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3 years ago

A 12 kg bowling ball would require what force to accelerate down and alleyway at a rate of 3 m/s²?

1 answer:

defon3 years ago

8 0

Part 1. F= m*a
M=12kg
A= 3 m/s^2

So F = 12kg * 3 m/s^2 which means
F=36 Newton’s

Part 2. F=12000N
M= F/A
A=4 m/s^2

So M= 12000N / 4 m/s^2
Which means M=3000 kg

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Sound waves are classified as which type of wave?answer

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3 years ago

Although all sports require a person to be in shape to some degree, some

Alexandra [31]

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Explanation:

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2 years ago

A block of wood has a mass of 120 g and a volume of 200cm what is the density of the word

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3 years ago

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C

Alex_Xolod [135]

Answer:

26.466cm³/min

Explanation:

Given:

Volume 'V'= 320cm³

P= 95kPa

dP/dt = -11 kPa/minute

pressure P and volume V are related by the equation

P $V^{1.4}$ =C

we need to find dV/dt, so we will differentiate the above equation

$V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt} = \frac{d[C]}{dt}$

$\frac{dP}{dt} V^{1.4} + P(1.4)V^{0.4} \frac{dV}{dt} =0$

lets solve for dV/dt, we will have

$\frac{dV}{dt} =\frac{-\frac{dP}{dt} V^{1.4} }{P(1.4)V^{0.4} } \\\frac{dV}{dt} =- \frac{-\frac{dP}{dt} V}{P(1.4)}$

$\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}$ (plugged in all the values at the instant)

$\frac{dV}{dt}$ = 26.466

Therefore, the volume increasing at the rate of 26.466cm³/min at this instant

6 0

3 years ago

Read 2 more answers

Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum in

Nikitich [7]

Answer:22 m/s

Explanation:

Given

launch angle $\theta =60^{\circ}$

height of goal $h=4.55\ m$

and horizontal distance $x=40\ m$

Suppose initial speed is $u$

Trajectory of a Projectile is given by

$y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }$

substituting the values we get

$4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }$

$4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}$

$\frac{31,360}{u^2}=69.28-4.55$

$\frac{31,360}{64.73}=u^2$

$u^2=484.47$

$u=22.01\ m/s$

So, initial launch speed is $22\ m/s$

4 0

3 years ago