

6 0
3 years ago
Bert is playing on his school's basketball team and one of the players on the other team continues to push him off balance when
allsm [11]

Answer:

Talk to his team captain and ask him to alert the referee to keep a better eye on the player.

4 0
3 years ago
A cyclist is travelling at 4 metres per second for 100 metres,<br> how long does it take?
Slav-nsk [51]

Answer:

25 seconds!

Explanation:

I just had to find what number to multiply 4 to get 100 and it was 25

4x25=100

Or you could do the easier way by dividing 100 by 4 which has the same answer: 25

100÷4=25

7 0
4 years ago
Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi
laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s

For second drop time is given by

t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s

Distance from second drop

s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m

Distance from second drop is 0.83999 m

Distance from third drop

s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m

Distance from third drop is 0.20999 m

6 0
4 years ago
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