Answer:
The time taken is 
Explanation:
From the question we are told that
The speed of first car is 
The speed of second car is 
The initial distance of separation is 
The distance covered by first car is mathematically represented as
Here 
is the initial distance which is 0 m/s
and 
is the final distance covered which is evaluated as 
So
The distance covered by second car is mathematically represented as
Here is the initial distance which is 119 m
and is the final distance covered which is evaluated as 
Given that the two car are now in the same position we have that
Answer: 90 m/s
Explanation:
Given
mass of racecar 
velocity of racecar 
mass of still honeybadger 
after collision race car is traveling at a speed of 
conserving linear momentum
![Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]](https://tex.z-dn.net/?f=Mu%2Bm%5Ctimes0%3DMv_1%2B%20mv_2%5Cquad%5Bv_2%3D%5Ctext%7Bvelocity%20of%20honeybadger%20after%20colllision%7D%5D)
The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.
Answer:
-0.55m/s
Explanation:
Given that: For the boy
Weight = 745N
Velocity = +0.35 m/s
Mass of the boy = ?
g = 9.81m/s^2
W = mg
745 = m×9.81
m = 75.94kg
For the girl
Given that:
Weight = 477 N
g = 9.81m/s^2
m = ?
W = mg
477 = m×9.81m/s^2
m = 48.62kg
To solve for the v of the girl, the two has to add up
48.62kg×v + 75.94kg×+0.35 m/s = 0
48.62v + 26.579 = 0
48.62v = - 26.579
v = -26.579/48.62
v = -0.5466
v = -0.55m/s
Hence, the velocity of the girl is -0.55m/s.
The negative sign is as a result of the two of them moving is opposite direction.
Answer:
The answers to the questions have been solved in the attachment.
Explanation:
The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.
