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3 years ago

10

As a ray of light becomes brighter, its corresponding light wave, shown here, will increase in amplitude, making it look MOST li

ke which wave?

2 answers:

grandymaker [24]3 years ago

6 0

Answer:

the answer is C

Explanation:

i did the usatestprep

andreev551 [17]3 years ago

5 0

Answer:

c usa test prep

Explanation:

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A bullet is shot downward out of a high building with an initial speed of 10 m/s. How fast is the bullet traveling after 5 secon

serious [3.7K]

Answer:

-59

Explanation:

6 0

3 years ago

A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the

lisabon 2012 [21]

Answer:

We are given x= bt +ct²

So

A. bxt= m

Because m/s*s= m

So b= m/s and c= m/s²

B.

x= bt-ct²

So at x=0 t=0

x=0 t= 2

We have

bt = ct² so t = b/c at x= 0

So b-2ct= 0

B. To find velocity we use

dx / dt = b - 2 Ct

C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

dv / dt = - 2C

3 0

4 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

3 years ago

Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, th

Yuliya22 [10]

Answer:

E_total = 3 N / A

Explanation:

The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.

E_total = E + E₁

the expression for the field of a point charge is

E₁ = k q₁ / r²

for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative

E_total = E -k q₁ / r₂

we substitute

0 = E - k q₁ / r²

q₁ = $\frac{E r^2}{k}$

let's calculate

q₁ = $\frac{4 \ 2^2}{9 \ 10^{-9}}$

q₁ = 1.78 10⁻⁹ C

now we can calculate the field for position x = 4 m

E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2

E_total = 3 N / A

8 0

3 years ago

Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind

IrinaVladis [17]

Hello!

Most ocean waves obtain their energy and motion from the wind.

Ocean waves are surface waves that move across the surface of the ocean. When wind touches the surface of the water, there is friction in the contact zone. This friction causes a drag effect, that makes wrinkles on the surface of the water. As the wrinkles get bigger, they transform into full-blown waves, and the taller the wave, the more energy it can extract from the wind, making them even bigger and to move longer distances.

Have a nice day!

3 0

3 years ago