W = 1/2k*x^2.
k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).
W = 1/2(2500)(0.04)^2 = 2J.
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :

So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
This question deals with the volume of different shapes.
a) volume of the sphere is "33.51 m³".
b) volume of the cylinder is "25.13 m³".
a)
The volume of a sphere is given by the following formula:

<u>Volume = 33.51 m³</u>
<u />
b)
The volume of a cylinder is given by the following formula:

<u>Volume = 25.13 m³</u>
<u />
Learn more about <em>volume </em>here:
brainly.com/question/16686115?referrer=searchResults
The attached picture shows the formulae of the <em>volume</em> of different shapes.
Answer:
t₁ = 0.95 s
Explanation:
In this chaos we must use the definition of Newton's second law
F = m a = m dv / dt
dv = F dt / m
Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive
dv = ∫ (3 + 2t) dt / m
v = (3 t + 2 t²/ 2) /m
Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit t = t and v = 0
0 - (-6) = (3 (t- 0) + (t² -0)) / m
t² + 3t -6m = 0
Let's look for the mass
W = mg
m = W / g
m = 20/32
m = 0.625 slug
Let's solve the second degree equation
t² + 3t -3.75 = 0
t = (-3 ± √ (32 + 4 1 3.75)) / 2
t = (-3 ± 4,899) / 2
t₁ = 0.95 s
t₂ = -3.95 s
We take the positive time
Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.