Answer:
Solution given:
height [H]=25m
initial velocity [u]=8.25m/s
g=9.8m/s
now;
a. How long is the ball in flight before striking the ground?
Time of flight =?
Now
Time of flight=![\sqrt{\frac{2H}{g}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2H%7D%7Bg%7D%7D)
substituting value
- =
![\sqrt{\frac{2*25}{9.8}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2%2A25%7D%7B9.8%7D%7D)
- =2.26seconds
<h3>
<u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>
b. How far from the building does the ball strike the ground?
<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?
we have
Horizontal range=u*![\sqrt{\frac{2H}{g}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2H%7D%7Bg%7D%7D)
<h3>
<u>The ball strikes 18.63m far from building</u>. </h3>
Answer:
that best describes the process is C
Explanation:
This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.
Heat absorbed by the smallest container
Q_c = m ce (
-T₀)
Heat released by the largest container is
Q_a = M ce (T_{i}-T_{f})
how
Q_c = Q_a
m (T_{f}-T₀) = M (T_{i} - T_{f})
Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.
Of the final statements, the one that best describes the process is C
since it talks about the thermal energy and the heat that is transferred in the process
Answer: 750Kg
Explanation:
Recall that force is the product of the mass M, of an object moving at a uniform acceleration.
i.e Force = Mass x Acceleration
In this case, Mass = ?
Force = 3.00 x 10^3 N = (3.00 x 1000N)
= 3000N
Uniform acceleration = 4.00m/s^2
Force = Mass x Acceleration
3000N = Mass x 4.00m/s^2
Mass = (3000N/4.00m/s^2)
Mass = 750Kg (The SI unit of mass is kilograms)
Thus, the mass of the car is 750Kg
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Answer:
![F = 505.13 N](https://tex.z-dn.net/?f=F%20%3D%20505.13%20N)
Yes it is better to pull the rope rather than push it
Explanation:
Let the force is applied at an angle of 60 degree
so we will have net vertical force on the crate is given as
![F_n + Fsin60 = mg](https://tex.z-dn.net/?f=F_n%20%2B%20Fsin60%20%3D%20mg)
here we know
![m = 180 lb](https://tex.z-dn.net/?f=m%20%3D%20180%20lb)
![m = 81.65 kg](https://tex.z-dn.net/?f=m%20%3D%2081.65%20kg)
![F_n = 81.65(9.81) - Fsin60](https://tex.z-dn.net/?f=F_n%20%3D%2081.65%289.81%29%20-%20Fsin60)
![F_n = 801 - 0.866 F](https://tex.z-dn.net/?f=F_n%20%3D%20801%20-%200.866%20F)
now friction force on the crate is given as
![F_x = \mu F_n](https://tex.z-dn.net/?f=F_x%20%3D%20%5Cmu%20F_n)
![Fcos60 = 0.7(801 - 0.866 F)](https://tex.z-dn.net/?f=Fcos60%20%3D%200.7%28801%20-%200.866%20F%29)
![0.5F + 0.61F = 560.7](https://tex.z-dn.net/?f=0.5F%20%2B%200.61F%20%3D%20560.7)
![F = 505.13 N](https://tex.z-dn.net/?f=F%20%3D%20505.13%20N)
Yes it is better to pull the rope rather than push it