Answer:

Explanation:
Given,
For the first rocket,
- Initial velocity of the first rocket A =

- Acceleration of the first rocket =

For the second rocket,
- Initial velocity of the second rocket B =

- Displacement of both the rockets A and B = s = 0 m
Fro the first rocket,
Let 't' be the time taken by the first rocket A for whole the displacement

Let
be the acceleration of the second rocket B for the same time interval
from the kinematics,


Hence the acceleration of the second rocket B is -33.65\ m/s^2.
Answer:
a= 2.7 m/s^2
Explanation:
acceleration: a
speed: V
Vf = final speed
Vi= initial speed (initial = beginning)
100 km/hour --> m/s
divide the speed value by 3.6
100/3.6= 27.8 m/s




a= 2.7 m/s^2
Answer: Given: h₀=5cm
p=30cm
hi
f=45cm
require: q=?
hi=?
formulas:
1/f=1/p+1/q
hi/h₀=q/p
calculations:1/q=1/f-1/p
1/q=1/45-1/30
1/q=0.022=0.033
1/q= = -0.011
q=1/-0.011
q= -90.91cm
hi=p × h₀/q
hi=30ₓ5/-90.91
hi=150/-90.91
hi= -1.65cm
image is virtual , erect and diminished
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Answer:
B. 0.16 m
Explanation:
The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.
Since the dart is thrown horizontally, the initial vertical velocity is zero:

While the horizontal velocity is

The horizontal distance covered is

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

Answer:
Explanation:
Kinetic energy of the block
= 1/2 m v²
= .5 x 3 x 4 x 4
= 24 J
Negative work of - 24 J is required to be done on this object to bring it to rest.
magnitude of acceleration due to frictional force
= force / mass
2 / 3
= 0 .67 m /s²
Let the body slide by distance d before coming to rest so work done by force = Kinetic energy
= 2 x d = 24
d = 12 m