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3 years ago

Which BEST describes what happens when light reaches opaque materials?

Physics

2 answers:

STatiana [176]3 years ago

8 0

Answer:B

Explanation:No light passes through l did it on usa testprep snd l also searched it up .hope this helped

kati45 [8]3 years ago

4 0

B) no light pass through

Explanation:

An opaque material is a material that does not allow light to pass through. Opaque materials are the most common materials, they do not allow light to pass through.

All opaque materials will cast shadows along their path when there is a screen.
Transparent materials allows light to pass through for example glasses.
Translucent materials allows light to go through partially.

Learn more:

Transmission of light brainly.com/question/8032392

#learnwithBrainly

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Two hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20 × 10−3 m2

Novosadov [1.4K]

Answer:

The charge flows through a point in the circuit during the change is 0.044 C.

Explanation:

Given that,

Number of turns in the copper wire, N = 200

Area of cross section, $A=1.2\times 10^{-3}\ m^2$

Resistance of the circuit, R = 118 ohms

If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.

We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :

$\epsilon=\dfrac{d\phi}{dt}$

Using Ohm's law :

$\epsilon=IR$

$IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}$

Electric current is equal to the rate of change of electric charge. So,

$dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C$

So, the charge flows through a point in the circuit during the change is 0.044 C.

4 0

3 years ago

A train travelled from city a to city b. For the first 14hours,it travelled at an average speed of 120km/h. For the 10 hours, it

Nataly [62]

Explanation:

distance covered in the first part of journey : 120×14 = 1680km

distance covered in second part of journey :

90×10 = 900 km

total distance traveled : 1680+900 = 2580km.

total time taken to complete the journey : 14+10= 24 hrs.

therefore, the average speed of the journey = 2580km/24hr = 107.5 km/h.

hope this helps you.

4 0

3 years ago

A 0.23-f capacitor is desired. What area must the plates have if they are to be separated by a 3.8-mm air gap?

tia_tia [17]

The area of the plates must have is(A)= 9.91×10⁷ m²

<h3 /><h3>How can we calculate the value of a area of a capacitor?</h3>

To calculate the the value of a area of the plates of a capacitor, we are using the formula,

C= $\frac{\epsilon_0 A}{d}$

Or, A= $\frac{C\times d}{\epsilon_0}$

Here we are given,

C= The desired capacitance of a capacitor.

= 0.23F

d=distance of separation between the plates.

=3.8mm= 0.0038m.

$\epsilon_0$ = permittivity of the vacuum.

=8.854×10⁻¹²F/m

We have to calculate the area of the plates must have = A m².

Now we put the known values in the above equation, we can get

A= $\frac{C\times d}{\epsilon_0}$

Or, A= $\frac{0.23\times 0.0038}{8.854\times 10^{-12}}$

Or, A= 9.91×10⁷ m²

From the above calculation, we can conclude that the area of the plates must have is(A)= 9.91×10⁷m²

Learn more about Capacitor:

brainly.com/question/13578522

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5 0

2 years ago

If a 1,000-kg car is traveling at 20.0 m/s when it reaches the bottom of a circular hill of radius 8.00 m, what normal force doe

scoray [572]

Answer:

i don't know this one.

Explanation:

6 0

3 years ago

An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 5 N, F2 = 8 N, and F3 = 5

DIA [1.3K]

The figure is missing: find it in attachment.

a) (3i - 5j) N

First of all, we have to write each force as a vector with a direction:

- Force F1 points downward, so along the negative y-direction, so we can write it as

$F_1 = -5 j$

- Force F2 points to the right, so along the positive x-direction, so we can write it as

$F_2 = +8 i$

- Force F3 points to the left, so along the negative x-direction, so we can write it as

$F_3 = -5 i$

Now we can write the net force, by adding the three vectors and separating the x-component from the y-component:

$F=F_1+F_2+F_3 = -5j+8i-5i = (8-5)i+(-5)j=(3i-5j)N$

b) 5.8 N

The magnitude of the net force can be calculated by applying Pythagorean's theorem on the components of the net force:

$F=\sqrt{F_x^2+F_y^2}$

where

$F_x = 3 N$ is the x-component

$F_y = -5 N$ is the y-component

Substituting into the equation,

$F=\sqrt{(3)^2+(-5)^2}=5.8 N$

c) $-59.0^{\circ}$

The angle of the net force, measured with respect to the positive x-axis (counterclockwise), can be calculated by using the formula

$\theta=tan^{-1} (\frac{F_y}{F_x})$

where

$F_x = 3 N$ is the x-component

$F_y = -5 N$ is the y-component

Substituting into the equation, we find

$\theta = tan^{-1} (\frac{-5}{3})=-59.0^{\circ}$

d) $0.84 m/s^2$

The acceleration can be found by using Newton's second law:

$F=ma$

where

F is the net force on an object

m is its mass

a is the acceleration

For the object in the problem, we know

F = 5.8 N

m = 6.9 kg

Solving the equation for a, we find the magnitude of the acceleration:

$a=\frac{F}{m}=\frac{5.8}{6.9}=0.84 m/s^2$

4 0

3 years ago