Determine the torque applied to the shaft of a car that transmits 225 hp

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of object(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²

Response:

a) , b)

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

b) The force applied by the water is:

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

Answer:

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

Now, the potential for the smaller droplets is given as:

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

Segment B

Segment C

Segment D

Sorted from the lowest to the highest acceleration: