Answer:
He traveled 9km
Explanation:
To do this problem you need to use the equation which is Speed= distance/time and this problem gives you the speed which is 18 km/h and it gives you the time 1/2 hour so you write the equation 18= d/ 1/2 which his distance is 9km
Answer:
82.25 moles of He
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 10 L
Mass of He = 0.329 Kg
Temperature (T) = 28.0 °C
Molar mass of He = 4 g/mol
Mole of He =?
Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
0.329 Kg = 0.329 Kg × 1000 g / 1 Kg
0.329 Kg = 329 g
Thus, 0.329 Kg is equivalent to 329 g.
Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:
Mass of He = 329 g
Molar mass of He = 4 g/mol
Mole of He =?
Mole = mass / molar mass
Mole of He = 329 / 4
Mole of He = 82.25 moles
Therefore, there are 82.25 moles of He in the tank.
I believe that it is electric field
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.
hope this helps :)
