We have to remember a point , which is ' the cart or spring rest on a smooth horizontal track ' , i.e., any frictional force doesn't take place.
<u>Explanation:</u>
a) According to the question the cart is pulled to position A and released, i.e., the velocity of the cart at A initially (say time,t=0) is 0 m/s ,then moves toward position E, where it reverses direction and returns again to position A , in the 2nd phase cart moves along A to E , the cart's velocity increase and again goes to zero at point E and again change the direction, hence
( File has been attached)
b) Let's , the distance between two consecutive points is x meter and the spring constant is k N.m
c) ( File has been attached)
d) Movinf Right
e) Moving left
Answer:
- 5436 J
Explanation:
mass of car, m = 120 kg
radius of loop, r = 12 m
velocity at the bottom (A) = Va = 25 m/s
Velocity at the top(B) = Vb = 8 m/s
Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m
by use of Work energy theorem
Work done by all the forces = change in kinetic energy of the body
Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A
- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)
- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)
- 28224 + Work done by friction = - 33660
Work done by friction = -33660 + 28224 = - 5436 J
I want to say its cooled by reflection because of the foil, sun reflects off of the foil back into the atmosphere. I don't think it's conduction because I have the foil on my windows and it's never warm to the touch. it's not a liquid so I don't believe it's convection. The foil reflects the radiation so I don't think it's b, c or d. so I wanna say A but I'm not 100% sure
Answer:
= 17º C
Explanation:
This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)
Let's reduce the magnitudes to the SI system
Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg
L = 3.33 105 J / kg
Water M = 860 g = 0.860 kg
= 4186 J / kg ºC
Q₁ = m L
Q₁ = 0.080 3.33 10⁵
Q₁ = 2,664 10⁴ J
Now let's see what this liquid water temperature is when this heat is released
Q = M ΔT = M (T₀₁ -
)
Q₁ = Q
= T₀₁ - Q / M ce
= 26.0 - 2,664 10⁴ / (0.860 4186)
= 26.0 - 7.40
= 18.6 ° C
The initial temperature of water that has just melted is T₀₂ = 0ª
The initial temperature of the liquid water is T₀₁= 18.6
m + M 
= M T₀₁ - m T₀₂o2
= (M To1 - m To2) / (m + M)
= (0.860 18.6 - 0.080 0) / (0.080 + 0.860)
= 17º C
gg
