Answer:
Explanation:
mass of the ball = 146 g = 146 / 1000 = 0.146 kg
initial speed of the ball = 40.6 m/s
final speed of the ball = - 45.1 m/s
time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s
impulse, Ft = change in momentum = mv - mu = m (v-u)
F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N
Answer:
the final temperature of the tea is 7.39⁰C.
Explanation:
Given;
mass of the tea, m = 375 g
specific heat capacity of the tea, C = 4.184 JJ/g°C
initial temperature of the tea, t₁ = 95°C
the final temperature of the tea, t₂ = ?
Energy lost by the refrigerator, Q = 137,460 J
The energy lost by the refrigerator is given by the following formula;
-Q = mc(t₂ - t₁)
-137,460 =375 x 4.184(t₂ - 95°C)
-137,460 = 1569(t₂ - 95°C)

Therefore, the final temperature of the tea is 7.39⁰C.
<u><em>The question doesn't provide enough data to be solved, but I'm assuming some magnitudes to help you to solve your own problem</em></u>
Answer:
<em>The maximum height is 0.10 meters</em>
Explanation:
<u>Energy Transformation</u>
It's referred to as the change of one energy from one form to another or others. If we compress a spring and then release it with an object being launched on top of it, all the spring (elastic) potential energy is transformed into kinetic and gravitational energies. When the object stops in the air, all the initial energy is now gravitational potential energy.
If a spring of constant K is compressed a distance x, its potential energy is

When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as

We have then,


Solving for h

We have little data to work on the problem, so we'll assume some values to answer the question and help to solve the problem at hand
Let's say: x=0.2 m (given), K=100 N/m, m=2 kg
Computing the maximum height


The maximum height is 0.10 meters
Answer:
i dont think anything will happen so maybe A: the wall is larger than you.
Explanation:
Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m