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Ne4ueva [31]
3 years ago
12

in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo

ve together. What is the total momentum after the collision? 55 kg*m/s 275 kg * m/s -55kg * m/s -275 kg m/2
Physics
1 answer:
guajiro [1.7K]3 years ago
4 0

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

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A toboggan approaches a snowy hill moving at 12.4 m/s. The coefficients of static and kinetic friction between the snow and the
Harrizon [31]

Answer:

a) Acceleration while the toboggan is going uphill = 8.85 m/s²

b) Acceleration of the toboggan is going downhill = 4.76 m/s²

Explanation:

Resolving the normal reaction into vertical and horizontal component

Nₓ = mg sin θ

Nᵧ = mg cos θ

The frictional force on the toboggan = Fr = μmg cos θ (note that μ = μ(k) = the coefficient of kinetic friction)

Doing a Force balance on the x component taken in the axis parallel to the inclined plane.

The net force that has to accelerate the toboggan has to match the frictional force acting downwards of the plane (In the opposite direction to motion) and x-component of the Normal reaction

ma = Nₓ + Fr

ma = (mg sin θ) + (μmg cos θ)

a = g[(sin 44°) + (0.29 cos 44°)

a = 9.8(0.6947 + 0.2086) = 8.85 m/s²

b) While coming downhill,

The frictional force is acting uphill now (In the direction opposite to motion),

The force balance is

ma = (mg sin θ) - (μmg cos θ)

a = g[(sin 44°) - (0.29 cos 44°)]

a = 9.8(0.6947 - 0.2086)

a = 4.76 m/s²

3 0
3 years ago
An automobile is traveling on a long, straight highway at a steady 76.0 mi/h when the driver sees a wreck 180 m ahead. at that i
masha68 [24]

a) The car’s speed just after leaving the icy portion of the road is the first part

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

Here s = 130 m, a = -1.2 m/s^2, u = 76 mile/hr = 33.975 m/s

             130 = 0*t - 0.5*(-1.2)*t^2\\ \\ \\ t=14.72 seconds

After 14.72 seconds cars speed is given by v =u +at

                                                          u = 33.975m/s, a = -1.2 m/s^2

                                                      v = 33.975-1.2*14.72 = 16.31 m/s

   So velocity of car on leaving icy surface = 16.31 m/s

b) Time taken after icy surface

              0 = 16.31 - 7.10*t

              t = 2.23 seconds

         Distance traveled during this time

                   v^2 = u^2+2as\\ \\ 0 = 16.31^2 - 2*7.10*s\\ \\ s = 18.74 m

   Total distance traveled = 130+18.74 = 148.74 m

c) Total time taken = 14.72+2.23=16.95 seconds

3 0
3 years ago
A bike travels at a constant speed of 4.0m/s for 5 seconds. How far it goes ?
alex41 [277]
The speed of 4m/s means that the bike goes 4 meters for each second of movement. Multiplying this by 5 seconds we have a total distance travelled of 20m.
3 0
3 years ago
Read 2 more answers
For a satellite of mass mS in a circular orbit of radius rS around the Earth, determine its kinetic energy, K . Express your ans
agasfer [191]

Since it is asking you to find the kinetic energy in relation to the mass, radius, mechanical energy (total energy), and constants, you will need to setup an equation first to "find" the Mechanical Energy, so that we can then solve for the kinetic energy, as from my experience with high school physics, there is only the graviational potential energy equation and force in relation to celestial bodies.

Knowing the ME is the total energy, we add up the energies of the system. Since it is being influenced by the Earth, as per the problem stating the satellite has circular orbit around the Earth, we know there is gravitational potential. Since it is orbiting, we can assume some type of velocity. Nothing else that we need to worry about should be occuring at this level of physics, leaving you with

ME= Ug+K

from here we solve for K, as plugging in could get confusing and messy at the moment.

ME-Ug=K

now using the equations presumably given in class, if not then using this equation, we can find the Ug

Ug=(-(Gm*M)/r)     note that M is the mass of the Earth and m is the satellite

this should give us

ME-(-(GmM)/r)=K

since there is a negative being subracted, we can change that to

ME+(GmM)/r=K

I believe this should be fine, as the Earth's mass is constant, but if not, then all you need to figure left is how to get rid of the M in the equation, as the rest of the terms and constants are for sure within the requirements.

8 0
3 years ago
While watching a movie a spaceship explodes and there is a loud bang and flash of light. What is wrong with this scene? Explain
Ksenya-84 [330]
Well if the ship was in space their shouldn’t be a loud bang. Because you can’t hear anything in space
4 0
3 years ago
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