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jarptica [38.1K]
2 years ago
12

Plz answer this asap first answer gets brainliest and 100 points

Physics
2 answers:
Ratling [72]2 years ago
4 0

The answer is C. From the yolk to the white.

tankabanditka [31]2 years ago
4 0

ANSWER:

the answer is c my batfriend

~batmans wife dun dun dun...aka ~serenitybella

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How are different wavelengths of light affected by entering a prism?
NeX [460]
What happens in the prism stays in the prism. When the light emerges, it has the same frequency and wavelength as when it entered. The prism permanently alters nothing but the angle.

<span>Reference https://www.physicsforums.com/threads/how-does-a-prism-affect-wavelength.489768/ by caseytrimble

Sorry this probably doesn't help


</span>
3 0
3 years ago
What is the answer?
Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE    =   KE

mgh = 1/2 mv^2     divide both sides of the equation by 'm'

gh    = 1/2 v^2             multiply both sides by 2

2 gh = v^2               take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

3 0
2 years ago
When you park your car uphill next to a curb, the right front wheel should be:?
Cloud [144]
<span> When headed uphill at a </span>curb<span>, turn the front </span>wheels<span> away from the </span>curb<span> and let </span>your vehicle<span> roll backwards slowly until the rear part of the front </span>wheel<span> rests against the </span>curb<span> using it as a block.</span>
6 0
2 years ago
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
Help will mark brain list
Hatshy [7]
A jagged line represents a resistor .
3 0
2 years ago
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