Answer/solution:
Given :
Mass =5kg
T 1 =20 C,T 2 =100 ∘C
ΔT=100−20=80 ∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
Answer:
The objects kinetic energy increases as it falls from some height.
Answer:
(a) 0.2618 J
(b) 0.1558 J
(c) 0 J
Explanation:
from Hook's Law,
The energy stored in a stretched spring = 1/2ke²
Ep = 1/2ke² ......................... Equation 1
Where k = spring constant, e = extension, E p = potential energy stored in the spring.
(a) When The spring is stretched to 4.11 cm,
Given: k = 310 N/m, e = 4.11 cm = 0.0411 m
Substituting these values into equation 1
Ep = 1/2(310)(0.0411)²
Ep = 155(0.0016892)
Ep =155×0.0016892
Ep = 0.2618 J.
(b) When the spring is stretched 3.17 cm
e = 3.17 cm = 0.0317 m.
Ep = 1/2(310)(0.0317)²
Ep = 155(0.0317)²
Ep = 155(0.0010049)
Ep = 0.155758 J
Ep ≈ 0.1558 J.
(c) When the spring is unstretched,
e = 0 m, k = 310 N/m
Ep = 1/2(310)(0)²
Ep = 0 J.
Answer:
2 secs; 65 feet
Explanation:
The function guiding the water bottle is given as:
f(x) = -16t² + 64t + 1
The bottle will reach maximum height when velocity, df/dt (velocity is the first derivative of distance) = 0.
Therefore:
df/dt = 0 = -32t + 64
=> 32t = 64
t = 64/32 = 2 seconds
This is the time it will take to reach the maximum height.
To find this height, we insert t = 2 into the function f(x):
f = -16(2)² + 64(2) + 1
f = -(16 * 4) + 128 + 1
f = -64 + 128 + 1
f = 65 ft
Its maximum height is 65 ft.
<u>We are given:</u>
the initial height of the object (h) = 100 m
initial velocity (u) = 0 m/s
we will let the value of g = 10 m/s/s
<u>Speed of the object just before hitting the ground:</u>
From the third equation of motion:
v² - u² = 2ah (where v is the final velocity)
replacing the variables, we get:
v² - (0)² = 2(10)(100)
v² = 2000
v = 10√20 = 44.7 m/s
Therefore, the speed of the object just before hitting the ground is 44.7 m/s
