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3 years ago

Earth orbits the sun once every

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

3 0

Answer:1 trip around the earth is an angular displacement of 2*pi

3.6525*10^2 days

Explanation:24 h/1 day * 3.600*10^3 s/1h = 3.156*10^7 s

Angular speed = angular displacement / time

Angular speed = 2*pi rads / 3.156*10^7 s = 1.9910*10^-7 rad/s

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How much heat is needed to increase the internal energy of a gas in a piston by 4,736J if the gas does 750J of work on the envir

kykrilka [37]

The heat absorbed by the gas is 5486 J

Explanation:

We can solve this problem by using the 1st law of thermodynamics, which states that the change in internal energy of a gas is given by:

$\Delta U = Q-W$

where:

$\Delta U$ is the change in internal energy

Q is the heat absorbed by the gas

W is the work done by the gas

In this problem, we have:

$\Delta U = +4736 J$ is the increase in internal energy of the gas

$W=+750 J$ is the work done by the gas (positive because the gas is doing work on the surroundings, by expanding)

Solving for Q, we find the heat needed:

$Q=\Delta U + W = 4736+750=5486 J$

And the positive sign tells us that this heat is absorbed by the gas.

Learn more about thermodynamics:

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#LearnwithBrainly

6 0

4 years ago

A point charge q1=+2.40μC is held stationary at the origin. A second point charge q2=−4.30μC moves from the point x=0.125m , y=0

algol [13]

Answer:

Explanation:

Potential energy of a pair of charges

= k q₁q₂ /r₁₂

At r₁₂ = .125 m

= - 9 x 10⁹ x 2.4 x 4.3 x 10⁻¹² / .125

= - 743 x 10⁻³J

At r₁₂ = √2 x .275 m

- 9 x 10⁹ x 2.4 x 4.3 x 10⁻¹² / √2 x .275

= - 238.85 x 10⁻³J

= 239 x 10⁻³ J

Change = - 238.85 x 10⁻³ +743 x 10⁻³

= 504.15 x 10⁻³ J

B )

Work done by electric force will be same but in negative sign

= - 504.15 x 10⁻³ J

= - 504 x 10⁻³ J

3 0

3 years ago

Can anybody help with this GCSE physics question?

Serjik [45]

Answer:

$volume = 18.45 \times 18.45 \times 18.45 \\ = 0.00000628 \: {m}^{3} \\ mass = density \times volume \\ = 8.0 \times {10}^{3} \times 0.00000628 \\ = 0.05\: kg$

$\small \boxed{ \boxed { becker}}$

8 0

3 years ago

A cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of

Harlamova29_29 [7]

Answer:

(a) 147 j (b) 73.5 J (C) 220.5 J

Explanation:

Mass of the cylinder m =6 kg

velocity = 7 m/sec

(a) transnational energy $KE =\frac{1}{2}mv^2=0.5\times 6\times 7^2=147 \ j$

(b) Rotational kinetic energy is given by $KE =\frac{1}{2}I\omega ^2$ where I is moment of inertia and ω is the angular velocity

Moment of inertia $I=\frac{1}{2}mr^2$

Putting the value of I in rotational kinetic energy formula

$KE=\frac{1}{2}\frac{1}{2}mr^2\omega ^2$

$KE=\frac{1}{4}m(rw)^2$

$KE=\frac{1}{4}m(v)^2$ as v=ωr

So rotational kinetic energy $=\frac{1}{2}\times translational\ kinetic \ energy=0.5\times 147=73.5j$

4 0

4 years ago

Saturn is moving with uniform angular speed of 2πf along the circumference of it's orbit around the sun with radius R, having ce

Luda [366]

Answer:Sir haymo knows

Explanation:

he gave as homework

8 0

2 years ago