All categories

2 months ago

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy.

1 answer:

5 0

This question is excellent for clearing up any confusion about the work-energy theorem.
When a force acts perpendicular to an object's motion, the work done by that force equals zero.
According to the work-energy theorem,
the net work performed by all forces equals the change in kinetic energy.

In this instance, since work done = 0,
therefore,
0 = change in kinetic energy.
This indicates that the kinetic energy remains unchanged.
I hope this clarifies things!

You might be interested in

A small crack occurs at the base of a 15.0-m-high dam. The effective area through which water leaves is 2.30 × 10-3 m2. (a) Igno

Ostrovityanka [3204]

Answer

Given data:

height of the dam = 15 m

effective area for water flow = 2.3 x 10⁻³ m²

Applying the principle of energy conservation:

$m g h = \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v= \sqrt{2\times 9.8 \times 15}$

$v= \sqrt{294}$

v = 17.15 m/s

water discharge

Q = A V

Q = 2.3 x 10⁻³ x 17.15

Q = 0.039 m³/s

3 0

2 months ago

A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 245 Hz. A person on the pl

Keith_Richards [3271]

through the Doppler effect. The formula for apparent frequency is derived as F apparent = F real x (Vair ± Vobserver) / (Vair ± Vsource). In this scenario, should the observer move towards the source—place a positive sign in the numerator and a negative in the denominator. Since the observer approaches the wall, we apply the formula to derive the necessary speed.

4 0

2 months ago

A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en

Yuliya22 [3333]

Answer:

The energy delivered is E = 0.18 J

Explanation:

Given,

Battery voltage, V = 9 V

Charge in the circuit, Q = 20 mC

= 20 x 10³ C

Energy supplied in the circuit can be computed as

E = Q V

E = 20 x 10⁻³ x 9

E = 180 x 10⁻³

E equals 0.18 J.

The energy delivered in the circuit is therefore E = 0.18 J

7 0

1 month ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Softa [3030]

The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "Doppler's effect."

The standard formula for the Doppler effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the observer relative to the medium =?.
$v_{s}$ = Velocity of the source in relation to the medium = 0 m/s.
$f_{o}$ = Frequency emitted from the source = 400 Hz.
$f$ = Frequency recognized by the observer = 408 Hz.

Substituting the given values into equation (A) will yield:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving the above will result in,
$v_{r}$ = 6.8 m/s

The correct result = 6.8m/s

7 0

3 months ago