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2 years ago

Practice questions, will mark brainliest!

Physics

1 answer:

andrew-mc [135]2 years ago

4 0

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

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If a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right, wh

mr Goodwill [35]

The net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N. Details about net force can be found below.

<h3>How to calculate net force?</h3>

The net force of a body can be calculated by multiplying the mass of the body by its acceleration as follows:

Force = mass × acceleration

According to this question, a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right.

Net force = 8kg × 4m/s²

Net force = 32N

Therefore, the net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N.

Learn more about net force at: brainly.com/question/18031889

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1 year ago

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Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

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3 years ago

How does displacement, acceleration, time, and velocity affect motion?

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Answer:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.

~Hoped this helped~

~Brainiliest?~

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Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, howe

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Answer:

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Explanation:

We know that after 0.510 s, the ball is level with Bob again. We can use this to find the vertical component of the initial velocity.

y = y₀ + v₀ᵧ t + ½ gt²

h+2 = h+2 + v₀ᵧ (0.510) + ½ (-9.8) (0.510)²

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v₀² = v₀ₓ² + v₀ᵧ²

(34.1)² = v₀ₓ² + (2.50)²

v₀ₓ = 34.0 m/s

And since we know the ball lands 126 m from the base of the cliff, we can find the time it takes to land:

x = x₀ + v₀ₓ t + ½ at²

126 = 0 + (34.0) t + ½ (0) t²

t = 3.71 s

Finally, we can now find the height of the cliff:

y = y₀ + v₀ᵧ t + ½ gt²

0 = h+2 + (2.50) (3.71) + ½ (-9.8) (3.71)²

h = 56.0 m

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I hope this helps.<span />

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3 years ago

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