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sweet-ann [11.9K]
3 years ago
5

Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an

additional horizontal acceleration during its descent, it
a. strikes the plane at the same time as the other body
b. strikes the plane earlier than the other body
c. has the vertical component of its acceleration altered
d. has the vertical component of its velocity altered
e. follows a straight line path along the resultant acceleration vector
Physics
2 answers:
Elodia [21]3 years ago
8 0

Answer:

option (a) is correct.

Explanation:

one body has only vertical acceleration while the other has vertical and horizontal both the accelerations.

The time of fall of both the bodies is same

h = 0.5gt²

as initial vertical velocity for both the bodies is zero and the height from which they fall is same, so the time of fall for both the bodies is same.

Option (A) is correct.

borishaifa [10]3 years ago
4 0

Answer:a

Explanation:

Given

two bodies are falling with negligible air Resistance and one of the body is given a slight horizontal acceleration

since there is no change in vertical acceleration therefore time taken by both bodies is same irrespective of change in horizontal direction

suppose both start from rest from a height of h

time taken

t=\sqrt{\frac{2h}{g}}

The only difference is that body with horizontal acceleration will be some distance away from first                  

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Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu
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Answer:

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b) Again,  

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2 years ago
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A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially
Drupady [299]

Answer:

x(t) = - 6 cos 2t

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Force of spring = - kx

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x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

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General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

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at time = 0 sec

x (0) = 0 m

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from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

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B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

4 0
3 years ago
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