As per Newton's III law we can say that
Force applied by object 1 on object 2 is always equal in magnitude and opposite in direction of the force that object 2 apply on object 1.
So we can say it as

now here above question is based upon the same
if a bag of vegetables applied a force F = 22.5 N of the surface stand the the same surface will apply same magnitude of force in opposite direction on the vegetables bag
So our answer will be F = 22.5 N (upwards).
Answer:
The work done by the gravel to stop the truck is 520.44 kJ
Explanation:
<u>Step 1</u>: Data given
Mass of the truck = 3047.8 kg
The ramp has an angle of 9.5 °
Velocity of the truck = 20.68 m/s
distance = 26.6 meters
<u>Step 2:</u> Calculate initial kinetic energy
sin 9.5° = 0.165
h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m
Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy
<u>Step 3: </u>Calculate potential energy
Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ
<u>Step 4:</u> What work is done by the truck on the gravel?
Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ
Answer:
SI derived units
Other quantities, called derived quantities, are defined in terms of the seven base quantities via a system of quantity equations. The SI derived units for these derived quantities are obtained from these equations and the seven SI base units.
Explanation:
Hope this helps :D
It’s cause by the fact that the earth is tilted 23.5 degrees
Answer:
Part 1) Time of travel equals 61 seconds
Part 2) Maximum speed equals 39.66 m/s.
Explanation:
The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

where
'v' is the final speed
'u' is initial speed
'a' is acceleration of the body
's' is the distance covered
Applying the given values we get

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance
Thus total time of journey equals
Part b)
the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 