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anastassius [24]
2 years ago
13

Which diagram best shows the field lines around two bar magnets that repel each other?

Physics
2 answers:
Lina20 [59]2 years ago
7 0
The answer is c. i’m 90% on it
maria [59]2 years ago
5 0

Answer:

c im pretty sure

Explanation:

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Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

f = 4.76 \times 10^{14} Hz

Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

8 0
2 years ago
The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength
VARVARA [1.3K]
So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6
8 0
3 years ago
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms
inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) 6.81*10^5 N/c

b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

E_r_m_s = sqrt*(S / CE_o)

Where

C = 3.00 * 10^-^8 m/s;

E_o = 8.85*10^-^1^2 C^2/N.m^2;

S = 1.23*10^9 W/m^2

Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

E_r_m_s= 6.81 *10^5N/c

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

B_r_m_s = E_r_m_s / C;

= 6.81*10^5 N/c / 3*10^8m/s;

B_r_m_s = 2.27*10^3 T

8 0
3 years ago
Read 2 more answers
A paper airplane used to study wing designs is an example of a ?
kow [346]
Model I'm guessing. Coz that's using an object to explain
7 0
3 years ago
Read 2 more answers
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³
Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, \rho_o = 800 kg/m³

density of water, \rho_w = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0
3 years ago
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