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JulsSmile [24]
3 years ago
12

Two forces 5N and 10N are acting at "O" and "P" respectively on a uniform rod of length 100 cm suspended at the position of cent

er of gravity 50cm mark as shown in figure .​

Physics
1 answer:
omeli [17]3 years ago
5 0

Answer:

Option C. 75 cm

Explanation:

To obtain the position of P, let us calculate the value of y as shown in the attached photo.

The value of y can be obtained as follow:

Anticlockwise moment = clockwise moment

Anticlockwise moment = 5 × 50

Anticlockwise moment = 250

Clockwise moment = y × 10

Anticlockwise moment = clockwise moment

250 = y × 10

Divide both side by 10

y = 250/10

y = 25 cm

Finally, we shall obtain the value of P as follow:

Since P lies after the pivot (i.e 50 cm), therefore,

P = 50 + y

y = 25

P = 50 + 25

P = 75 cm

Therefore, the position of P on the metre rod is 75 cm

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0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a
Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

PV=nRT

P_{i}=\dfrac{nRT}{V}

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}

P_{i}=5.75\times10^{6}\ Pa

P_{i}=56.827\ atm

We need to calculate the final temperature

Using relation pressure and volume

P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}

P_{2}=\dfrac{56.827\times70}{400}

P_{2}=9.94\ atm

Hence, The final pressure of the gas is 9.94 atm.

3 0
2 years ago
A sealed container filled with gas is heated. What happens?
Luda [366]

The internal pressure increases as the gas is heated

5 0
3 years ago
Read 2 more answers
.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric
maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0
3 years ago
1. A Zambeef delivery track travels 18 km north, 10 km east, and 16 km south. What is its final displacement from the origin?
VARVARA [1.3K]

Answer:

2km

Explanation:

Given data

We are told that the direction traveled are

North>>>East>>>South

Hence the displacement is defined as the distance away from the initial position is

Initial position =18km

FInal position = 16km

The displacement = 18-16= 2km

Hence the displacement is 2km

6 0
2 years ago
An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a
vaieri [72.5K]

363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

      Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}

      v=\lambda \times f   --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,

         L=\frac{n \lambda}{2}

For first harmonic, n =1

         L=\frac{\lambda}{2}

For second harmonic, n =2

         L=\frac{2 \lambda}{2}=\lambda

For third harmonic, n =3

         L=\frac{3 \lambda}{2}

         \lambda=\frac{2 L}{3}   -------> eq. 2

Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get

   v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}

3 0
3 years ago
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