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3 years ago

A particle vibrates in Simple Harmonic Motion with amplitude. What will be its displacement in one time-period? A 2A 4A 0

1 answer:

3 0

'Displacement' is the distance and direction between the starting point and
ending point, regardless of the path followed to get there.

A particle that's executing simple harmonic motion is always in the same place
where it was one time period ago, and where it will be later after another time
period has passed.

So its displacement during exactly one time period is exactly zero.

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Paul lifts a sack weighing 245 newtons vertically from the ground and places it on a platform at a height of 0.7 meters. If he t

d1i1m1o1n [39]

Answer:

B. 17.15 watts

Explanation:

Given that

Time = 10 seconds

height = distance = 0.7 meters

weight of sack = mg = F = 245 newtons

Power = work done/ time taken

Where work done = force × distance

Substituting the given parameters into the formula

Work done = 245 newton × 0.7 meters

Work done = 171.5 J

Recall,

Power = work done/time

Power = 171.5 J ÷ 10

Power = 17.15 watts

Hence the power expended is B. 17.15 watts

6 0

3 years ago

During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

$I=2.50\cdot 10^4 A$ is the current

$t=40.0 \mu s = 40.0\cdot 10^{-6} s$ is the time interval

Solving the formula for q, we find the amount of charge transferred:

$q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C$

6 0

3 years ago

A solid ball with a mass of 4.25 kg and a

Liono4ka [1.6K]

7.20 units long yea

8 0

2 years ago

You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building.

DanielleElmas [232]

Answer:

t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

v = v₀ + a t₁

the elevator starts with zero speed

v = a t₁

t₁ = v / a

t₁ = 18.8 / 2.40

t₁ = 7.833 s

in this time he runs

y₁ = v₀ t₁ + ½ a t₁²

y₁ = ½ a t₁²

y₁ = ½ 2.40 7.833²

y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

x_total = x₁ + x₂

x₂ = x_total - x₁

x₂ = 373 - 73,627

x₂ = 299.373 m

this distance travels at constant speed,

v = x₂ / t₂

t₂ = x₂ / v

t₂ = 299.373 / 18.8

t₂ = 15.92 s

therefore the total travel time is

t_total = t₁ + t₂

t_total = 7.833 + 15.92

t_total = 23.757 s

6 0

3 years ago

A car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The diameter of a tire is 80.2 cm. Find the numb

BaLLatris [955]

By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.

ANGULAR MOTION

Since the car accelerate from rest, initial velocity will be zero.

Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters

Initial velocity U = 0

Final velocity V = 24.3 m/s

Time t = 9.1 s

If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.

First equation

V = U + at

Substitute all necessary parameters into the equation.

24.3 = 0 + 9.1a

a = 24.3/9.1

a = 2.67 m/ $s^{2}$

Third Equation of motion

$V^{2} = U^{2} + 2aS$

Substitute all the necessary parameters

$24.3^{2}$ = 0 + 2 x 2.67 x S

590.49 = 5.34S

S = 590.49 / 5.34

S = 110.58 m.

Given that the diameter of a tire is 80.2 cm,

the radius (r) will be 80.2/2 = 40.1 cm

convert it to meter

r = 40.1/100 = 0.401 m

The Circumference of the tire = 2 $\pi$ r

Circumference = 2 x 3.143 x 0.401

Circumference = 2.52 m

Assuming no slipping, number of revolutions = 110.58/2.52

Number of revolutions = 43.89 rev.

Number of revolutions = 43 rev.

Therefore, the number of revolutions the tire makes during this motion is 43 rev.

Learn more about circular motion here: brainly.com/question/6860269

4 0

2 years ago