Answer:
Explanation:
Given:
Initial θ = 0 rad (from rest)
Final θ = 14.3 rad
Time, t = 5 s
B.
Angular velocity, w = dθ / dt
= (14.3 - 0)/5
= 2.86 rad/s
A.
Acceleration, ao = dw/dt
Initial angular velocity, wi = 0 rad/s (from rest)
Final angular velocity, wf = 2.86 rad/s
a = (2.86 - 0)/5
= 0.572 rad/s^2
Answer:
The mass goes down
Explanation:
Because mass is the quantity of matter contained in a substance And Volume is the space occupied by a substance. So when the volume is less the mass decreases
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)
=> 
Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>
Answer:
The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.
Explanation:
Let us call the mass of the balloon
and the mass of the basket
, then according to newton's second law:
,
where
is the upward acceleration, and
is the net propelling force (counts the gravitational force).
Also, the tension
in the rope is 79.8 N more than the basket's weight; therefore,

and this tension must equal


Combining equations (2) and (3) we get:

since
, we have

Putting this into equation (1) and substituting the numerical values of
and
, we get:


Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.