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liberstina [14]
3 years ago
11

Newton's first law of motion is sometimes called the law of _________.

Physics
1 answer:
Gre4nikov [31]3 years ago
3 0
Newtons first law of motion is also known as the law of inertia
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You have to run 2.2 miles in track. How far is this in feet? Note: There are
Romashka [77]

Answer: B

Explanation: This can be easily done by inputting 5280 * 2.2.

3 0
2 years ago
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Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant
Cerrena [4.2K]

Answer:

the  force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81  + \frac{70 \times 120^2}{100}\\\\= 10766.7 N

Hence, the  force exerted by the seat on the pilot is 10766.7 N

4 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
6/10
g100num [7]

Answer:pounds

Explanation:

7 0
2 years ago
Which of these is emitted during beta decay ?
satela [25.4K]

Answer:

C. a small charged particle.

Explanation:

typically beta radiation emits an electron which is a small negativity charged particle.

hope it helps. :)

4 0
2 years ago
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