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jekas [21]
3 years ago
10

PLEASE HELP

Physics
1 answer:
hichkok12 [17]3 years ago
7 0

The correct answer is 195.6 N

Explanation:

Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:

F = 120 kg x 1.63 m/s2

F= 195.6 N

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A sphere of radius R = 0.295 m and uniform charge density -151 nC/m^3 lies at the center of a spherical, conducting shell of inn
cupoosta [38]

Answer:

a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C

Explanation:

a) r = 0.76R

As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:

Q = ρ*V(r) = ρ*\frac{4}{3} *\pi *r^{3}

where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³

Q = -151e-9 *\frac{4}{3} *\pi *0.224m^{3} = -7.11e-9 C

Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:

E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-7.11e-9C}{(0.76*0.295m)^{2}} =-1.27e3 N/C

⇒ E = -1.27*10³ N/C

b) r= 3.90 R

As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0

c) r = 2.8 R

As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.

First we need to find the total charge of the sphere, as follows:

Q = ρ*V =

Q = -151e-9 *\frac{4}{3} *\pi *0.295m^{3} = -16.2e-9 C

In the same way that for a) we can write the following expression:

E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-16.2e-9C}{(2.8*0.295m)^{2}} =-0.21e3 N/C

⇒ E = -0.21*10³ N/C

d) r= 7.30 R

In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.

As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.

This leaves an excess charge on the outer surface of the shell as follows:

Qsh = 66.7 nC - 16.2 nC = 50.5 nC

Now, we can repeat the same process than for a) and c) as follows:

E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{50.5e-9C}{(7.3*0.295m)^{2}} =0.1e3 N/C

⇒ E = 0.1*10⁻³ N/C

6 0
3 years ago
A ball is dropped from rest on a cliff what is the speed of the ball 5 seconds later
diamong [38]

Answer:

The velocity of the ball after 5 seconds will be 49 m/s

Explanation:

<em>v = final velocity</em>

<em>u = initial velocity</em>

<em>g = acceleration due to gravity</em>

<em>t = time</em>

Initial velocity of the ball = 0 (As the ball is dropped from rest )

Acceleration due to gravity = 9.8 m/s

Time taken = 5 sec

As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question

Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v

Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.

Velocity of the ball after 5 seconds

Distance covered by the ball in 5 sec

4 0
2 years ago
The diagram below shows a golf ball being struck by a club. The ball leaves the club with a speed of 40 meters per second at an
hjlf

Answer:

61.22m

Explanation:

Maximum height in projectile is expressed as;

H = u²sin²θ/2g

u is the speed

g is the acceleration due to gravity

θ is the angle of projection

Substitute the given values into the formula;

H = u²sin²θ/2g

H = 40²sin²60/2g

H = 1600(0.8660)²/2(9.8)

H = 1600(0.8660)²/2(9.8)

H = 1,199.9/19.6

H = 61.22m

Hence the maximum height achieved by the ball is 61.22m

4 0
2 years ago
The normal eye, myopic eye and old age
Leviafan [203]

Answer:

length f′0 when the eye is presbyopic and by a screen (the retina) at the distance d from the

Explanation:

7 0
3 years ago
PLEASE HELP MEEEEEE!!!!!!!!!!!!!!!!!!
Simora [160]
You would need to freeze it in a freezer. Hope this helps if it does could I have brainlist thanks
5 0
2 years ago
Read 2 more answers
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