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svetoff [14.1K]

3 years ago

8

Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them

1 answer:

denpristay [2]3 years ago

3 0

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

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A pen contains a spring with a spring constant of 257 N/m. When the tip of the pen is in its retracted position, the spring is c

Mandarinka [93]

Explanation:

Below is an attachment containing the solution.

3 0

3 years ago

A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston

AleksandrR [38]

Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure $P=4.63\times10^{4}\ Pa$

Remove heat $\Delta U= -1.95\times10^{4}\ J$

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

$dU=Q-W$ ...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

$dU=Q-PdV$

Where, W = PdV

Put the value in the equation

$dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))$

$dU=-17746.78\ J$

Hence, The change in internal energy of the system is -17746.78 J

3 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s

DedPeter [7]

Answer:

$212.8 m/s^{2}$

Explanation:

Time taken by stone to cover horizontal distance

$t=\sqrt{\frac {2h}{g}}$ where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

$t=\sqrt{\frac {2*2.1}{9.81}}$ = 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

$a=\frac {v^{2}}{r}$ where r is radius of circle, substituting r with 1.1m

$a=\frac {15.3^{2}}{1.1}$

$a=212.8 m/s^{2}$

Therefore, centripetal acceleration is $212.8 m/s^{2}$

8 0

3 years ago

If a bean of mass 2.0g jumps 1.0cm from your hand into the air, how much potential energy has it gained in reaching its highest

kirill [66]

$PE = mg\Delta h = 0.002 \, kg \cdot 9.8 \, m/s^2 \cdot 0.01 \, m = ~2 \cdot 10^{-4} \, J$

6 0

3 years ago