Find the angles which the vecotr <span><span><span><span><span>v </span><span>⃗ </span></span>=3i−6j+2k</span> </span><span><span>v→</span>=3i−6j+2k</span></span>
makes with the coordinate axes.
If the angles are <span><span><span>α,β,θ</span> </span><span>α,β,θ</span></span>
, show that for any 3-dimensional vector:
. . . <span><span><span><span><span>cos </span><span>2 </span></span>α+co<span><span>s </span><span>2 </span></span>β+<span><span>cos </span><span>2 </span></span>θ=1</span> </span><span><span>cos2</span>α+co<span>s2</span>β+<span>cos2</span>θ=1</span></span>
It is a part of the sugary liquid in the plant that uses photosynthesis to produce.
Answer :
556.59 m/s.
Explanation:
Given that,
An airplane is traveling at an altitude of 15,490 meters.
A box of supplies is dropped from its cargo hold. We need to find the velocity of cargo when it hits the ground.
The initial velocity of the box is 0 as it as at rest. Let v is the velocity of cargo when it hits the ground. We can find it using third equation of motion as follows :
Put u = 0 and a = g
So, when it hits the ground its velocity is 556.59 m/s.
The person driving the truck was killed
the wall was destroyed
Answer:
So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.
Now,
We have F=ma,
and a=v/t,
so we can have another equation for it as,
Now, providing the required data to, it; ∴t =2 sec,
F=(1)×(20/2),
- So, the car would be acting the force,F of about 10 N while the car is present on the lower region of the track.
