Answer:
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Explanation:
We can answer this question by using Kepler's second law of planetary motion, which states that:
"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"
This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).
In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
To solve this problem, we should recall the law of
conservation of energy. That is, the heat lost by the aluminium must be equal
to the heat gained by the cold water. This is expressed in change in enthalpies
therefore:
- ΔH aluminium = ΔH water
where ΔH = m Cp (T2 – T1)
The negative sign simply means heat is lost. Therefore we
calculate for the mass of water (m):
- 0.5 (900) (20 – 200) = m (4186) (20 – 0)
m = 0.9675 kg
Using same mass of water and initial temperature, the final
temperature T of a 1.0 kg aluminium block is:
- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)
- 900 T + 180,000 = 4050 T
4950 T = 180,000
T = 36.36°C
The final temperature of the water and block is 36.36°C
The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, 
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:
where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s
Substituting,
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
