Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:
c
Explanation:
cuz its informing the length of 5 and weight on 20N
Answer:19ohms
Explanation:
equivalent resistance=5+2+12
equivalent resistance=19ohms
Explanation:
V=u+at
where,
v=final speed
u=initial speed,(starting speed)
a=acceleration
t=time
- v=u+at = 6=2+a*2
6=2+2a
2a=6-2
2a=4
a=4/2 = 2
a =2
2. to find time taken
v=u+at
25=5*2t
2t=25-5
2t=20
t=20/2
t=10sec
3. finding final speed
v=u+at
v=4+10*2
=4+20
v=24m/sec
5.v=u+at
=5+8*10
=5+80
V=85m/sev
6. v=u+at
8=u+4*2
8=u+8
U=8/8
u=1
these are your missing values
Answer:
t = 12,105.96 sec
Explanation:
Given data:
weight of spacecraft is 2000 kg
circular orbit distance to saturn = 180 km
specific impulse = 300 sec
saturn orbit around the sun R_2 = 1.43 *10^9 km
earth orbit around the sun R_1= 149.6 * 10^ 6 km
time required for the mission is given as t
![t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cmu_sun%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28R_1%20%2B%20R_2%29%5D%5E%7B3%2F2%7D)
where
is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2.![t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%201.32712%20x%2010%5E%7B20%7D%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28149.6%20%2A%2010%5E%206%20%2B1.43%20%2A10%5E9%20%29%5D%5E%7B3%2F2%7D)
t = 12,105.96 sec