Answer:
P.E = mgh
K.E = 1/2mv²
Explanation:
P.E = mgh
That is m= mass in kilograms, g=acceleration due to gravity and h= height in meters.
K.E = 1/2mv²
That is 1/2 of an object's mass multiplied by the velocity squared.
Answer:
momentum= mass x velocity = 0.35x15= 5.25kgm/s
Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,
Then applying the new Cartesian sign convention to it,
Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i) 
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
Answer:
d) 700 m/s
Explanation:
if k is the force constant and x is the maximum compression distance, then:
the potential energy the spring can acquire is given by:
U = 1/2×k×(x^2)
and, the kinetic energy system is given by:
K = 1/2×m×(v^2)
if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:
initial energy = final energy
Ki +Ui = Kf +Uf
Ui = 0 J and Kf = 0J
Ki = Uf
1/2×m×(v^2) = 1/2×k×(x^2)
m×(v^2) = k×(x^2)
v^2 = k×(x^2)/m
= (500)×((21×10^-2)^2)/(19×10^-3 + 8)
= 2.75
v = 1.66 m/s
the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:
m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.
m1×v1 = v×(m1 + M)
v1 = v×(m + M)/(m1)
= (1.66)×(19×10^-3 + 8)/(19×10^-3)
= 699.86 m/s
≈ 700 m/s
Therefore, the velocity of the bullet just before it hits the block is 700 m/s.
