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2 years ago

A piano wire has a length of 81 cm and a mass of 2.0

1 answer:

4 0

<span>Frequency = 394 Hz
Length of the string L = 81 cm = 0.81 m
Mass of the string = 0.002 kg
Tension T = ?
Wave length of the string is two times the length.
n x lambda = 2L, we also have lambda = vt = v / f, t is time period and given n = 1.
Therefore L = v / 2f => v = 2fL
Deriving form force equation, force here is tension T so
v = squareroot of (TL/m) hence
2fL = squareroot of (TL/m) => 4 x f^2 x L^2 = (T x L) / m => T = 4 x f^2 x L x m
T = 4 x 0.81 x (394)^2 x 0.002 = 4 x 0.81 x 155236 x 0.002
T = 1005.9 N = 1.006 x 10^3 N</span>

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan

OlgaM077 [116]

Answer:

a) I = 13.04 A

b) R = 8.82 ohms

c) 1291.87 kilocalories are generated an hour.

Explanation:

let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.

a) we know that:

P = I×V

I = P/V

= (1500)/(115)

= 13.04 A

Therefore, the current of the heater is 13.04 A

b) we now have voltage and current, according to Ohm's law:

R = V/I

= (115)/(13.04)

= 8.82 ohms

Therefore, the resistance of the heating coil is 8.82 ohms.

c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:

E = P×t

= (1500)(1×60×60)

= 5400000 J

since 1 calorie = 4.81 J

1 kilocalorie = 0.001 calories

E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories

Therefore, 1291.87 kilocalories are produced/generated in one hour.

8 0

2 years ago

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0

2 years ago

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t

arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vector $n=-0.60\hat{i}-0.080\hat{j}$

We need to calculate the torque on the loop

Using formula of torque

$\tau=NIAB\sin\theta$

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

$\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}$

$\tau=4.7\times10^{-4}\ N-m$

Hence, The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

5 0

2 years ago

A negative oil droplet is held motionless in a millikan oil drop experiment. What happens if the switch is opened?

scoray [572]

In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.

Explanation:

Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.

Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.

By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.

After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.

0 0

2 years ago