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4 years ago

What is needed to give a larger boulder a large acceleration?

Physics

2 answers:

Ronch [10]4 years ago

7 0

More force needs to be applied

kolbaska11 [484]4 years ago

4 0

Answer:

Huge amount of force

Explanation:

As per Newton's 2nd law of motion the acceleration of an object is directly proportional to the force applied on it. it can be expressed as

$F = mass \times acceleration$

Here the mass of the boulder is a lot. To give it large acceleration a huge amount of force will be needed to apply on it.

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Which phrase describes a scientific law?

olga_2 [115]

Answer:

a scientific law must be universally correct there could be no contradictions regarding the law anywhere

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4 years ago

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Identical cannonballs are fired with the same force, one each from four cannons having respective bore lengths of 1.0 meter, 2.0

ANEK [815]

I think its options 2 idk man

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4 years ago

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

Which characteristic should a good scientific question have? A) It should lead to a hypothesis that is not testable. B) it shoul

yarga [219]

Answer:

B. IT should have a very broad focus with many variables.

4 0

3 years ago

A typical tire for a compact car is 22 inches in diameter. If the car is traveling at a speed of 60 mi/hr, find the number of re

Softa [21]

Answer:

rpm= 916.7436 rev/min

Explanation:

First determine the perimeter of the wheel, to know the horizontal distance it travels in a revolution:

perimeter= π×diameter= π × 22 inches × 0.0254(m/inche)= 1.7555m

Time we divide the speed of the car, which is the distance traveled horizontally over time unit, by the perimeter of the wheel that is the horizontal distance traveled in a revolution, this dates us the revolutions over the time unit:

revolutions per time= velocity/perimeter

velocity= (60 mi/hr) × (1609.34m/mi) = 96560m/h

revolutions per time= (96560.6m/h) / (1.7555m)= 55004.614 rev/hr

rpm= (55004.614 rev/hr) × (hr/60min)= 916.7436 rev/min

3 0

4 years ago