Answer:
Explanation:
The torque of a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation of the system
is the angle between the direction of the force and d
In this problem, we have:
, the force
, the distance of application of the force from the centre (0,0)
, the angle between the direction of the force and a
Therefore, the torque is
Three 40w lamp for 6 hours
Answer:
that's your 1st cousin which is your mom or dad's sibling's child sooo when you're that close and that related i would say that that would be the reason why.
Explanation:
hoped that helped!!
Answer:
Explanation:
We know that,
Neptune is 4.5×10^9 km from the sun
And given that,
Earth is 1.5×10^8km from sun
Then,
Let P be the orbital period and
Let a be the semi-major axis
Using Keplers third law
Then, the relation between the orbital period and the semi major axis is
P² ∝ a³
Then,
P² = ka³
P²/a³ = k
So,
P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³
Period of earth P(earth) =1year
Semi major axis of earth is
a(earth) = 1.5×10^8km
The semi major axis of Neptune is
a (Neptune) = 4.5×10^9km
So,
P(E)²/a(E)³ = P(N)² / a(N)³
1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³
Cross multiply
P(N)² = (4.5×10^9)³ / (1.5×10^8)³
P(N)² = 27000
P(N) =√27000
P(N) = 164.32years
The period of Neptune is 164.32years
<em><u>One</u></em>
Givens
- delta B = 0.20 T/s
- A = 0.07 m^2
- R = 3.5 ohms
Formula
Φ = ΔB*A
e = Φ
Solution (first part)
e = 0.2 * 0.07
e = 0.014 emf
Solution (second part)
i = e/R
i = 0.014 / 3.5
i = 4 * 10^-3
i = 4 ma
Answer
A
<em><u>Two</u></em>
Givens
N = 200 turns
Φ = 30 degrees
Delta B = 0.45 T/s
phi = 30 degrees
r = 0.06 meters
Formula
e = -N * delta B * A * Cos(phi)
Solution
e = -200 * 0.45 (pi r^2) * Cos(30)
e = - 200 * 0.45 * (3.14 * 0.06^2) * cos(30)
e = 0.881 emf
Answer
A
