Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
Mobility
We can find the drift velocity replacing,
The electric field is given by,
Answer:
77J
Explanation:
Not really an explanation to this, I just had this lesson last year and remembered it.
Hope I helped! ☺
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction - The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get
and the direction is the same as the direction of the two forces.
Learn more about forces:
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Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
Answer:
The ball is dropped at a height of 9.71 m above the top of the window.
Explanation:
<u>Given:</u>
- Height of the window=1.5 m
- Time taken by ball to cover the window height=0.15
Now using equation of motion in one dimension we have
Let u be the velocity of the ball when it reaches the top of the window
then
Now u is the final velocity of the ball with respect to the top of the building
so let t be the time taken for it to reach the top of the window with this velocity
Let h be the height above the top of the window
