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GarryVolchara [31]
3 years ago
9

If i jump on a trampoline. is it an inelastic or elastic collision?

Physics
2 answers:
Stells [14]3 years ago
8 0
It would be an inelastic collision.

hope this helps you.
Sladkaya [172]3 years ago
5 0
Inelastic Collision :P
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When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
3 years ago
Recent findings on the topic of brain based research indicate all of the following except
Gre4nikov [31]
The answer is D, the brain actually stops growing around age 18
5 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
35. What would a ship's position be if that ship started at
Tcecarenko [31]

Answer:

The final position of the ship after the given time period is  42 km West of B.

Explanation:

Given;

average velocity of the ship, v = 35 km/h

time taken for the ship to reach point D, t = 1.2 hours

The position of the ship after the given time period is calculated as follows;

x = v x t

x = (35 km/h) x 1.2 h = 42 km

x  = 42 km West of B.

Therefore, the final position of the ship after the given time period is  42 km West of B.

5 0
2 years ago
a kid is on a skate board going 14kph and trows a set of keys on the ground at 8kph. the spped of the keys relitive to the gound
Vanyuwa [196]
To determine the speed relative to the ground, since the ground is our reference frame, it would be v = 0, for the kid on the skateboard, you would need to take into account the speed that he/she is going and the speed of the keys thrown at.

I believe.
5 0
3 years ago
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