PHYSICS HIGH SCHOOL

A very small 100-g object is attached to one end of a massless 10-cm rod that is pivoted without friction about the opposite end. The rod is held vertical, with the object at the top, and released, allowing the rod to swing. What is the speed of the object at the instant that the rod is horizontal?

Answers

Answer 1

Answer:

Answer:

1.40071 m/s

Explanation:

m = Mass of object = 100 g

v = Velocity of object

h = Height of drop = 10 cm

g = Acceleration due to gravity = 9.81 m/s²

Here, the loss of potential energy is equal to the gain in kinetic energy as energy is conserved

$mgh=(1)/(2)mv^2\n\Rightarrow gh=(1)/(2)v^2\n\Rightarrow v^2=2gh\n\Rightarrow v=√(2gh)\n\Rightarrow v=√(2* 9.81* 0.1)\n\Rightarrow v=1.40071\ m/s$

The speed of the object at the instant that the rod is horizontal is 1.40071 m/s

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B. The planets formed at the center of a collapsing spinning cloud of gas and dust

C. The sun and planets formed from a collapsing spinning cloud of gas and dust.

D. A comet hit the sun and sent off fragments the became the planets.

I'm pretty sure it's C?

Answers

Answer: C. The sun and planets formed from a collapsing spinning cloud of gas and dust.

Explanation:

The entire solar system was formed from solar nebula. A nebula is a cloud of dust and gases. When this cloud spins and contracts, a protostar is formed. The temperature rises and nuclear fusion kicks in forming a star. The rest of the matter forms planets and other celestial bodies of the star system. Similarly, the Sun was formed in solar nebula from collapsing of spinning cloud of dust and gas.

You're right, Answer C

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Which value from his calculation most strongly affects the number of significant figures in the answer?

Answers

The value from his calculation what most strongly affects the number of significant figures in the answer is : the one with the least significant figures
No matter what calculation that he does, this condition will always apply

Hope this helps

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force of static friction f on the block?A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Answers

When the block is at rest, the static frictional force is equal to the horizontal component of the block's weight (F = mgsin(θ)).

The static frictionalforce on the body at rests is determined by applying Newton's second law of motion.

F = ma

where;

F is applied force on a body
m is the mass of the body
a is the acceleration of the body

If the block is at rest, then the net horizontal force on the block is zero.

$\Sigma F_x = 0\n\nmg sin(\theta ) - F_s = 0\n\nF_s = mg sin(\theta)$

Thus, when the block is at rest, the static frictional force is equal to the horizontal component of the block's weight.

Learn more here:brainly.com/question/13758352

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block. The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that

[tex]f=mgsin \theta[tex]

Therefore, option B is the correct option.

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Answers

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Answers

Explanation:

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