Answer:

Explanation:
Given that,
Radius, r = 2 m
Velocity, v = 1 m/s
We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

So, the magnitude of centripetal acceleration is
.
Answer:
40 N/m
Explanation:
F = -kx (This is the Hooke's Law equation)
F is the force the spring exerts = 8 N
-k = spring constant
x = displacement (The distance stretched past it's natural length) = 20cm
x needs to be in meters, and 20 cm is = to 0.2 meters
Finally:
8N = -k (0.2m)
-k = 8N / 0.2 m
k = -40 N/m
Answer:
Explanation:
impedance z=(XL^2+R^2)^1/2
power across te resistor ==i^2r
286/300
I=.976
C. The bowling ball and the bicycle
p = mv
P of bike = 12x5 = 60
P of rock = 2x20 = 40
P of ball = 5 x 10 = 60
Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.