Answer:
the third one is incorrect
Explanation:
10 x 10³= 10^1 x 10^3 = 10^4
We can answer this using one of the equations of linear
motion:
v = d / t
where:
v = velocity
d = distance
t = time
<span>In the problem, we are asked to find for the time in
which Driver B will catch up to Driver A. Therefore, find the time when dA = dB. Rearranging the
equation and equation dA and dB will result in:</span>
<span>vA * tA = vB * tB
---> 1</span>
It was given that:
vA = 68 mph
tA = tB + 3 (since person A was travelling 3 hours
earlier)
vB = 85 mph
tB = unknown
Substituting into equation 1:
68 * (tB + 3) = 85 * tB
68 tB + 204 = 85 tB
tB = 12 hrs
Therefore driver B would catch up to driver A after 12
hrs.
<span> </span>
The mosquito exerts the same force on the car as the car exerts on the mosquito.
F = M (V^2/r) = 20 x 100 /.15 =13333.33 N<span>convert units as needed to wind up with Newtons Kg-m/s^2 hope it helps :)</span>
Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of 
When the hand is 
above the ground
Using the equation of motion we can write
Substitute the values
![\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]](https://tex.z-dn.net/?f=%5CRightarrow%202.5%3D-19t%2B0.5%5Ctimes%209.8t%5E2%5C%5C%5CRightarrow%204.9t%5E2-19t-2.5%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19%5Cpm%20%5Csqrt%7B%28-19%29%5E2-4%5Ctimes%204.9%5Ctimes%20%28-2.5%29%7D%7D%7B2%5Ctimes%2019%7D%5C%5C%5CRightarrow%20t%3D4.0049%5Cquad%20%5B%5Ctext%7BNeglecting%20the%20negative%20value%20of%20%7Dt%5D)
Thus, the ball will take 4 s to hit the ground.
