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lianna [129]
3 years ago
6

in an isolated system bicyclist 1 and bicyclist 2 each with a mass of 80 kg, collide. bicyclist 1 was moving to the right at 8 m

/s, while bicyclist 2 was moving to the left at 2 m/s. what is the magnitude of their combined momentum before the crash
Physics
1 answer:
Reika [66]3 years ago
3 0
Consider speed of bicyclist 1 as V1=+8m/s and bicyclist 2 as V2=-2m/s  m=80kg 

Since momentum has opposite signs moving to right or left   

Therefore the combined momentum , P is given by :   

P=m*V1+m*V2 =80*(8-2)kg*m/s=+480 kg m/s
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A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
You have discovered a planet that is one-quarter the radius of Earth (Rp = 1/4R⊕) and one-half as massive (Mp = 1/2M⊕). How does
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It's 8 times as much as before.
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At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at
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3 years ago
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride
marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

\Delta x = 6.17 cm = 0.0617 m

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

2mg = k'\Delta x (1)

where

m = 76.2 kg is the mass of each children

g=9.8 m/s^2 is the acceleration of gravity

k' is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

k'=k+k=2k

Substituting into (1) and solving for k, we find:

2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m

b)

The period of the oscillating system is given by

T=2\pi \sqrt{\frac{m}{k'}}

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

k'=2k=2(12,103)=24,206 N/m is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

f=\frac{1}{T}

where

f is the frequency

T is the period

In  this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

f=\frac{1}{2.09}=0.478 Hz

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

t=10T=10(2.09)=20.9 s

#LearnwithBrainly

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Answer:

Yes

Explanation:

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