1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec
Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s
In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:




Answer:
a) 600 meters
b) between 0 and 10 seconds, and between 30 and 40 seconds.
c) the average of the magnitude of the velocity function is 15 m/s
Explanation:
a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:
![Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m](https://tex.z-dn.net/?f=Area%20%5C%2C%5C%2CTrapezoid%3D%28%5Cleft%5BB%2Bb%5D%5C%2C%28H%2F2%29%5C%5Cdisplacement%3D%20%5Cleft%5B%2840-0%29%2B%2830-10%29%5Cright%5D%20%5C%2C%2820%2F2%29%3D600%5C%2C%5C%2Cm)
b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.
Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.
c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s
Answer:
John Felix Anthony Cena is an American professional wrestler, actor, and television presenter. He is currently signed to WWE. Born and raised in West Newbury, Massachusetts, Cena moved to California in 1998 to pursue a career as a bodybuilder.
Explanation: