Given required solution
M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
Answer:
v₂ = v/1.5= 0.667 v
Explanation:
For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.
Initial moment before pushing
p₀ = 0
Final moment after they have been pushed
= m₁ v₁ + m₂ v₂
p₀ = 
0 = m₁ v₁ + m₂ v₂
m₁ v₁ = - m₂ v₂
Let's replace
M (-v) = -1.5M v₂
v₂ = v / 1.5
v₂ = 0.667 v
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
Answer:
Check the explanation
Explanation:
To tackle situations like the one above, the rate of gravity of that star must be equal to the rate of power output but we don’t have radius of that star. Also temp is not mentioned. And emissivity of star is also not mentioned. So the only possible way is like Einstein mass energy relationship E=mc^2=6.5380e40
power =E/Time so this energy is transferred per sec.
The car would go from zero to 58.0 mph in 2.6 sec.
Since the force on the car is constant, therefore the acceleration of the car would also be constant.
Now for constant acceleration we can use the equation of motion
Using first equation of motion to calculate the acceleration of the car
v=u+at
29=0+a×1.30 ...... Eq. (1)
Again using the first equation of motion
58=0+a*t ....... Eq. (2)
Dividing eq. (2) with equation 1
t=2×1.3
t=2.6 sec