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DanielleElmas [232]
3 years ago
6

Open Meet and enter this code: jnw-xodp-yij​

Physics
2 answers:
nikdorinn [45]3 years ago
8 0

Answer:

nrjfnvsncjanxsafkemfksevnjenf thx for the points i guess

Explanation:

Arada [10]3 years ago
5 0

Answer:

Is this for zoommmmmmmmmm?

Explanation:

It had to be 20 letters I had to do something

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Write one general properties of non contact force​
Gekata [30.6K]

Answer:    Gravitational force between the two masses does not depend on the medium separating two masses.

3 0
2 years ago
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0
3 years ago
A car moves around a circular track at a constant rate. What must change?
suter [353]

B. only its velocity should change

4 0
2 years ago
Read 2 more answers
What is the rate of acceleration due to gravity on Earth?
Katena32 [7]

Answer:

Negative 9.8 meters per second squared

Explanation:

The negative is for the direction (down, towards the center of the earth). Often this can be estimated as -10 m/s^2 to make calculations easier.

3 0
3 years ago
Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to
Tomtit [17]

Answer:

P=42.075W

Explanation:

The power provided by a resistor (wire in this case) is given by:

P=\frac{V^2}{R}.

The resistance of a wire is given by:

R=\frac{\rho L}{A}

Where for the resistivity the one of the copper should be used: \rho=1.68\times10^{-8}\Omega m.

The area A is that of a circle, which written in terms of its diameter is:

A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}

Putting all together:

P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}

Which for our values is:

P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W

7 0
3 years ago
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