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3 years ago

WILL MARK BRAINLIEST IF GOTTEN RIGHT!

Physics

1 answer:

castortr0y [4]3 years ago

8 0

Answer:

Angle Angle Similarity Postulate

Explanation:

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A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

4 years ago

figure 2 shows a charged ball of mass m = 1.0 g and charage q = -24*10^-8 c suspended by massless string in the presence of a un

Vlad [161]

Answer:

E = 307667 N/C

Explanation:

Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.

This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:

0.0098 N = T * cos(37)

then T = 0.0098/cos(37) N = 0.01227 N

Knowing the tension's magnitude, we can find its horizontal component:

T * sin(37) = 0.007384 N

and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:

0.007384 N = E * 2.4 * 10^(-8) C

Then E = 0.007384/2.4 * 10^(-8) N/C

E = 307667 N/C

8 0

3 years ago

What is the name of the line drawn perpendicular to the surface where a light ray strikes?

AnnyKZ [126]

It is the normal line

7 0

3 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)