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2 years ago

A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the

Physics

1 answer:

zlopas [31]2 years ago

8 0

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

$a=2.5 m/s^2$

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

$d=0+\frac{1}{2}(2.5)(35)^2=1531 m$

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2 years ago

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

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Let h = distance (m) to the water surface.

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2 years ago

Read 2 more answers

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

A 10kg sphere hits a stationary 8kg sphere. After the collision the 8kg sphere moves off in the positive direction at 4m/s. If t

Bumek [7]

M = mass of the first sphere = 10 kg

m = mass of the second sphere = 8 kg

V = initial velocity of the first sphere before collision = 10 m/s

v = initial velocity of the second sphere before collision = 0 m/s

V' = final velocity of the first sphere after collision = ?

v' = final velocity of the second sphere after collision = 4 m/s

using conservation of momentum

M V + m v = M V' + m v'

(10) (10) + (8) (0) = (10) V' + (8) (4)

100 = (10) V' + 32

(10) V' = 68

V' = 6.8 m/s

7 0

2 years ago