Compared to the pucks given, the pair of pucks will rotate at the same rate.
Answer: Option A
<u>Explanation:</u>
The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.
The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.
By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since , the turning rate of the two-puck framework must stay unaltered.
Velocity is height , speed is an object being propelled by force
Answer
given,
Speed of car A = 95 Km/h
= 95 x 0.278 = 26.41 m/s
Speed of Car B = 121 Km/h
= 121 x 0.278 = 33.64 m/s
Distance between Car A and B at t=0 = 41 Km
a) Distance travel by car B
d = 26.41 t + 41000
speed of the car A = 33.64 m/s
distance = s x t
26.41 t + 41000 = 33.64 x t
7.23 t = 41000
t = 5670.82 s
time taken by Car B to cross Car A is equal to t = 5670.82 s
distance traveled by car A
D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km
b) distance travel by the car B in 30 s after overtaking car A
D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km
Explanation:
<em>Hi</em><em>,</em><em> </em><em>there</em><em>!</em><em>!</em>
<em>Energy</em><em> </em><em>is</em><em> </em><em>defined</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>capacity</em><em> </em><em>or</em><em> </em><em>ability</em><em> </em><em>to</em><em> </em><em>do</em><em> </em><em>work</em><em>.</em><em> </em><em>It's</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>is</em><em> </em><em>Joule</em><em>.</em>
<em>here</em><em>,</em>
<em>Joule</em><em> </em><em>=</em><em> </em><em>(</em><em>kg</em><em>×</em><em>m</em><em>×</em><em>m</em><em>)</em><em>/</em><em>(</em><em>s</em><em>×</em><em>s</em><em>)</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>kg</em><em>×</em><em>m</em><em>^</em><em>2</em><em>/</em><em>s</em><em>^</em><em>2</em><em>.</em>
<em>Therefore</em><em>, </em><em> </em><em>the</em><em> </em><em>derived</em><em> </em><em>unit</em><em> </em><em>is</em><em> </em><em>kg</em><em>.</em><em>m</em><em>^</em><em>2</em><em> </em><em>by</em><em> </em><em>s</em><em>^</em><em>2</em><em>.</em>
<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>
Answer:
part (a) towards north east direction.
part (b) s = 46.60 m
Explanation:
Given,
- velocity of the river due to east =
- velocity of the boat due to the north =
part (a)
River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,
Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are
Let 'v' be the velocity of the boat relative to the shore.
Let be the angle of the velocity of the boat relative to the shore with the horizontal axis.
Direction of the velocity of the boat relative to the shore.
part (b)
- Width of the shore = w = 300m
total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken
Let 't' be the total time taken by the boat to cross the width of the river.
Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river