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Fantom [35]
3 years ago
15

If a 12 V battery is connected to a light bulb with a resistance of 2 ohms, how much current will flow through the light bulb?

Physics
1 answer:
Free_Kalibri [48]3 years ago
4 0

Answer:

6amp

Explanation:

From Ohm's Law

V = IR

Where V = 12volts, R = 2ohms

12volts = I × 2ohms

I = 12/2

I = 6amp

The current flowing through the light bulb is 6amp

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A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. A) cm/h B) cm/min C) m/h
Crazy boy [7]
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
7 0
3 years ago
A hypothetical planet has a mass of 1.66 times that of Earth, but the same radius. What is gravitiy near its surface?
baherus [9]

Answer: 16.22 m/s^2

Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so

((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2

6 0
3 years ago
Read 2 more answers
How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
3 years ago
(b) The distance of mass from mass A if there is no gravitational force acted on C
shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ =  6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}

By plugging in the values and removing like terms, we get;

\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}}  = \dfrac{2 \times 0.05}{x^2}

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0
3 years ago
Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
laila [671]

Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

7 0
3 years ago
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