Answer:
A. 30.7cm
B. ![1.7*10^{-10}C](https://tex.z-dn.net/?f=1.7%2A10%5E%7B-10%7DC)
C. The electric field is directed away from the point of charge
Explanation:
A.
![because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm](https://tex.z-dn.net/?f=because%2C%20E%3D%5Cfrac%7Bv%7D%7Bd%7D%20%5C%5C%5C%5Cd%3D%20%5Cfrac%7B4.98%7D%7B16.2%7D%5C%5C%5C%5C%20d%20%3D%200.307m%5C%5C%5C%5Cd%20%3D%2030.7%20cm)
B.
Considering Gauss's law
![EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C](https://tex.z-dn.net/?f=EA%20%3D%20%5Cfrac%7BQ%7D%7Be%7D%5C%5C%5C%5C%20where%2C%20e%20%3D%20pertittivity.%20space%3D%208.85%2A%2010%5E%7B-12%7D%20Fm%5E%7B-1%7D%20%5C%5C%5C%5CA%20%3D%20surface.%20area.%20with.radius%200.307m%5C%5CQ%3D%20eEA%20%3D%20%288.85%2A10%5E%7B-12%7D%29%2816.2%29%284%5Cpi%29%280.307%29%5E%7B2%7D%5C%5C%5C%5C%3D%201.7%2A10%5E%7B-10%7DC)
C. The electric field directed away from the point of charge when the charge is positive.
Answer:
a)The parachutist in the air for 12.63 seconds.
b)The parachutist falls from a height of 293 meter.
Explanation:
Vertical motion of parachutist:
Initial speed, u = 0m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 50
v = 31.32 m/s
Time taken for this
31.32 = 0 + 9.81 x t
t = 3.19 s
After 50m we have
Initial speed, u = 31.32m/s
Acceleration, a = -2 m/s²
Final speed , v = 3 m/s
We have equation of motion, v² = u² + 2as
Substituting
3² = 31.32² - 2 x 2 x s
s = 243 m
Time taken for this
3 = 31.32 - 2 x t
t = 9.44 s
a) Total time = 3.19 + 9.44 = 12.63 s
The parachutist in the air for 12.63 seconds.
b) Total height = 50 + 243 = 293 m
The parachutist falls from a height of 293 meter.
This question was apprently selected from the "Sneaky Questions" category.
The store is 3 km from his home, and he walks there with a speed of 6 km/hr. So it takes him (3 km) / (6 km/hr) = 1/2 hour to get to the store.
That's 30 minutes. So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.
a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.
Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>
Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>
There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).
For circular motion.
Centripetal acceleration = mv²/r = mω²r
Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m
m = mass = 175g = 0.175kg.
Angular speed, ω = Angle covered / time
= 2 revolutions / 1 second
= 2 * 2π radians / 1 second
= 4π radians / second
Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5 Use a calculator
≈13.817 m/s²
The magnitude of acceleration ≈13.817 m/s² and it is directed towards the center of rotation.
Tension in the string = m*a
= 0.175*13.817
= 2.418 N
The design of rutherfords experiment show what he was trying to find out is by detect charged particles. He shot positively charged alpha particles at foil containing gold atoms, what they did showed what was in them neutrons etc Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.