Explanation:
Take F=ma
a = F/m
For a higher, F higher or m lower
Means higher horse power for engine or lower mass for the car
Answer:
Explanation:
Let initial extension in the spring= x₀
Force on the spring = F₀
Let spring constant = k
Fo = k x₀
Fn = 3k x₀
Fn /Fo = 3
PEs0 ( ORIGINAL) =1/2 k x₀²
PEsn ( NEW) =1/2 k (3x₀)²
PEsn / PEs0 = 9
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
v=5.02 m/s
c.Work done by friction on the incline,
Answer:
129900
Explanation:
Given that
Mass of the particle, m = 1 g = 1*10^-3 kg
Speed of the particle, u = ½c
Speed of light, c = 3*10^8
To solve this, we will use the formula
p = ymu, where
y = √[1 - (u²/c²)]
Let's solve for y, first. We have
y = √[1 - (1.5*10^8²/3*10^8²)]
y = √(1 - ½²)
y = √(1 - ¼)
y = √0.75
y = 0.8660, using our newly gotten y, we use it to solve the final equation
p = ymu
p = 0.866 * 1*10^-3 * 1.5*10^8
p = 129900 kgm/s
thus, we have found that the momentum of the particle is 129900 kgm/s
Answer:
Rebounce angle is 345°
Rebounce speed is 989.95m/s
Explanation:
Calculate the x component of the velocity of the bullet before impact by using the following relation:
Vbx= Vb Cos thetha
Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°
Substituting
Vbx = Cos15 ×1400 = 1352.30m/s
Calculate the y component using the relation:
Vby = Vo Sin theta
Vby = sin 15° × 1400
Vby = 362.35m/s
The rebounce angle = 360 - incidence angle
Rebounce angle =( 360 - 15)° = 345°
The rebound speed V' = Vby - Vbx
V' = (1352.30 - 362.35)m/s
V' = 989.95 m/s
