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mel-nik [20]
2 years ago
15

An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje

ct is dropped from the plane. If air resistance is negligible, what happens to the vertical distance between the two objects while they are both falling?
Physics
1 answer:
Delvig [45]2 years ago
8 0

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

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Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor
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156.8 Joules of energy is in the box's gravitational potential energy store

<u>Explanation</u>:

<em>Given:</em>

Mass of the box Dane is holding = 8 Kilograms

Height at which Dane is holding the box above the ground= 2 metres

<em>To Find:</em>

Gravitational potential energy in the box=?

<em>Solution:</em>

gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J

Thus  Gravitational potential energy is  represented as,

PE_g=mgh

where

PE_g is the gravitational potential energy

m is the mass

h is the height

g is the gravitational force( 9.8 m/s^2)

Now substituting the given values,

PE_g=8\times 9.8\times 2

PE_g=156.8 Joules

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2 years ago
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
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Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

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2 years ago
A coworker did not clean his work area before going home this could cause an accident so you quickly clean up the next day you s
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Answer:

THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM

Explanation:

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Answer:

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Explanation:

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Answer:

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