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tigry1 [53]
2 years ago
5

A car accelerates from rest at 4.7m/s^2. How much time does it need to attain a speed of 6m/s?

Physics
1 answer:
lesantik [10]2 years ago
8 0

Answer:

100mm per second =25 +29

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After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
3 years ago
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for
xxTIMURxx [149]

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

4 0
2 years ago
Compare relative age with absolute age.
oee [108]

Answer & Explanation:

Relative age is the age of a rock layer (or the fossils it contains) compared to other layers. It can be determined by looking at the position of rock layers. Absolute age is the numeric age of a layer of rocks or fossils.

8 0
3 years ago
Lexi is balancing equations. She is finding one equation to be very difficult to balance. Which explains how to balance the equa
Rina8888 [55]

Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer

Equation is already balanced.

Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4

6 0
2 years ago
A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0
2 years ago
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