The driver of a 2.0 × 10³ kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car
1 answer:
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Answer:
L_max= 0
Explanation:
The formula for magnitude of maximum orbital angular momentum is given by
l= orbital quantum number
l= n-1
n= shell number or principal quantum number
for n=1 , l=0
therefore,
L_max= 0
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".
Solution:
The fundamental frequency in a standing wave is given by
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
The wavelength of the standing wave is instead twice the length of the string:
So the speed of the wave is
And the time the pulse takes to reach the shop is the distance covered divided by the speed:
Solution:
The fundamental frequency in a standing wave is given by
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
The wavelength of the standing wave is instead twice the length of the string:
So the speed of the wave is
And the time the pulse takes to reach the shop is the distance covered divided by the speed:
Answer:
It takes the train 7.5 seconds to reach top speed
The train travels 112.5 meters before it reaches its top speed
Explanation:
The Metro Rail makes a stop at NRG Stadium to pick up some Texans
fans after the game
After all the passengers board, the train accelerates at a rate of 4 m/s²
to a top speed of 30 m/s
We need to find the train takes how many seconds to reach the
top speed
Given is:
→ The train start from rest, then its initial speed u = 0
→ It acceleration a = 4 m/s²
→ Its top speed v = 30 m/s
To find the time we can use this rule;
→
→ seconds
<em>It takes the train 7.5 seconds to reach top speed</em>
<em></em>
Now we need to find how far the train travels before reaches top speed
We can use this rule;
→ v² = u² + 2as, where s is the distance
→ v = 30 m/s , u = 0 , a = 4 m/s²
Substitute these values in the rule
→ (30)² = 0 + 2(4) s
→ 900 = 8 s
Divide both sides by 8
→ s = 112.5 m
<em>The train travels 112.5 meters before it reaches its top speed </em>
A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
b) The work done by the frictional force against the motion of the block is equal to:
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (
), and part of it becomes increase of thermal energy of the floor (
). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
So, we have
b) The work done by the frictional force against the motion of the block is equal to:
Part of these 105.1 Joules of work becomes increase of thermal energy of the block (
c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
So, we have