Answer:
Explanation:
This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is
which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):
and get
12.1 m
Answer:
-6.6 km/h
Explanation:
In 7hr plane travelled 2020km;
For the first 4hr the average speed was 310km/h;
d=st, s=d/t;
Distance covered in first 4h is d = 310km/h×4h = 1240km;
See the image attached for further solution
<em>To determine the y component of velocity of a projectile </em><u><em>sine </em></u><em>operation is performed on the angle of launch.</em>
<u>Answer:</u> <em>sine</em>
<u>Explanation:</u>
Thus 
The initial velocity u can be resolved along two directions.
Along the X direction initial velocity = u cos θ
Along y direction initial velocity= u sin θ
From the equation of motion 
Thus velocity along x direction
=u cos θ
Velocity along y direction
= u sinθ -gt
Sign of g is negative.
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :


p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.