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Eduardwww [97]
3 years ago
8

What is pulling force? Give any two examples.,

Physics
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

Push or Pull Forces - example

When you push against a wall the force that you exert is an example of a push force. When you pull a trolley car the force that you exert is an example of pull force.

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4 years ago
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Consider an ideal gas of 7 moles that is in contact with a thermal reservoir of temperature 475 K. The gas is enclosed in a cont
kozerog [31]

Answer:

(a). The initial pressure is 5.5\times10^{4}\ Pa

(b). The final pressure is 1.8\times10^{4}\ Pa

Explanation:

Given that,

Number of moles = 7

Temperature = 475 K

Initial volume = 0.50 m³

Expanded volume = 1.50 m³

We need to calculate the initial pressure

Using formula of pressure

P_{i}=\dfrac{nRT_{i}}{V_{i}}

Put the value into the formula

P_{i}=\dfrac{7\times8.31\times475}{0.50}

P_{i}=55261.5\ Pa

P_{i}=5.5\times10^{4}\ Pa

We need to calculate the final pressure

Using formula of pressure

P_{f}V_{f}=nRT_{f}

After expansion,

\dfrac{P_{f}V_{f}}{P_{i}V_{i}}=\dfrac{nRT_{f}}{nRT_{i}}

P_{f}=\dfrac{T_{f}}{T_{i}}\times\dfrac{P_{i}V_{i}}{V_{f}}

Put the value into the formula

For thermal process,

T_{i}=T_{f}

P_{f}=\dfrac{5.5\times10^{4}\times0.50}{1.50}

P_{f}=18333.33\ Pa

P_{f}=1.8\times10^{4}\ Pa

Hence, (a). The initial pressure is 5.5\times10^{4}\ Pa

(b). The final pressure is 1.8\times10^{4}\ Pa

3 0
4 years ago
An object carries a +15.5 uC charge.
abruzzese [7]

Answer:

3.67 N

Explanation:

From the question given above, the following data were obtained:

Charge of 1st object (q₁) = +15.5 μC

Charge of 2nd object (q₂) = –7.25 μC

Distance apart (r) = 0.525 m

Force (F) =?

Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:

For the 1st object

1 μC = 1×10¯⁶ C

Therefore,

15.5 μC = 15.5 × 1×10¯⁶

15.5 μC = 15.5×10¯⁶ C

For the 2nd object:

1 μC = 1×10¯⁶ C

Therefore,

–7.25 μC = –7.25 × 1×10¯⁶

–7.25 μC = –7.25×10¯⁶ C

Finally, we shall determine the force. This can be obtained as follow:

Charge of 1st object (q₁) = +15.5×10¯⁶ C

Charge of 2nd object (q₂) = –7.25×10¯⁶ C

Distance apart (r) = 0.525 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²

F = 3.67 N

Therefore, the force on the object is 3.67 N

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