Answer:
Q=1005 J
t= 0.67 sec
Explanation:
Lets take condition of room is 1 atm and 25°C.
Heat capacity ,c = 21 J /K.mol
If we assume that air is ideal gas that
P V = n R T



V= 107250 L
At STP number of moles given as

V=22.4 L at S.T.P.

n=4787.94 moles
n= 4.784 Kmoles
So heat required to raise 10°C temperature
Q = n x c x ΔT
Q = 4.78794 x 21 x 10
Q=1004.64 J
Time t
t= Q/P
P= 1.5 KW
t = 1.004.64 /1.5
t= 0.66 sec
Answer:
Explanation:
Given
Force P is acting upward
C is vertical contact Force
W is the weight of the crate
As P is unable to move the Block therefore Normal reaction keeps on acting on block
thus we can say that
P-W+C=0
P=W-C
<span>Hello!
We have the following data:
</span>
Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)
Formula of the consumption of electric energy:

Solving:




How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

<span>Product of extremes equals product of means
</span>




I hope this helps! =)
<span>
</span>
Answer:
= 1,386 m / s
Explanation:
Rocket propulsion is a moment process that described by the expression
- v₀ =
ln (M₀ / Mf)
Where v are the velocities, final, initial and relative and M the masses
The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)
We consider that the rocket starts from rest (v₀ = 0)
At the time of burning half of the fuel the mass ratio is that the current mass is
M = 2.5 Mf
- 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2
= 1,386 m / s
At r = 2R> R The expression for the electric field will be given by: (2R)^2*E=kQ. Where, k=(9*10^9)N.m/C^2, Q=(8*10^-10)C and R=0.025m. So substituting and clearing, we have that the magnitude of the electric field will be: E=(9*10^9)*(8*10^-10)/((2*0.025)^2)=2880 N / C.