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3 years ago

13

g Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an an

gle of 55.7 degrees with respect to the normal to the surface. What is the angle of reflection from the second surface

1 answer:

Mamont248 [21]3 years ago

5 0

Answer:

add the angles

HOPE you understand

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An 870 N firefighter, F_ff, stands on a ladder that is 8.00 m long and has a weight of 355 N, F_ladder. The weight of the ladder

castortr0y [4]

Answer:

Explanation:

Weight of the ladder is 355N

WL = 355N

The weight of the ladder acts at center of the ladder I.e at 4m from the bottom

Weight of firefighters is 870N

Wf = 870N

The fire fighter is at 6.3m from the bottom of the ladder.

N1 is the normal force exerted by the wall

N2 is the normal force exerted by the ground

Using Newton law

Check attachment

ΣFy = 0, since body is in equilibrium

N₂ - WL - Wf = 0

N₂ = WL + Wf

N₂ = 870 + 355

N₂ = 1225 N

This the normal force exerted by the ground on the wall.

Now to get N₁, let take moment about point A

So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.

ΣMa = 0

Clockwise moment is equal to anti-clockwise moment

Moment Is the produce of force and perpendicular distance.

M = F×r

So,

WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0

355•Cos50 × 4 + 870•Cos50 × 6.3 =

N₁•Sin50 × 8

1825.52 + 3523.12 = 6.13N₁

6.13N₁ = 5348.64

N₁ = 5348.64/6.13

N₁ = 872.54 N.

The normal force exerted by the wall is 872.54N

5 0

3 years ago

HELP ME PLEASE. i need to turn this in by tomorrow, also pls use the words above^ thank you i’ll mark u the brainliest.

Vlad [161]

Answer:

the corrdct answer is core

4 0

3 years ago

After three half-lives, one-ninth of an original radioactive parent isotope remains, and eight-ninths has decayed into the daugh

BartSMP [9]

Answer:

B. False

The true statement will be:

After three half-lives, one-eighth of an original radioactive parent isotope remains, and seven-eighth has decayed into daughter isotopes.

Explanation:

Half life means the time after which the amount of the given substance becomes half of its initial amount, by radiation.

Thus after each successive half life the material is halved or half of it is decayed, which leads to a general formula:

Remaining Amount = 1/2^n of parent isotope

where, n is the number of half lives passed. So, for three half-lives:

Remaining Amount = 1/2^3 of parent isotope

<u>Remaining Amount = 1/8 of parent isotope</u>

Thus, the decayed amount will be:

Decayed Amount = (1-1/8) of parent isotope

<u>Decayed Amount = 7/8 of parent isotope</u>

Thus, the given statement is <u>B.False</u>

8 0

3 years ago

Compare the momentums of a 50 kg dolphin swimming at 16.4 m/s and a 4100 kg elephant walking 0.20 m/s?

Mamont248 [21]

Their momentum is the same.
momentum of dolphin=50kg*16.4m/s=820kg*m/s
momentum of elephant=4100kg*0.20m/s=820kg*m/s

6 0

3 years ago

Read 2 more answers

A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to s

Komok [63]

1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
$2aS = v_f^2-v_i^2 = -v_i^2$
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
$a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2$

2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
$F=ma$
This force will be constant, and since m is always the same, then a is the same even in the second situation.

3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
$2aS=-v_i^2$
$S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m$

4 0

3 years ago