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3 years ago

Kepler's second law: as a planet moves around its orbit, it sweeps out areas in times.

Physics

1 answer:

natali 33 [55]3 years ago

6 0

Kepler's second law: as a planet moves around its orbit, it sweeps out EQUAL areas in EQUAL times.

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If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0

2 years ago

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

2 years ago

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2. We want

lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

$\Delta x=v_ot+\dfrac{1}{2}at^2$

t is time taken by the ball to hit the ground

$v_o$ is initial speed of the ball

So, the correct option is (A).

8 0

3 years ago

as your roller coaster climbs to the top of the steepest hill on its track when does the first car have the greatest potential e

pochemuha

...the potential energy that you build while going up the hill on the roller coaster could be let go as kinetic energy -- the energy of motion that takes you down the hill of the roller coaster.

4 0

3 years ago

The shape of an atom.is Spherical Square Pyramifal

zvonat [6]

Spherical because it’s more like clouds

7 0

3 years ago

Read 2 more answers

Kepler's second law: as a planet moves around its orbit, it sweeps out ______areas in ______ times.

Kepler's second law: as a planet moves around its orbit, it sweeps out areas in times.