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Drupady [299]
3 years ago
5

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above

the ground. Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations?
Physics
1 answer:
ryzh [129]3 years ago
6 0

Answer:

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v  that is

v² = u² - 2gh      ............1        for upward speed

v² = u² + 2gh     ............2        for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9  = 26.76

v² = 8² + 2(9.8) 1.9   = 101.24

v = 5.173  m/s    ..............3

v = 10.061 m/s   ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173  = 2.827 m/s     .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s     ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

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3 0
3 years ago
A 100 kg box as showsn above is being pulled along the x axis by a student. the box slides across a rough surface, and its posit
Anit [1.1K]

Answer:

a) 2 m/s

b) i) K.E = 50 (1.5t^2 + 2) ^2\\

ii) F = 3tm

Explanation:

The function for distance is x = 0.5t ^3 + 2t

We know that:

Velocity = v= \frac{d}{dt} x

Acceleration = a= \frac{d}{dt}v

To find speed at time t = 0, we derivate the distance function:

x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2

Substitute t = 0 in velocity function:

v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2

Velocity at t = 0 will be 2 m/s.

To find the function for Kinetic Energy of the box at any time, t.

Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\

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F = m \times a\\F= m \times 3t\\F = 3tm

6 0
3 years ago
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