Answer:
525 kg.m/s
Explanation:
★ Momentum = Mass× Velocity
→ P = (7.5 × 70) kg.m/s
→ P = (75 × 7) kg.m/s
→ <u>P</u><u> </u><u>=</u><u> </u><u>5</u><u>2</u><u>5</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>
Answer:
The minimum frequency required to ionize the photon is 111.31 × Hertz
Given:
Energy = 378
To find:
Minimum frequency of light required to ionize magnesium = ?
Formula used:
The energy of photon of light is given by,
E = h v
Where E = Energy of magnesium
h = planks constant
v = minimum frequency of photon
Solution:
The energy of photon of light is given by,
E = h v
Where E = Energy of magnesium
h = planks constant
v = minimum frequency of photon
738 × = 6.63 × × v
v = 111.31 × Hertz
The minimum frequency required to ionize the photon is 111.31 × Hertz
Answer:
The frequency of the wheel is the number of revolutions per second:
f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz
And now we can calculate the angular speed, which is given by:
\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.
Explanation:
Answer:
The scrum master can suggest that:
Postpone the work on system integration for a while until the train is in sustain and enhance the section of road map of implementation
Explanation:
System Demo:
It is such a software that is used to measure the progress of Agile Release Train.
As the teams didn't expect that the integration of system will be a difficult task and they ultimately failed to accomplish the goal of integration so that scrum master asked them to stop working on the system integration until the Agile Release Train's sustainability and to improve and optimize the implementation of road map.
Answer:
Explanation:
Assuming this problem: "Part (a) of the figure attached shows a non-conducting rod with a uniformly distributed charge +Q. The rod forms a half circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P (Figure b), by what factor is the magnitude of the electric field at P multiplied?"
On this case the charge density is given by this formula:
assuming a half circle
We can find the force acting on the x axis with this:
We can cnvert the integral using the symmetrical property:
And we can find the electric field like this:
And the electric field just by the charge is given by:
And if we find the ratio for the two electrical fields we got: