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3 years ago

5

A 2.0 kg arrow is fired at 10 m/s into a stationary 10kg block of ice what is the velocity of the arrow / ice combination after

the collision?

1 answer:

PilotLPTM [1.2K]3 years ago

6 0

Answer:

1.67 m/s

Explanation:

Since this is an inelastic collision (the block and the arrow stick together after the collision), we can solve the problem by using the law of conservation of momentum:

$p_i = p_f\\m_a v_a + m_b v_b = (m_a + m_b) v$

where

$m_a = 2.0 kg$ is the mass of the arrow

$v_a = 10 m/s$ is the initial speed of the arrow

$m_b = 10 kg$ is the mass of the block

$v_b = 0$ is the initial speed of the block

$v = ?$ is the final speed of the arrow+block

Substituting and re-arranging the equation, we find:

$v=\frac{m_a v_a + m_b v_b}{m_a +m_b}=\frac{(2.0 kg)(10 m/s)+0}{2.0 kg+10.0 kg}=1.67 m/s$

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A bus travel with an average velocity of 60km/h. How long does it take to cover a distance of 500km

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Around 8 hours and 20 minutes

Explanation:

I divided 500 by 60 and got 8.3333333333 and i round it up to 8.20, so it is 8 hours and 20 minutes.

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If you were designing a room in a house, where would be the better place to put a heater, near the floor or near the ceiling? Wh

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Answer: The correct explanation is 2.

Explanation: The warm air is less dense (it expands) and thus it is lighter than the cold air so it will rise up to the floor. Therefore, when you place the heater on the floor it will warm the cold air which would then rise and be replaced by more cold air which would again get warm and rise and so on until the room is heated. This means that the correct explanation is 2.

On the other hand, if you put the heater at the ceiling, it will warm the cold air near the ceiling which would stay up there (it is lighter than the cold air under it). This means that the only way for the heat to spread from this ceiling level warm air to the lower levels is via conduction which is slow.

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2 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

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