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bulgar [2K]
3 years ago
14

Two workers push on a wooden crate. One worker push with a force of 543 N and the other with a force of 333 N. The mass of the w

ooden crate is 206 kg. What is the acceleration of the crate?
Physics
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

a=4.2524\ m.s^{-2}

Explanation:

Given:

two workers push a crate.

  • Force by first worker, F_1=543\ N
  • force by second worker, F_2=333\ N
  • mass of crate, m=206\ kg

(Assuming that both the workers push in the same direction.)

We know that,

<u>Acceleration is given as:</u>

a=\frac{F}{m}

a=\frac{F_1+F_2}{m}

a=\frac{543+333}{206}

a=4.2524\ m.s^{-2}

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Which item(s) would be sufficient to make a circuit?
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Explanation :

A circuit is the representation of the path of the flow of current. The circuit can be either closed or open.

When the switch is off the circuit is closed circuit and when the switch is not connected the circuit is open.

The items that are sufficient to make a circuit are as follows :

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Other components can be ammeter, voltmeter, ac source, variable resistors etc.

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A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener
defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s

(a) the force constant of the spring

\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m

7 0
3 years ago
In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot
Fittoniya [83]

Answer:

Q=7.9\times 10^{-10}\ C

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

A=5\times 10^{-3}\ m^2

We Capacitance given as

For air

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}

C_1=2.2\times 10^{-11}\ F

C_2=\dfrac{K\varepsilon _oA}{d}

C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F

C_2=6.6\times 10^{-11}\ F

Net capacitance

C=C₁+C₂

C=8.8\times 10^{-11}\ F

We know that charge Q given as

Q= C V

Q=12\times 6.6\times 10^{-11}\ C

Q=7.9\times 10^{-10}\ C

6 0
3 years ago
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