The magnitude of the resultant is
√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .
The direction of the resultant is
tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .
Answer:
Option (c).
Explanation:
An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,
is the angle of projection
The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.
As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same
.
By definition, the speed of an object is given by:
Where,
dr/dt: derived from the position with respect to time
Therefore, speed has units of length over units of time.
Thus, speed is a derived quantity, since it depends on the value of two other quantities.
Answer:
a derived quantity is:
C. Speed
I believe that the loss of Phyllis' recollections is thought to be the most heartbreaking side effect of her ailment in light of the fact that once a man's memory scatters then piece of the individual begins to vanish with them. A memory holds a considerable measure of essential data, for example, people's identity, where They have lived, and their connections that they have had with individuals.
Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
