Answer:
T = 1010 degree Celsius
Explanation:
mass of ball (Mb) = 100 g
mass of water (Mw) = 400 g
temp of water = 0 degree
specific heat of platinum (C) = 0.04 cal/g degree celsius
we can calculate the temperature of the furnace from the equation before
Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)
100 x 0.04 x ( T - 10) = 400 x (10 - 0)
4 (T - 10) = 4000
T - 10 = 1000
T = 1010 degree Celsius
Gravitational potential energy can be described as m*g*h (mass times gravity times height).
Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.
After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.
To find how much energy was added, we subtract final energy from initial energy:
147 J - 44.1 J = 102.9 Joules.
The correct answer would be C. it moves at a constant speed. The troposphere(the layer our weather is in) is not nearly high enough for gravity to be different at different altitudes.
(a)
consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.
m = mass of the tennis ball = 60 g = 0.060 kg
v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s
v = final velocity of the tennis ball after being hit by racket = - 39 m/s
ΔP = change in momentum of the ball
change in momentum of the ball is given as
ΔP = m (v - v₀)
inserting the above values
ΔP = (0.060) (- 39 - 20)
ΔP = - 3.54 kgm/s
hence , magnitude of change in momentum : 3.54 kgm/s
Answer:
t_{out} = 
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
