Answer:
Δx = 39.1 m
Explanation:
- Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
- Solving (1) for Δx, we have:
Answer:
The answer is
D. 1230j
Explanation:
When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force
given data
Mass of gun M=10kg
Mass of bullet m=20g----kg=20/1000 =0.02kg
Propulsive speed of bullet = 350m/s
Hence the moment of the bullet will be equal and opposite to that of the gun
mv=MV
where V is the recoil velocity which we are solving for
V=mv/M
V=0.02*350/10
V=7/10
V=0.7m/s
The energy contained in the bullet can be gotten using
KE=1/2m(v-V)²
KE=1/2*0.02(350-0.7)²
KE=1/2*0.02(349.3)²
KE=1/2*0.02*122010.49
KE=1/2*2440.20
KE=1220.1J
roughly the energy is 1230J
Answer with Explanation:
From work energy theorem we have
Now since it is given that the work done on the object by the force is positive hence we conclude that the term on the right hand of the above relation is positive hence
Part a)
Hence we conclude that the mechanical energy of the particle will increase.
Part b) Since the mechanical energy of the particle is the sum of it's kinetic and potential energies, we can write
Now since the sum of the 2 energies ( Kinetic and potential ) is positive we cannot be conclusive of the individual values since 2 cases may arise:
1) Kinetic energy increases while as potential energy decreases.
2) Potential energy increases while as kinetic energy decreases.
Hence these two cases are possible and we cannot find using only the given information which case will hold
Hence no conclusion can be formed regarding the individual energies.
Answer:
The efficiency is just 0.016
Explanation:
The efficiency is given by the useful energy ÷ by the total energy.
1 W is the same that 1 joule per second
If you do the math:
1.6/100=0.016
Answer:
The free fall acceleration on the surface of this planet is 4.35 m/s²
Explanation:
Given that,
Mass of planet
Radius of planet
We need to calculate the free fall acceleration on the surface of this planet
Using formula of gravity
Where, = mass of planet
= radius of plane
Put the value into the formula
Hence, The free fall acceleration on the surface of this planet is 4.35 m/s²