Answer:
The value is 
Explanation:
From the question we are told that
The mass of the bullet is 
The mass of the wood is 
The height attained by the combined mass is 
Generally according to the law of energy conservation
Here 
is the kinetic energy of the bullet before collision.
and 
is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as
![PE_m = [m_b + m_w] * g * h](https://tex.z-dn.net/?f=PE_m%20%20%3D%20%20%5Bm_b%20%20%2B%20m_w%5D%20%2A%20%20g%20%2A%20%20h)
So
![KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08](https://tex.z-dn.net/?f=KE_b%20%3DPE_c%20%20%20%3D%20%5B0.005%20%20%2B%200.90%5D%20%2A%209.8%20%2A0.08)
=>
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.
Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²
Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2
