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Makovka662 [10]
3 years ago
5

A 2,294N force is applied to a 408kg mass. What is the acceleration of the mass?

Physics
1 answer:
Nataly_w [17]3 years ago
8 0

Answer:

5.622

Explanation:

a=F/M

a=2294/408

a=5.622

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A capacitor is connected to an ac power supply. The maximum current is 4.0 A and the reactance of the capacitor is Xc = 8.0 Ω. W
lions [1.4K]

Answer:

option (b)

Explanation:

Io = 4 A

Xc = 8 ohm

The formula for the average power dissipated is given by

P = Vrms x Irms x cos Ф

Where, Ф be the phase difference between Vrms and Irms .

As we know that in case of capacitor, the current leads the voltage by 90 degree, it means the value of Ф is 90 degree.

Thus, Average power = Vrms x Irms x Cos 90

Average power = 0 Watt

6 0
3 years ago
What element is in group 1 peroid 6
Ahat [919]

Answer:

Group 1: Hydrogen is a gas and every other element in the group is a solid at room temperature. Period 6: They are all solids except for Radon and Mercury.

Explanation:

6 0
2 years ago
Which has the greater density, 5 kg of lead or 10 kg of gold?
NARA [144]
Gold is much heavier than lead. It is very dense. ... Therefore gold weighs 19.3 times as lead
5 0
2 years ago
The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the
wel

Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C

5 0
3 years ago
A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Deter
qaws [65]

Answer:

Impedance = 93.75 ohms

Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = X_{L}= = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms  

Impedance = Z = \sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =

= \sqrt{8788.511} = 93.747 ohms  

Voltage = \sqrt{2} × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

8 0
3 years ago
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