Answer:
D) ¼ as large
Explanation:
The magnitude of the electric force between two charges is
where
k is the Coulomb's constant
q1 and q2 are the magnitudes of the two charges
r is the separation between them
From the formula we see that the magnitude of the force is inversely proportional to the square of the distance.
In this problem, the distance is doubled:
So the new force will be
So the force will decrease to 1/4 of its original value.
Answer:
.
Explanation:
This problem can be solved using Newton's law of universal gravitation: ,
where F is the gravitational force between two masses and
,
is the distance between the masses (their center of mass), and
is the gravitational constant.
We know the weight of the astronout on the surface, with this we can find his mass. Letting be the weight on the surface:
,
,
,
since we now that we get that the mass is
.
Now we can use Newton's law of universal gravitation
,
where is the mass of the astronaut and
is the mass of the earth. From Newton's second law we know that
,
in this case the acceleration is the gravity so
, (<u>becarefull, gravity at this point is no longer</u>
<u>because we are not in the surface anymore</u>)
and this get us to
, where
is his new weight.
We need to remember that the mass of the earth is and its radius is
.
The total distance between the astronaut and the earth is
meters.
Now we can compute his weigh:
,
,
.
Answer:
Graph B represents correct graph for velocity time
Explanation:
Initially the object is dropped under the influence of constant force due to gravity
So here we can say its acceleration is constant due to gravity
hence here the velocity will increase downwards due to gravity and it is given as
now when it enters into the pool an upward force of same magnitude will act on it
so net force on it is
so here we will have no acceleration and it will move with constant speed after that
so here correct graph is
Graph B
Answer:
(A) 570 rad
(B) 10 s
(C) 12.5 rad/s²
Explanation:
The equations of motion for circular motions are used.
(A)
At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by
After this time, it decelerates through an angular displacement of 440 rad.
Total angular displacement = 130 + 440 rad = 570 rad
(B)
At the time the circuit breaker tips, the angular velocity is given by
This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:
Here, (the angular displacement during deceleration)
The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.
This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.
Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.
(C)
The deceleration is given by
The negative sign appears because it is a deceleration.