c. The Moon is positioned directly between the earth and the sun is the statement that does not result in an eclipse.
Explanation:
- The Sun is completely blocked in a solar eclipse because the Moon passes between Earth and the Sun.
- it is just the right distance away from Earth, the Moon can fully blocks the Sun's light from Earth's perspective
Absolutely ! If you have two vectors with equal magnitudes and opposite
directions, then one of them is the negative of the other. Their correct
vector sum is zero, and that's exactly the magnitude of the resultant vector.
(Think of fifty football players pulling on each end of the rope in a tug-of-war.
Their forces are equal in magnitude but opposite in sign, and the flag that
hangs from the middle of the rope goes nowhere, because the resultant
force on it is zero.)
This gross, messy explanation is completely applicable when you're totaling up
the x-components or the y-components.
Answer:
D
Explanation:
Ball A is a non positively charged non metal while ball B is metal ball.
Given: The ball B positive charge of small magnitude
To prove: Balls will attract each other
IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
Hence the correct answer will be D .
You are correct earth science is studied to predict planetery changes
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
