Car A will have highest speed is 83.3m/s .
<h3>What is speed ? </h3>
The rate of change of position of an object in any direction.
The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction
.
speed = distance /time
In case of Car A ,
We have given distance 150Km in 3 min ,
First we have convert the distance km to m
150×1000m
then conversion of min to sec
38×60sec
speed = 15000/180
speed = 83.3m/sec
In case of Car B
we have given 800m in 150 min
lets convert the time into second
150×60
Speed = 800/150×60
speed = 0.88m/ s
In case of Car C
We have given here distance 250 Km and time in 8 hours
convert km to m
25000
and time into sec
88×60×60
speed = 0.86m/ s
Hence ,Car A has highest speed amongst them .
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Answer:
20 ms¯¹
Explanation:
3. Determination of the final velocity
From the question given above, the following data were obtained:
Time (t) = 4 s
Acceleration (a) = 5 ms¯²
Initial velocity (u) = 0 ms¯¹
Final velocity (v) =?
Acceleration is simply defined as the change in velocity per unit time.
Mathematically, it can be expressed as:
Acceleration (a) = final velocity – Initial velocity / time
a = v – u / t
With the above formula, we can obtain the final velocity of the car as follow:
Time (t) = 4 s
Acceleration (a) = 5 ms¯²
Initial velocity (u) = 0 ms¯¹
Final velocity (v) =?
a = v – u / t
5 = v – 0 / 4
5 = v / 4
Cross multiply
v = 5 × 4
v = 20 ms¯¹
Thus, the final velocity of the car is 20 ms¯¹
Answer:
option C
Explanation:
The correct answer is option C
Fire cut of fireman cut is diagonal cut which is provided at the end of the beam to prevent the fall of masonry wall if a fire breaks out in the building.
Fire cut allows joist to leave if it fails without affecting the masonry wall standing.
Without fire cut, the burnt beam will rotate downward affecting the connection of beam and wall and leading to damage it.
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
