Answer:
5.56 A
Explanation:
From the question,
Q = it.............. Equation 1
Where Q = charges, i = current, t = time.
Make i the subject of the equation
i = Q/t.............. Equation 2
Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds
Substitite these values into equation 2
i = 200/(0.6×60)
i = 5.56 A
Hence the magnitude of the current flowing through the circuit is 5.56 A
Answer:
a) 1321.45 N
b) 1321.45 N
c) 2.66 m/s^2
d) 2.21*10^-22 m/s^2
Explanation:
Hello!
First of all, we need to remember the gravitational law:
Were
G = 6.67428*10^-11 N(m/kg)^2
m1 and m2 are the masses of the objects
r is the distance between the objects.
In the present case
m1 = earth's mass = 5.9742*10^24 kg
m2 = 497 kg
r = 1.92 earth radii = 1.92 * (6378140 m) = 1.2246*10^7 m
Replacing all these values on the gravitational law, we get:
F = 1321.45 N
a) and b)
Both bodies will feel a force with the same magnitude 1321.45 N but directed in opposite directions.
The acceleration can be calculated dividing the force by the mass of the object
c)
a_satellite = F/m_satellite = ( 1321.45 N)/(497 kg)
a_satellite = 2.66 m/s^2
d)
a_earth = F/earth's mass = (1321.45 N)/( 5.9742*10^24 kg)
a_earth = 2.21*10^-22 m/s^2
24 knots is the maximum wind velocity for a 30 crosswind if the maximum crosswind component for the airplane is 12 knots.
What is crosswind component?
Although you should always prepare to land the aircraft into the wind, in practice the wind only occasionally blows directly down the runway. You will typically be landing with at least a small amount of crosswind. There is a limit to how much direct crosswind an airplane can land in. Only a portion of the wind is operating as a direct crosswind when it is blowing at a shallow (acute) angle to the runway.
The following are the variables needed to solve for the crosswind component:
Runway and wind direction
Wind speed
To learn more about crosswind component click the given link
brainly.com/question/27960498
#SPJ4
Answer:
a) 24.692 m/s
b) 19.4 m
Explanation:
To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:
We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:
a) The velocity (V2) is:
Substituting the known values we can get the velocity at the out:
Atmospheric pressure= 101000 Pa
Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3
b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:
We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:
Substituting the known values, the height (Z3) is:
Z3=Maximum Height=19.376=19.4 m
Answer:
77.08 C
Explanation:
= mass of the water = 500 g = 0.5 kg
= specific heat of water = 4186 J/(kg °C)
= Rate of change of temperature = 3 °C /min = (3/60 ) °C /s = 0.05 °C /s
= thermal conductivity of glass = 0.84
= Area of the element = 0.090 m²
= thickness of the element = 1.5 mm = 0.0015 m
= Temperature inside = 75 °C
= Temperature outside = ?
Using conservation of energy
Heat gained by water = Heat transferred through glass