One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
That is of course adding both tmepratures. There is one more that is a lil bit more complex
</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
Answer:
(a). Z = 54.54 ohm
(b). R = 36 ohm
(c). The circuit will be Capacitive.
Explanation:
Given data
I = 2.75 A
Voltage = 150 V
rad = 48.72°
(a). Impedance of the circuit is given by
Z = 54.54 ohm
(b). We know that resistance of the circuit is given by
Put the values of Z & in above formula we get
R = 36 ohm
(c). Since the phase angle is negative so the circuit will be Capacitive.
If you write down the formula for friction, you will get an answer.
Ff = u * N Where N is a push down force that an object experiences.
u (mu) is a constant and has no units
It may not be accelerating and still experience friction. A is not correct.
Color and Density will not affect the frictional force. B is not so.
Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.
The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s