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olga nikolaevna [1]
3 years ago
6

Q1 what is the direction of current?​

Physics
1 answer:
Oduvanchick [21]3 years ago
4 0

Answer:

the external circuit is directed away from the positive terminal and toward the negative terminal of the battery

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What do you mean you need to be more specific
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3 years ago
Pls say this a doubt i have
hodyreva [135]
Question four bulbs A,B,C and D are connected in a circuit shown in the figure below, the letters X, Y and Z represent three switches. Which switch is used to operate switch A separately?
Answer: x
4 0
3 years ago
A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest
Ymorist [56]

Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

3 0
3 years ago
La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C
Lina20 [59]

Answer:

Frequency, f=4.41\times 10^{14}\ Hz

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, \lambda=6.8\times 10^{-7}\ m

Speed of light is, c=3\times 10^8\ m/s

We need to find the frequency of visible red light. It can be calculated using below relation.

c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz

So, the frequency of visible red light is 4.41\times 10^{14}\ Hz.

3 0
3 years ago
A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The
VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

\dfrac{1}{2}mu^2 = m g d sin \theta

u^2 = 2 g d sin30^0

u= \sqrt{2\times 9.8 \times 10 sin30^0}

u = 9.89 m/s

Using second law of motion  

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,  

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0
3 years ago
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