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Pie
3 years ago
6

A string is attached to a vibrating machine that has a frequency of 120 Hz. The other end of the string is passed over a pulley

of negligible mass and friction and is attached to a weight hanger which holds a mass m=0.5kg. What is the tension in newtons?
Physics
1 answer:
BabaBlast [244]3 years ago
4 0

Answer:

Explanation:

Given that,

The frequency of vibration is 120Hz

The mass of object attached is 0.5kg.

We want to find the tension In the string

The tension in the string is the weight of the object, it is how much gravity is pulling the object to the centre of the earth

Using newton second law.

F_net = m•a_y

The body is not accelerating the y-direction, then, a_y = 0

F_net = 0

Force acting on the string is the weight of the object and the tension in the string

T - W = 0

T = W

Where weight is mass × gravity

W = mg

Then,

T = W = mg

T = mg = 0.5 × 9.81

T = 4.905 N

The tension in the string is 4.905 N

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A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and
N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t    where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec =  1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r  in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0
3 years ago
Read 2 more answers
Gravity on Jupiter is 25 m/s/s, what is the weight of a 12 kg object on Jupiter?
stiv31 [10]

Answer:

300 Newtons

Explanation:

Weight is the force of attraction between two bodies, one usually larger (like a planet), and one smaller (like a person). Force can be calculated using the formula: Force = mass × acceleration.

The mass here is 12kg, the acceleration, which in this case, is the acceleration due to gravity is 25m/s/s, by plugging in our values, we have

Force = 12 × 25 = 300 Newtons or 300 N for short.

3 0
2 years ago
4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o
goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a  capacitor chosen to provide a peak-to-peak ripple voltage of  (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through  a 10-to-1 step-down transformer. It uses a silicon diode  that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

 (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0
3 years ago
What happens when you compress a gas?
8090 [49]
The density increases.
When gases are compressed, their volume decreases, and the resulting pressure increases. The temperature will change if either P or V are held constant. Since the volume decreases, then density, or m/V, increases.
P×V ~ T
8 0
3 years ago
Read 2 more answers
If a distant galaxy has a substantial redshift (as viewed from our galaxy), then anyone living in that galaxy would see a substa
mario62 [17]

Answer:

Option A

Explanation:

The statement makes sense since it's already explained that the galaxy is moving away from us and unlike option C which depicts that the galaxy is moving to us.

This statement makes sense. The redshift means that we see the galaxy moving away from us, so observers in that galaxy must also see us moving away from them—which means they see us redshifted as well

3 0
3 years ago
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