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3 years ago

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Two objects, one having three times the mass of the other, are dropped from the same height in a vacuum. At the end of their fal

l, their velocities are equal because the force of gravity is the same for both objects. Is this not true?

2 answers:

Feliz [49]3 years ago

7 0

Answer:

For two or more bodies of different mass released from height in a vacuum have the same velocity but varying force

Explanation:

Consider a body H with initial velocity (u) and final velocity V undergoing acceleration a and covering a distance( s)

From Network equation of motion it can be seen that

V^2=u^2+2as

From this it can be seen that velocity is not dependent on the the masses of the body.

Rather it depends on acceleration due to gravity which is a constant for both of the body

gogolik [260]3 years ago

5 0

Answer:

Two objects will have the equal velocities but the forces on both of them will not be equal. The equal velocities of these objects are due to their equal acceleration.

Explanation:

From the newton's equation

$v^{2} -u^{2} = 2as$

so here we can say that velocity does not depends on the mass.

The acceleration of both objects will be same but not the forces because

F = Ma

As the force is depending on the mass so it will not be the same for both objects.

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

Light of a given wavelength is used to illuminate the surfacce of a metal, however, no photoelectrons are emitted. In order to c

Jlenok [28]

Answer:

B. use light of a shorter wavelength.

Explanation:

We know that

$E= \frac{hc}{\lambda}$

h= plank's constant

c= speed of light

λ= wavelength of the incident light

so, in order to have sufficient energy for for the emission of electron, the incident's radiation must have λ small enough.

B. use light of a shorter wavelength.

6 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

What is (9x10^9)(2.6x10^-6)(1.4x10^-6) / 36

Natali [406]

Answer:

0.00091

Explanation:

(9x10^9) (2.6x10^-6) (1.4x10^-6) / 36

(9,000,000,000) (0.0000026) (0.0000014) /36

|

23,400(0.0000014) /36

|

0.03276 /36

|

0.00091

3 0

3 years ago

1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the

skelet666 [1.2K]

Answer:

Explanation:

Blue: 10/30

Red: 5/30

Yellow: 15/30

3 0

3 years ago