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
Explanation:
Initial speed of the rocket, u = 0
Acceleration of the rocket, 
Time taken, t = 3.39 s
Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

Let x is the initial position of the rocket. Using third equation of kinematics as :


Let
is the position at the maximum height. Again using equation of motion as :

Now
and v and u will interchange



x = 524.14 meters
Hence, this is the required solution.
Answer: [B]: The letter, "<em /> I " ; for current; in units of "Amps" .
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It would be Joules.
Workdone is measured in Joules.
Workdone = Force * distance
Force = mass * acceleration
= kg * ms⁻²
= kgms⁻²
Distance = m
So, Force * distance
kgms⁻² * m
Apply laws of indices that says
x² * x³ = x²⁺³ = x⁵
Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2
Answer:
37.42 m/s
Explanation:
We know that apparent frequency,
is given by
where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and
is the speed of car which is unknown, \bar f is given as 440 Hz
