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3 years ago

100 POINTS AND BRAINLIEST! What were the Magdeburg Hemispheres?

2 answers:

amm18123 years ago

8 0

The Magdeburg hemispheres are a pair of large copper hemispheres, with mating rims. They were used to demonstrate the power of atmospheric pressure. When the rims were sealed with grease and the air was pumped out, the sphere contained a vacuum and could not be pulled apart by teams of horses.

pogonyaev3 years ago

7 0

Answer:

Magdeburg hemispheres are two half-spheres of equal size. Placing them together traps air between them. This air is merely trapped, and not compressed, so the pressure inside is the same as the pressure of the atmosphere outside the spheres. The spheres thus pull apart with nearly no resistance.

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Any 3 differences between telescope and microscope

Keith_Richards [23]

An instrument used to observe or imagine very small object using an optical mangifier
mirco cell.
Telescope is a magnifer of distance object

4 0

3 years ago

What is the volume of an object that has Mass=100g and Density= 745g/mL.

lana66690 [7]

Answer:0.13ml

Explanation:

6 0

3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers

I am in plato cuzz wauup what u doing

Irina-Kira [14]

Yoohoo dudio!!! whatz up??!!

MARK AS THE BRAINLIEST SWEETY!!

8 0

3 years ago

A rope is vibrating so as to form the standing wave pattern shown. How many antinodes are present in the rope?a- 5b- 4c- 8d- 10e

denpristay [2]

c) 8

Explanation

When you shake a rope, the particles in the rope move up and down, and the wave moves forward or away from the source of energy. The rope moves in a direction that is perpendicular

Nodes are the places where the rope doesn't move at all; antinodes occur where the motion is greatest.

Step 1

let's check the graph:

a) Nodes

and Antinodes

so, in the ripe there are 8 antinodes

therefore, the answer is

c) 8

4 0

1 year ago