Answer:
Hope this help you!!
Explanation:
Crust : The crust is the thinnest layer of the Earth. It has an average thickness of about 18 miles below land, and around 6 miles below the oceans. The crust is the layer that makes up the Earth's surface and it lies on top of a harder layer, called the mantle.
Mantle : The mantle is the mostly-solid bulk of Earth's interior. The mantle lies between Earth's dense, super-heated core and its thin outer layer, the crust. The mantle is about 1,802 miles thick, and makes up a whopping 84% of Earth's total volume
Outer Core : The outer core is the third layer of the Earth. It is the only liquid layer, and is mainly made up of the metals iron and nickel, as well as small amounts of other substances. The outer core is responsible for Earth's magnetic field. As Earth spins on its axis, the iron inside the liquid outer core moves around.
Inner Core : It's Almost The Size of the Moon. The Earth's inner core is surprisingly large, measuring 1,516 miles across. It's Mostly Made of Iron. It Spins Faster Than the Surface of the Earth. It Creates a Magnetic Field.
Answer:
The orbital speed of this second satellite is 5195.16 m/s.
Explanation:
Given that,
Orbital radius of first satellite 
Orbital radius of second satellite 
Mass of first satellite 
Mass of second satellite 
Orbital speed of first satellite = 4800 m/s
We need to calculate the orbital speed of this second satellite
Using formula of orbital speed

From this relation,

Now, 

Put the value into the formula


Hence, The orbital speed of this second satellite is 5195.16 m/s.
Answer:
speed of electrons = 3.25 ×
m/s
acceleration in term g is 3.9 ×
g.
radius of circular orbit is 2.76 ×
m
Explanation:
given data
voltage = 3 kV
magnetic field = 0.66 T
solution
law of conservation of energy
PE = KE
qV = 0.5 × m × v²
v =
v =
v = 3.25 ×
m/s
and
magnetic force on particle movie in magnetic field
F = Bqv
ma = Bqv
a =
a =
a = 3.82 ×
m/s²
and acceleration in term g
a =
a = 3.9 ×
g
acceleration in term g is 3.9 ×
g.
and
electron moving in circular orbit has centripetal force
F =
Bqv =
r =
r =
r = 2.76 ×
m
radius of circular orbit is 2.76 ×
m
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.