9. What is the equivalent resistance of the circuit shown in the figure below?

Now it's your turn

eduard

Answer:

The average force the golf club exerts on the ball is 600 N

Explanation:

Newton's second law of motion states that force, F, is directly proportional to the rate of change of momentum produced

F = m× (v₂ - v₁)/(Δt)

The given parameters of the motion of the ball are;

The mass of the ball, m = 45 g = 0.045 kg

The initial velocity of the ball, v₁ = 0 m/s

The speed with which the ball was hit by the golfer, v₂ = 40 m/s

The duration of contact between the golf club and the ball, Δt = 3 ms = 0.003 seconds (s)

By Newton's law of motion, the average force, 'F', which the golf club exerts on the ball is therefore, given as follows;

F = 0.045 kg × (40 m/s - 0 m/s)/(0.003 s) = 600 N

The average force the golf club exerts on the ball = F = 600 N.

5 0

3 years ago

True or false nodes and antinodes occur in longitudinal waves

Vlad [161]

I believe it is false because longitudinal waves travel parallel to the vibration

8 0

3 years ago

During the first stage of a multistage rocket, a satellite is:

lianna [129]

B. sent through the atmosphere

7 0

4 years ago

The wavelength of red light is given as 6.5 × 10-7 m. If another wave of an unknown type has a frequency of 5.3 × 1015 Hz. What

lakkis [162]

V = fλ
3 x 10⁸ = f x 6.5 x 10⁻⁷
f(red) = 4.6 x 10¹⁴ Hz
The frequency of the second wave is higher, meaning it has a lower wavelength and greater energy than that of red light.

7 0

4 years ago

A 0.54 kg particle has a speed of8.0 m/s at point A and kineticenergy of 7.5 J at point B.

dybincka [34]

Answer:

a) The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle at point B is approximately 5.270 meters per second.

c) The total work done on the particle as it moves from A to B is - 9.78 joules.

Explanation:

Let be this particle a conservative system, that is, that non-conservative forces (i.e. friction, viscosity) are negligible.

a) The translational kinetic energy of the particle ( $K$ ), measured in joules, is determined by the following formula:

$K = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Speed, measured in meters per second.

If we know that $m = 0.54\,kg$ and $v = 8\,\frac{m}{s}$ , the translational kinetic energy at point A is:

$K = \frac{1}{2}\cdot (0.54\,kg)\cdot \left(8\,\frac{m}{s} \right)^{2}$

$K = 17.28\,J$

The translational kinetic energy of the particle at point A is 17.28 joules.

b) The speed of the particle is clear in (1):

$v = \sqrt{\frac{2\cdot K}{m} }$

If we know that $K = 7.5\,J$ and $m = 0.54\,kg$ , then the speed of the particle at point B:

$v = \sqrt{\frac{2\cdot (7.5\,J)}{0.54\,kg} }$

$v\approx 5.270\,\frac{m}{s}$

The speed of the particle at point B is approximately 5.270 meters per second.

c) According to the Work-Energy Theorem, the total work done on the particle as it moves from A to B ( $W_{A\rightarrow B}$ ), measured in joules, is equal to the change in the translational kinetic energy of the particle. That is:

$W_{A\rightarrow B} = K_{B}-K_{A}$ (2)

If we know that $K_{A} = 17.28\,J$ and $K_{B} = 7.50\,J$ , then the change in the translational kinetic energy of the particle is:

$W_{A\rightarrow B} = 7.50\,J-17.28\,J$

$W_{A\rightarrow B} = -9.78\,J$

The total work done on the particle as it moves from A to B is - 9.78 joules.

7 0

3 years ago