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Arte-miy333 [17]

2 years ago

12

Two objects with masses of 2.75 kg and 4.45 kg are connected by a light string that passes over a light frictionless pulley to f

orm an Atwood machine.
Required:
a. Determine the tension in the string.
b. Determine the acceleration of each object.
c. Determine the distance each object will move in the first second of motion if they start from rest.

1 answer:

Llana [10]2 years ago

4 0

Answer:

Explanation:

F = ma

4.45g - 2.75g = (4.45 + 2.75)a

a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²

a)

T = 2.75(9.81 + 2.32) = 33.3 N

or

T = 4.45(9.81 - 2.32) = 33.3 N

b) 2.32 m/s² upward for 2.75 kg mass

2.32 m/s² downward for 4.45 kg mass

c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

3 years ago

An AC generator consists of 20 circular loops of wire with an area of 75 cm2. It has a maximum induced voltage of 24 V. If its a

Monica [59]

Faraday's law allows us to find the magnetic field that produces the emf in the rotating system is:

The magnetic field is: B = 0.424 T

Faraday's law of induction states that when the magnetic flux changes in time, an induced electromotive force is produced.

fem = $- \frac{d \Phi_B }{dt}$

where fem is the induced electromotive force and Ф the flux,

The magnetic flux is the scalar product of the field and the area.

$\Phi_B = B . A = B A \ cos \theta$

In this case we have several turns, so the expression remains.

fem = $- N B A \ \frac{d cos \theta}{dt}$

Indicate that the turns rotate at a constant frequency, therefore we can use the uniform rotational motion ratio.

θ = w t

We substitute

$fem = - N B A \ \frac{d \ cos \ wt}{dt}\\fem = N B A w sin \ wt$

the maximum induced electromotive force occurs when the sine function is ±1

fem = N B A w

They indicate that the fem = 24 V, the number of the turn is N = 20, the area is A = 75 cm² = 75 10⁻⁴ m² and the frequency f = 60 Hz

Frequency and angular velocity are related.

w = 2π f

We substitute.

fem = N B A 2π f

$B = \frac{fem }{2 \pi \ NA \ f}$

Let's calculate.

$B= \frac{24 }{2\pi \ 20 \ 75 \ 10^{-4} 60}$ B = 24 / 2pi 20 75 10-4 60

B = 0.424 T

In conclusion, using Faraday's law we can find the magnetic field that produces the emf in the rotating system is:

The magnetic field is; B = 0.424 T

Learn more about Faraday's law here: brainly.com/question/24617581

8 0

2 years ago

The angular speed of the hour hand of a clock, in rad/min, is:___________

lapo4ka [179]

The angular speed is defined as:

<h2> ω= $\frac{2\pi}{T}$ </h2>

where

$T=12*60=720min$

$\omega=\frac{2\pi}{720}$

$\omega=4.4**10^{-3} rad/min$

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3 years ago

Penny puts Fluffy in one wagon and pushes the wagon, once, as hard as she can. Penny puts Butch in the other wagon and pushes it

loris [4]

D; the wagons will both move in the same way

4 0

3 years ago

Read 2 more answers

If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in

motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

$F_{initial} = \frac{kq_1q_2}{r^2}$

Here,

k = Coulomb's constant

$q_{1,2}$ = Charge at each object

r = Distance between them

As the distance is doubled so,

$F_{final} = \frac{kq_1q_2}{( 2r )^2}$

$F_{final} = \frac{ kq_1q_2}{ 4r^2}$

$F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}$

$F_{final} = \frac{1}{4} F_{initial}$

$\frac{F_{final}}{ F_{initial}} = \frac{1}{4}$

Therefore the factor is 1/4

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2 years ago