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3 years ago

Who remebers paper poppers

Physics

1 answer:

Tcecarenko [31]3 years ago

4 0

YEAH OMG THEY WERE SO OMG OMG OMG no I acc don’t

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The eiffel tower is a steel structure whose height increases by 19.5 cm when the temperature changes from −8 to +42 °c. what is

Readme [11.4K]

The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m

6 0

4 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

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3 years ago

Un contratista colocará azulejo importado en la pared de una cocina, que mide 6 metros de ancho y 4 metros de alto.

Olin [163]

No se neta Srry pero si hablo espanol

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4 years ago

What cities (more than 1) has many fronts

Arisa [49]

Chattanooga - Chatype, London - Johnston, Berlin - BMF Change, Milan - Milano City, Eindhoven - Eindhoven, Stockholm - Stockholm Type, Minneapolis, and St. Paul - Twin.

6 0

3 years ago

I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT

ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR = E- V

IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

$R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms$

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0

3 years ago