Answer:
17304 J
Explanation:
Complete statement of the question is :
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .
Part A
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Solution :
= tension force in the tow rope = 350 N
= length of the incline surface = 120 m
= work done by tension force = ?
The tension force acts parallel to incline surface, hence work done by tension force is given as
= height gained by the rider = 30 m
= total mass of rider and tube = 84 kg
Potential energy gained is given as
= Thermal energy created
Using conservation of energy
Answer:
C
Explanation:
Gravity is the main reason that make our planets to pull each other
Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of aluminium = 
= specific heat of water = 
= mass of aluminum = 0.500 kg = 500 g
= mass of water = 0.250 kg = 250 g
= final temperature of mixture = ?
= initial temperature of aluminum = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:
Therefore, the temperature when the water and pan reach thermal equilibrium short time later is,
Answer:
47.8 °C
Explanation:
Use the heat equation:
q = mCΔT
where q is the heat absorbed/lost,
m is the mass of water,
C is the specific heat capacity,
and ΔT is the change in temperature.
Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.
100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT
ΔT = 47.8 °C
Answer:
The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
Explanation:
Given;
distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m
current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively
The magnitude of the magnetic field halfway between the wires can be calculated as;
where;
B is magnitude of the magnetic field halfway between the wires
I₁ is current in the first wire
I₂ is current the second wire
μ₀ is permeability of free space
r is distance half way between the wires
Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.
