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Area of Parallelogram = breadth x height⟂ = (A x B)
Here, A = i  j + k
And, B = 3i + 4j k
= i (14) j (3+1) +k (43)
=> (A x B) = 3i +2j +k
A body is projected horizontally from the top of a cliff with a velocity of 9.8m/s. What time elapses before horizontal and vertical velocities become equal? Take g = 9.8m/s^{2}
Horizontal velocity at any instant, v_{x} = u = 9.8m/s
Vertical velocity at any instant, v_{y} = 0 + gt = 9.8t
9.8 = 9.8t
t = 1s
Two vectors are equal if and only if they have the same magnitude and the same direction.
Hence, Option D is correct
A cyclist moves along a circular path of radius 70m. If he completes one round in 11s, calculate total length of path.
Radius of the circular path, r = 70m
Time takes to complete one round, t = 11s
Total length of the path, s = 2πr = 2*22/7*70 = 440m
Force, area and velocity all are a set of different or the same fundamental quantities multiplied or divided together while current itself is a fundamental quantity and can be expressed in terms of any other.
A projectile is the name given to any body which once thrown into space with some initial velocity, moves thereafter under the influence of gravity alone without being propelled by an engine or fuel. The path followed by a projectile is called its trajectory.
The vector product of two vectors depends upon the sine of the angle between the vectors and when the vectors are parallel the sine component would be zero hence the resultant is a null vector.
We know that for two vectors P and Q
P.Q = PQ cos a
And PxQ = PQ sin a
Where a is angle between them
When P.Q = PxQ
We get sin a = cos a
Thus a = 45^{0}
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10^{4} J by the end of the second revolution after the beginning of the motion?
kinetic energy =8×10−4J
or, (1/2)mv^{2}=8×10−4
or, (1/2)×10×10−3v²=8×10−4
or, v^{2}=16×10−2
=>v=0.4m/s
initial velocity of particle, u=0m/s
we have to find Tangential acceleration at the end of 2nd revolution.
total distance covered, s=2(2πr)=4πr
so, v^{2}=2as
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