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2 years ago

An object with a mass of 62.5 kg accelerates 12.3 m/s2 when an unknown force is applied to it. What is the amount of the force?

Physics

1 answer:

AlexFokin [52]2 years ago

4 0

Answer:768.75N

Explanation:

mass=62.5kg

acceleration=12.3m/s^2

Force=mass x acceleration

Force=62.5 x 12.3

Force=768.75N

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2 years ago

A 1280 kg car is moving 4.92 m/s. a 509 N force then pushes it forward for 28.7 m. what is its final KE?(unit=J)PLEASE HELP

laila [671]

Answer:

The answer to your question is: total energy = 30100.4 J

Explanation:

Kinetic energy (KE) is the energy due to the movement of and object, its units are joules (J)

Data

mass = 1280 kg

speed = 4.92 m/s

Force = 509 N

distance = 28.7 m

Formula

$KE = \frac{1}{2} mv^{2}$

Work = Fd

Process

- Calculate Kinetic energy

- Calculate work

- Add both results

KE = $\frac{1}{2} (1280)(4.92)^{2}$

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8 0

2 years ago

Read 2 more answers

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

3 years ago