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Dmitry_Shevchenko [17]
3 years ago
10

Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a ligh

t object?
Plss help!!!!
Physics
2 answers:
pantera1 [17]3 years ago
8 0

Answer:

The time of fall of an object is dependent only on the height of fall for a given acceleration due to gravity and it is independent of the object's mass

Explanation:

The gravitational force acting on an object on Earth can be observed as the weight, 'W', of the object which is a function of the mass, 'm', of the object and the relationship between the gravitational force and the mass of the object is given as follows;

W = m × g

Where;

g = The acceleration due to gravitational force = Constant

The relationship between the speed, 'u', and therefore, the time, 't', with which an object falls, the distance or height of fall, 'h', and the acceleration of the motion, 'g' is given as follows;

h = u·t + 1/2·g·t²

Given that the initial velocity of the object, 'u', of the object allowed to fall is u = 0 m/s, we have;

h = u·t + 1/2·g·t² = 0 × t + 1/2·g·t²

h = 0 × t + 1/2·g·t² = 1/2·g·t²

h = 1/2·g·t²

t² = 2·h/g

t = √(2·h/g)

Therefore, the time, 't', it takes every object to fall from a height, 'h', is dependent only on the height, 'h', for a given acceleration due to gravity, 'g' and it is independent of the mass, 'm', of the object.

guapka [62]3 years ago
4 0

Answer:

This is because the acceleration of objects due to gravity is independent of the mass of the object and is constant for all objects, therefore, all objects fall with the same speed.

Explanation:

The weight of an object or force of gravity acting on an object on the surface of earth is a product of its mass and acceleration due to gravity.

Mathematically, w = mg

where, m=mass of the object; g = acceleration due to gravity

Also, from newton's law of gravitation, gravitational force on the object ,F = GMm/r²

where G is the gravitational constant; M is mass of Earth; m is mass of object; r is the distance of separation between the object and the center of mass of the earth which is approximately the radius of earth.

Since the weight of an object is equal to the force of gravitation acting on it

W = F

mg = GMm/r²

g = GM/r²

The expression above is that of the relationship between the force of gravity acting on a body on the earth's surface, the weight of that body and the acceleration due to gravity, g.

It can be seen that the acceleration due to gravity g is independent of the mass of the object. Therefore, the acceleration of objects due to gravity is constant for all objects and all objects fall with the same speed.

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<h2>Answer:</h2>

<u>By wrapping the wire along a solenoid and connecting it to electricity</u>

<h2>Explanation:</h2>

If you wrap a copper wire into coils and run an electrical current through it, you will create a magnetic field. If you rotate a permanent magnet as opposed to an item that has been magnetized inside a coil of copper wire, you can create an electrical current. The strength of magnetic field generated is proportional to the amount of current through the winding.

5 0
3 years ago
A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate
ser-zykov [4K]

Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

= 800N/200N

= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

=80%

4. output work= load×load distance

= 800N × 5m

= 4 × 1000J

5. Efficiency = (output work/input work) ×100%

Or, 80% = (4000J/input work) ×100%

Or, 80%/100% = 4000J/inputwork

Or, 4/5 = 4000J/inputwork

Or, input work =4000J × 5/4

Input work = 5×1000J

I hope it helped! ;-)

8 0
3 years ago
Read 2 more answers
Just before hitting the ball to the ground the ball had a kinetic energy of 12.5 J. Calculate the velocity of which the ball hit
lesantik [10]

Answer:

v = √(25/m)

Explanation:

KE = ½mv²

v = √(2KE/m)

We know the kinetic energy, but not the mass

v = √(25/m)

3 0
3 years ago
A student wants to approximate the amount of work that the force due to gravity does on the student as the student walks up a se
Inessa [10]

Answer:

Options A and D

Explanation:

In this question the student needs to collect these measurements in order to approximate the work done

A. The mass of the student

D. The final vertical height above the initial vertical position.

Workdone = mgh

m = mass

g = gravity = 9.8m/s²

h = vertical height between the initial and the final positions.

The vertical height has to be known as gravity only acts straight down.

3 0
3 years ago
A mass of 0.56 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic mot
NemiM [27]

Answer:

(a) 0.42 m

(b) 20.16 N/m

(c) - 0.42 m

(d) - 0.21 m

(e) 17.3 s

Solution:

As per the question:

Mass, m = 0.56 kg

x(t) = (0.42 m)cos[cos(6 rad/s)t]

Now,

The general eqn is:

x(t) = Acos\omega t

where

A = Amplitude

\omega = angular frequency

t = time

Now, on comparing the given eqn with the general eqn:

(a) The amplitude of oscillation:

A = 0.42 m

(b) Spring constant k is given by:

\omega = \sqrt{k}{m}

\omega^{2} = \frac{k}{m}

Thus

k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m

(c) Position after one half period:

x(t) = 0.42cos\pi = - 0.42\ m

(d) After one third of the period:

x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m

(e) Time taken to get at x = - 0.10 m:

-0.10 = 0.42cos6t

6t = co^{- 1} \frac{- 0.10}{0.42}

t = 17.3 s

7 0
3 years ago
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