Answer:
29.6 kg
Explanation:
Centripetal force = mass × centripetal acceleration
F = m v² / r
88.0 N = m (2.50 m/s)² / 2.10 m
m = 29.6 kg
An energy crisis<span> is any significant (bottleneck; logistics; or price rise) in the supply of energy resources to an economy. In popular literature, it often refers to one of the energy sources used at a certain time and place, in particular those that supply national electricity grids or those used as fuel in vehicles.</span>
Answer:
D
Explanation:
1.) The reaction is at dynamic equilibrium.
A: Nitrogen and hydrogen combine at the same rate that ammonia breaks down.
2.) Which statement about the reaction is necessarily correct?
A: Both calcium carbonate and sodium carbonate are being produced.
3.) Both calcium carbonate and sodium carbonate are being produced.
A: The reaction is reversible.
4.) What is the fastest motion that can be measured in any frame of reference?
A: 300,000 km/s
5.) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north.
Why are we able to use these motions to describe the motion relative to the ground?
A: The people are moving much slower than the speed of light so the ground acts as a frame of reference.
Answer:
a). M = 20.392 kg
b). am = 0.56 
(block), aM = 0.28 (bucket)
Explanation:
a). We got N = mg cos θ,
f = 
= 
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
M = 20.392 kg
b). 
.............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
.....................(iv)
We got, N = mg cos θ
∴ 
................(v)
Mg - 2T = M
(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),
![$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bg%5Cleft%5B%5Cfrac%7BM%7D%7B2%7D-m%20%5Csin%20%5Ctheta-%5Cmu_K%20m%20%5Ccos%20%5Ctheta%5Cright%5D%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%2Bm%29%7D%3Da_m%24)
![$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7B9.8%5Cleft%5B%5Cfrac%7B20.392%7D%7B2%7D-10%28%5Csin%2030%2B0.5%20%5Ccos%2030%29%5Cright%5D%7D%7B%28%5Cfrac%7B20.392%7D%7B4%7D%2B10%29%7D%3Da_m%24)
Using equation (iv), we get,
Answer:
maximum speed 56 km/h
Explanation:
To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose
sin 10 = Nx / N
cos 10 = Ny / N
Ny = N cos 10
Nx = N sin 10
Let's develop Newton's equations on each axis
X axis
We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve
Nx + fr = m a
a = v2 / r
fr = mu N
N sin10 + mu N = m v² / r
N (sin10 + mu) = m v² / r
Y Axis
Ny -W = 0
N cos 10 = mg
Let's solve these two equations,
(mg / cos 10) (sin 10 + mu) = m v² / r
g (tan 10 + μ / cos 10) = v² / r
v² = r g (tan 10 + μ / cos 10)
They ask us for the maximum speed
v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)
v² = 294 (0.8364)
v = √(245.9)
v = 15.68 m / s
Let's reduce this to km / h
v = 15.68 m / s (1 km / 1000m) (3600s / 1h)
v = 56.45 km / h
This is the maximum speed so you don't skid
