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2 years ago

Someone solvw this for me please

Physics

1 answer:

soldi70 [24.7K]2 years ago

6 0

The velocity of the mass at time, t = 1 s, is determined as 20 m/s.

<h3>Magnitude of the force experienced by the body</h3>

The magnitude of the force experienced by the body is calculated as follows;

|F| = √(8² + 6²)

|F| = 10 N

<h3>Acceleration of the mass</h3>

From Newton's second law of motion;

F = ma

where;

m is mass
a is acceleration

a = F/m

a = 10 /0.5

a = 20 m/s²

<h3>Velocity of the mass after 1 second</h3>

v = at

v = 20 x 1

v = 20 m/s

Learn more about velocity here: brainly.com/question/6504879

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The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t

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Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

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Where, u = initial velocity

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3 years ago

For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

Mariulka [41]

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

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Where v is the final velocity

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and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

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h = ut + 1/2(gt²)

u = 0 m/s

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h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

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3 years ago

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sergey [27]

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7 0

3 years ago