What is the difference between an open and closed circuit?

A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2

saw5 [17]

Answer:

Half as large.

Explanation:

Using Newton's law of universal gravitation, if the mass of the planet is <em>M</em> and of the Moons 1 and 2 is <em>m</em>, them the force exerted by the planet on them will be:

$F_1=\frac{GMm}{r}$

$F_2=\frac{GMm}{2r}$

Which clearly shows that the force that the planet exerts on the Moon 2 is half the force it exerts on the Moon 1.

7 0

3 years ago

A mercury in barometer shows a height of 70cm.What height would be shown in the barometer at the same place if water density 1.0

Pie

Answer:

P = d g h pressure due to density of liquid of height h

d1 g h1 = d2 g h2 since P1 equals P2

d1 h1 = d2 h2 and the density of Hg is 13.6 times that of water

h2 = (d1 / d2) * h1 = 13.6 * 70 cm = 952 cm or 9.52 m

8 0

2 years ago

A 4kg block and a 2kg block can move on horizontal frictionless surface. The blocks are accelerated by a +12-N force that pushes

Stolb23 [73]

Answer:

a) -4 N

b) +4 N

Explanation:

Draw a free body diagram for each block.

For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.

For the small block, there is 1 force, F pushing to the right.

There are also weight and normal forces in the vertical direction, but we can ignore those.

Sum of forces on the large block in the x direction:

∑F = ma

12 − F = 4a

Sum of forces on the small block in the x direction:

∑F = ma

F = 2a

2F = 4a

Substitute:

12 − F = 2F

12 = 3F

F = 4

The small block pushes on the large block 4 N to the left (-4 N).

The large block pushes on the small block 4 N to the right (+4 N).

4 0

3 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

mrs_skeptik [129]

Answer:

surface charge density on each sphere is $440 \times 10^{-9}$ C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = $(\frac{1}{4\pi\epsilon })\times (\frac{Q_{1} }{R^{2} } )$ .................1

put here value

660000 = $9 \times 10^9 \times \frac{Q1}{0.12^2}$

Q1 = 1056 × $10^{-9}$

and

here field inside a conductor is zero so that electric potential ( V ) is constant

$\frac{Q{1} }{R} = \frac{Q{2} }{r}$ ..................2

so Q2 will be

Q2 = $\frac{5}{12} \times 1056 \times 10^{-9}$

Q2 = $440 \times 10^{-9}$ C

6 0

3 years ago

Question 6

Margaret [11]

Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

<h2>Formation of caves</h2>

The type of rocks that once existed in these locations are limestone and dolomite whereas the pH of the nearby groundwater is slightly acidic which is responsible for the formation of caves. Caves are formed by the dissolution of limestone due to acid rain.

Rainwater reacts with carbon dioxide from the air and percolates through the soil, which turns into a weak acid. This slowly dissolves out the limestone which become turn to form caves so we can conclude that Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

Learn more about caves here: brainly.com/question/7965722

Learn more: brainly.com/question/26111031

4 0

1 year ago