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Dmitry_Shevchenko [17]

3 years ago

14

The magnitude of the force involved in a certain collision (measured in newtons) is equal to the change in momentum (measured in

kilogram meters/second). What is the time interval of the collision?

1 answer:

Vinvika [58]3 years ago

4 0

Answer:

The time-interval of the collision is $t=\frac{{\Delta}p}{F}$

Explanation:

As given the force is equal to the rate of change of momentum. Mathemticaly this is:

$F=\frac{dp}{dt}.$

We can rearrange this equation to solve for ${\delta}t$ which gives

$t=\frac{{\Delta}p}{F}$

which is our answer.

In words this means the time interval is equal to the momentum change in that interval divided by the force applied that caused this momentum change.

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Answer:

It always takes energy to change the motion of an object.

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3 years ago

A vessel is filled with a liquid of density 1900 kg/m3 . There are two holes (one above the other) in the side of the vessel. Li

FromTheMoon [43]

Answer:

110 meters is the distance where they will intersect

Explanation:

given,

liquid density = 1900 kg/m³

distance of upper hole = 19 m

distance of lower hole = 117 m

acceleration due to gravity = 9.8 m/s²

the speed at each point

$v = \sqrt{2gh}$

for upper hole $v = \sqrt{2\times 9.8 \times 19}$

v = 19.29 m/s

lower hole $v = \sqrt{2\times 9.8 \times 117}$

v = 47.88 m/s

The path for each is parabolic

x = v t

$y = \dfrac{1}{2}gt^2$

$y = \dfrac{1}{2}g(\dfrac{x}{v})^2$

$y = \dfrac{gx^2}{2v^2}$

we get

upper hole

$y = \dfrac{9.8\times x^2}{2\times 19.29^2}= 0.0132 x^2$ lower hole

$y= \dfrac{9.8\times x^2}{2\times 47.88^2}=0.00214x^2$ y for upper hole = 80 + y for lower hole

$0.0132 x^2= 98 + 0.00214 x^2$

$0.0082 x^2 = 98$

x = 109.32 meters

110 meters is the distance where they will intersect

4 0

3 years ago

likoan [24]

Answer:

i think answer id leech because when it move it leave the liner

4 0

3 years ago

A projectile is fired horizontally from a gun that is 54.0 m above flat ground, emerging from the gun with a speed of 330 m/s. (

krok68 [10]

Explanation:

Given

height of building (h)=54 m

Projectile velocity=330 m/s

initial vertical velocity $(u_y)=0$

Initial horizontal velocity $(u_x)=330 m/s$

Time taken to cover vertical height of 54 m

$h=ut+\frac{gt^2}{2}$

$54=0+\frac{9.81\cdot t^2}{2}$

$t=\sqrt{\frac{54\times 2}{9.81}}$

t=3.31 s

Horizontal distance traveled in this time

$R_x=u_x\times t$

$R_x=330\times 3.31=1094.94 m$

Vertical component of velocity when it hits the ground

v=u+at

$v=0+9.81\times 3.31=32.47 m/s$

3 0

3 years ago