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3 years ago

9

The strength of the pull of gravity depends on _______. (1 point) speed and distance mass and distance mass and speed speed and

inertia

1 answer:

KIM [24]3 years ago

7 0

F_g=Gm_1m_2/r^2. mass and distance determines gravity.

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2. A 200.0-kg bear grasping a vertical tree slides down at constant velocity. What is the friction force between the tree and th

Neporo4naja [7]

Answer: 1960 N

Explanation:

The bear is sliding down at constant velocity: this means that its acceleration is zero, so the net force is also zero, according to Newton's second law:

$F_{net}=ma=0$

There are two forces acting on the bear: its weight W, pulling downward, and the frictional force Ff, pulling upward. Therefore, the net force is given by the difference between the two forces:

$F_{net}=W-F_f=0$

From the previous equation, we find that the frictional force is equal to the weight of the bear:

$F_f=W=mg=(200.0 kg)(9.8 m/s^2)=1960 N$

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3 years ago

URGENT!!!! WILL GIVE BRAINLIEST!!!!!! 15 POINTS

Artyom0805 [142]

It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)

3 0

3 years ago

Read 2 more answers

9. Kokio dydžo Archimedo jėga veikia 0,5 m² tūrio medinį rastą vandenyje?

ICE Princess25 [194]

Answer:

i dont knowwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

6 0

3 years ago

A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is fo

Alex787 [66]

Answer:

ρ = 1.6*10⁻⁸ Ω/m.

Explanation:

Applying Ohm's Law to the wire, assuming that it can be treated as a pure resistance, the resistance of the wire can be obtained as follows:

$R = \frac{V}{I} = \frac{9.11V}{36.0A} = 0.253 \Omega (1)$

At the same time, we know that there exists a relationship between the resistance, the resistivity ρ, the length L and the area A of the wire, that is given for the following expression:

$R = \rho* \frac{L}{A} (2)$

The area of the circular section of the wire, can be expressed as a function of the diameter d, as follows:

$A = \frac{\pi*d^{2} }{4} = \frac{\pi*(0.002m)^{2}}{4} = \pi*10e-6 (3)$

Replacing the left side of (2) by (1), and (3) on the right side, we can solve for the resistivity ρ as follows:

$\rho = \frac{R*A}{L} = \frac{0.253\Omega*\pi*10e-6}{50.0m} = 1.6e-8 \Omega/m$

ρ = 1.6*10⁻⁸ Ω/m

4 0

3 years ago

In which stage of a star's life cycle, does gravity and fusion become balanced, and an adult star forms?

V125BC [204]

Answer:

im guessing red giant-

8 0

3 years ago