Answer:
The elastic potential energy is zero.
The net force acting on the spring is zero.
Explanation:
The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.
Let's now analyze the different statements:
The spring constant is zero. --> false. The spring constant is never zero.
The elastic potential energy is at a maximum --> false. The elastic potential energy of a spring is given by
where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.
The elastic potential energy is zero. --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).
The displacement of the spring is at a maxi
num --> false, for what we said above
The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched
Answer:
18 W
Explanation:
Applying,
P = V²/R.................. Equation 1
Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs
Since: It is a series circuit,
Then,
R = R1+R2............. Equation 2
Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb
Given: R1 = R2 = 8 Ω
Substitute into equation 1
R = 8+8
R = 16 Ω
Also Given: V = 12 V
Substitute into equation 1
P = 12²/8
P = 144/8
P = 18 W
Answer:
please read the answer below
Explanation:
The angular momentum is given by
By taking into account the angles between the vectors r and v in each case we obtain:
a)
v=(2,0)
r=(0,1)
angle = 90°
b)
r=(0,-1)
angle = 90°
c)
r=(1,0)
angle = 0°
r and v are parallel
L = 0kgm/s
d)
r=(-1,0)
angle = 180°
r and v are parallel
L = 0kgm/s
e)
r=(1,1)
angle = 45°
f)
r=(-1,1)
angle = 45°
the same as e):
L = 5kgm/s
g)
r=(-1,-1)
angle = 135°
h)
r=(1,-1)
angle = 135°
the same as g):
L = 5kgm/s
hope this helps!!
Answer:
v = 38.73 m/s
Explanation:
Given
Extension of the bow, x = 50 cm = 0.5 m
Force of the arrow, F = 150 N
Mass of the arrow, m = 50 g = 0.05 kg
speed of arrow, v = ? m/s
We start by finding the spring constant
Remember, F = kx, so
k = F/x
k = 150 / 0.5
k = 300 N/m
the potential energy if the bow when pulled back is
E = 1/2kx²
E = 1/2 * 300 * 0.5²
E = 0.5 * 300 * 0.25
E = 37.5 J
The speed of the arrow will now be found by using the law of conservation of energy
1/2kx² = 1/2mv²
kx² = mv²
v² = kx²/m, on substituting, we have
v² = (300 * 0.5²) / 0.05
v² = 75 / 0.05
v² = 1500
v = √1500
v = 38.73 m/s
Answer:
Maximum height the atmosphere pressure can support the
water=10.336 m
Explanation:
We know that ,
Case 1 - Mercury in the tube
Case 2 - Water in the tube
Since atmospheric pressure is same
.
or, 
∴ 
Hence height of the water column =10.336 m
