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3 years ago

2. A wave front has the form of a A. circle. B. straight line. C. surface of a sphere. D. sine wave.

2 answers:

6 0

The answer to this question would be depending on what wave you are referring to. If it is a water wave, the wave would be propagating as a circle so option C.
Hope this helps!!

UNO [17]3 years ago

3 0

The answer to this question would be depending on what wave you are referring to. If it is a water wave, the wave would be propagating as a circle so option A. However, when you refer to light or sound waves it should be spherical so it should be option C.

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Current is constant at all points in a parallel circuit.<br><br> True<br> False

Sergio [31]

This is false. Current is the speed of the charge, 1 amp of current is 1 coulomb per second. So you can imagine the current of a circuit as the current of a river. In a parallel circuit, the river breaks into two separate streams. Some of the water goes down one river, some goes down the other. However, the total amount of water/coulombs never changes. This means that some of the total current will go down one river, and one the other. However, with less coulombs now the current will decrease.

Long story short, since there are two paths, the charge will split and depending on the resistance of each parallel stream a different amount of charge will go down each branch.

5 0

3 years ago

A laser emits light at power 6.20 mW and wavelength 633 nm. The laser beam is focused (narrowed) until its diameter matches the

Ipatiy [6.2K]

Answer:

a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa , c) F = 20.6 10⁻¹² N

Explanation:

a) The intensity defined as the energy per unit area

S = U / A

Area of a circle is

W = 6.2 mw = 6.2 10-3 W

R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m

A = π R2

A = π (1,080 10⁻⁶)²

A = 3.66 10 -12 m²

S = 6.2 10-3 / 3.66 10-12

S = 1.69 10⁹ W / m²

b) The radiation pressure

P = 1 / c (dU / dt) / A

S = (dU / dt) / A

P = S / c

P = 1.69 10 9 / 3. 108

P = 5.63 Pa

c) the definition of pressure is force over area

P = F / A

F = P A

F = 5.63 3.66 10⁻¹²

F = 20.6 10⁻¹² N

d) for this we use Newton's second law

F = ma

a = F / m

8 0

3 years ago

8) 8 resistor is connected in **parallel** with a 6 resistor. This

lyudmila [28]

Answer:

The equivalent resistance of the combination is 3.42 ohms.

Explanation:

We have,

8 ohms resistor is connected in parallel with a 6 ohms resistor. It is required to find the equivalent resistance of this combination.

For a parallel combination, the equivalent resistance is given by :

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

Plugging the values of R₁ and R₂, we get :

$\dfrac{1}{R_{eq}}=\dfrac{1}{8}+\dfrac{1}{6}\\\\R_{eq}=3.42\ \Omega$

So, the equivalent resistance of the combination is 3.42 ohms.

4 0

3 years ago

If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F ab

bazaltina [42]

Answer:

F_ab = 260.17 N

Explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

Point A = 0.2 k

Point B = 0.4 i + 0.2 j

Point C = 0.2 j + 0.4 k

Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)

= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

vec(CD) = D - C = 0.4 i - 0.2 j

unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

= [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

(M)_c = ( r_ac x vec(F_ab) ) :

$\left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]$

(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD) is as follows:

(M)_cd = unit(CD) . (M)_c :

(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)

(M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

57 = F_ab*(sqrt(30) / 25)

F_ab = 260.17 N

7 0

4 years ago

A certain spring stretches 15 cm when a load of 50.0 N is suspended from it. How much will the spring stretch if 78N is suspende

andriy [413]

Answer:

23.4 cm

Explanation:

15 : 50 = x : 78

15*78 = 50x

x = 15*78/50 = 23.4 cm

8 0

3 years ago