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katen-ka-za [31]
2 years ago
11

A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g

=10m/s2, the body will reach the surface of the earth in how much time?

Physics
1 answer:
kap26 [50]2 years ago
7 0
I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!

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Find your mass if a scale on earth reads 650 N when you stand on it.
netineya [11]

Weight = (mass) x (gravity)

Acceleration of gravity on Earth = 9.8 m/s²

                                           Weight on Earth = (mass) x (9.8 m/s²)

Divide each side by  (9.8 m/s²):          Mass = (weight) / (9.8 m/s²)

                                                            Mass = (650 N) / (9.8 m/s²)

                                                           Mass = 66.33 kg  (rounded)
 
7 0
2 years ago
Discuss how the hardness or softness of the landing surface is related to the time required to stop the egg
monitta

Answer

To understand this concept it is necessary to understand Newton's Second Law

According to Newtons Second law applied force is equal to rate of change of momentum of a body.

Mathematically,

F=\frac{dp}{dt}

here, \frac{dp}{dt} is rate of change of momentum with respect to time

It means If two eggs fall from same height,one on softer surface and other on hard surface that time the momentum of both eggs will remain equal at both the surfaces. But, impact time will be different. On hard surface egg will stop almost instantly so impact time will be small and hence the force on egg will be large therefore the egg will breakup.

On the other hand on the soft surface like a cotton, egg will not stop instantly but it will slow down for a few seconds and then stop due to which the time of impact will increase. Therefore the force on egg will be less and it won't break up.


3 0
2 years ago
Hurryyyyyy When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand p
masya89 [10]
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.

</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
6 0
2 years ago
Read 2 more answers
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

5 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
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