Answer:
The magnitude of the tension on the ends of the clothesline is 41.85 N.
Explanation:
Given that,
Poles = 2
Distance = 16 m
Mass = 3 kg
Sags distance = 3 m
We need to calculate the angle made with vertical by mass
Using formula of angle
We need to calculate the magnitude of the tension on the ends of the clothesline
Using formula of tension
Put the value into the formula
Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.
Answer:
The magnetic force points in the positive z-direction, which corresponds to the upward direction.
Option 2 is correct, the force points in the upwards direction.
Explanation:
The magnetic force on any charge is given as the cross product of qv and B
F = qv × B
where q = charge on the ball thrown = +q (Since it is positively charged)
v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)
B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)
F = qv × B = (+qvî) × (Bj)
F =
| î j k |
| qv 0 0|
| 0 B 0
F = i(0 - 0) - j(0 - 0) + k(qvB - 0)
F = (qvB)k N
The force is in the z-direction.
We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.
Hope this Helps!!!
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
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Answer:
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Explanation:
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?
It decreases in speed on its way down and increases in speed on its way down.
it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center
.It increases in speed on his way down because its under the influence of gravity
from newton's equation of motion we can check by
using V^2=u^2+2as
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Answer:
Tangential speed or Rotational speed
