Answer : The rate of heat transfer to the water is, 37.92 kJ/min
Explanation : Given,
Time = 10 min
Mass of water = 200 g
Latent heat of fusion of water = 334 J/g
Latent heat of vaporization of water = 2230 J/g
Now we have to calculate the rate of heat transfer to the water.

Now put all the given values in the above formula, we get:


Thus, the rate of heat transfer to the water is, 37.92 kJ/min
(D) The gravitational force between the astronaut and the asteroid.
Reason :
All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.
Answer:
neutral
Explanation:
3p - 3e = 0 and that leaves 2 neutrons so it will be neutral
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)
Incomplete question.The complete question is here
What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.
Answer:
F= 0.034 N
Explanation:
Given Data
Outer=2 cm
Edge of blade=21 cm
Mass=7.7 g
Length of blade=21 cm
The last 2cm is treated as if they were at a point 20cm from the center of rotation
To Find
Force=?
Solution
Convert the given frequency to angular frequency
ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)
ω= 3/2*π rad/sec
Now to find centripetal force.
F = m×v²/r
F= m×ω²×r
Put the data
F = 0.0077 kg × (3/2×π rad/sec
)²× 0.20 m
F= 0.034 N