Answer:
,
(Northeast)
Explanation:
Let consider that owner has the resultant direction of the momentums from the dog and the cat. First, the momentum of the owner is:
![\|\vec p \| = \sqrt{\left[(27.4\,kg)\cdot \left(2.19\,\frac{m}{s} \right)\right]^{2}+\left[(7.19\,kg)\cdot \left(2.78\,\frac{m}{s} \right)\right]^{2}}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20p%20%5C%7C%20%3D%20%5Csqrt%7B%5Cleft%5B%2827.4%5C%2Ckg%29%5Ccdot%20%5Cleft%282.19%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5Cright%5D%5E%7B2%7D%2B%5Cleft%5B%287.19%5C%2Ckg%29%5Ccdot%20%5Cleft%282.78%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5Cright%5D%5E%7B2%7D%7D)

The speed of the owner is:


Lastly, the direction of the owner is:
![\alpha = \tan^{-1}\left[\frac{(27.4\,kg)\cdot \left(2.19\,\frac{m}{s} \right)}{(7.19\,kg)\cdot \left(2.78\,\frac{m}{s} \right)} \right]](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B%2827.4%5C%2Ckg%29%5Ccdot%20%5Cleft%282.19%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%7D%7B%287.19%5C%2Ckg%29%5Ccdot%20%5Cleft%282.78%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%7D%20%5Cright%5D)
(Northeast)
The answer for the cars speed is tue
Answer:
S=720m anf vf=12m/s
Explanation:
acceleration=a=0.1m/s²
time taken=t=2minutes=2×60=120seconds
vi=0m/s
vf=?
distance covered=S=?
by using second equation of motion
S=vit+1/2at²
S=0m/s×120seconds+1/2(0.1m/s²)(120s)²
S=0m/s+1/2×1440m
S=720m
and now we have to find the vf
vf=vi+at
vf=0m/s+(0.1m/s)(120s)
vf=12m/s
i hope it will help you
Answer:
The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds
Explanation:
The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend
The direction in which the student tosses the ball = The horizontal direction
Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0
The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;
h = 1/2 × g × t²
Where;
g = The acceleration due gravity of the ball = 9.81 m/s²
t = The time of motion to cover height, h
Then height is already given as h = 3.8 m
Substituting gives;
3.8 = 1/2 × 9.81 × t²
t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²
∴ t = √0.775 ≈ 0.88 seconds
The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.