A capacitor with the capacitance constant C has a charge of Q and a potential difference of V.

The time taken for an object to fall from a height of h m from the earth'fs surface is t s.

telo118 [61]

Answer:

B. Longer than t s,

Explanation:

Gravitational accln on earth is 9.8 m/s^2,

and one you provided as on moon is 1.6 m/s^2

that mean on moon gr. accl. is lesser!

now the time taken on earth will be lesser cuz from the same height if you drop the object from rest!

since accln on earth is higher,the object will attain higher velocity as compare to that of on moon!

✌️:)

6 0

2 years ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

3 years ago

A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it

mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0

3 years ago

Masses m and 2m approach each other at the same speed v and collide head-on. Find the final speed of mass 2m, if mass rebounds a

antoniya [11.8K]

Answer:

<em>The rebound speed of the mass 2m is v/2</em>

Explanation:

I will designate the two masses as body A and body B.

mass of body A = m

mass of body B = 2m

velocity of body A = v

velocity of body B = -v since they both move in opposite direction

final speed of mass A = 2v

final speed of body B = ?

The equation of conservation of momentum for this system is

mv - 2mv = -2mv + x

where x is the final momentum of the mass B

x = mv - 2mv + 2mv

x = mv

to get the speed, we divide the momentum by the mass of mass B

x/2m = v = mv/2m

speed of mass B = <em>v/2</em>

6 0

3 years ago

An electron (m=9.11x10^-31 kg) moves in a circle whose radius is 2.00x10^-2 m. If the force acting on the electron is 4.60x10^-1

harina [27]

Answer:

The speed of the electron is v = 1.01 x 10¹⁵ m/s

Explanation:

Given data,

The mass of the electron, m = 9.11 x 10⁻³¹ kg

The radius of the circle, r = 2.00 x 10⁻² m

The force acting on electron, F = 4.60 x 10⁻¹⁴ N

The speed of the electron, v = ?

The centripetal force of the electron is given by

F = mv² / r

∴ v² = Fr/m

v =√(Fr/m)

Substituting the given values in the above equation,

v =√( 4.60 x 10⁻¹⁴ x 2.00 x 10⁻² / 9.11 x 10⁻³¹ )

v = 1.01 x 10¹⁵ m/s

Hence, the speed of the electron is v = 1.01 x 10¹⁵ m/s

4 0

3 years ago