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kozerog [31]
3 years ago
6

Could someone please help me

Physics
1 answer:
BaLLatris [955]3 years ago
3 0

Answer:

the size of the shadow will be smaller due to smaller hands

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Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds. ​
Andreas93 [3]
  • initial velocity=u=15m/s
  • Final velocity=v=75m/s
  • Time=t=5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}

\\ \sf\longmapsto Acceleration=\dfrac{60}{5}

\\ \sf\longmapsto Acceleration=12m/s^2

7 0
2 years ago
Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one
kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0
3 years ago
The time T in seconds for a pendulum of length L feet to make one swing is given by Upper T=2\pi \sqrt((L)/(36)). How long is a
Andreyy89

Answer:

3.6ft

Explanation:

Using= 2*π*sqrt(L/32)

To solve for L, first move 2*n over:

T/(2*π) = sqrt(L/32)

Next,eliminate the square root by squaring both sides

(T/(2*π))2 = L/32

or

T2/(4π2) = L/32

Lastly, multiply both sides by 32 to yield:

32T2/(4π2) = L

and simplify:

8T²/π²= L

Hence, L(T) = 8T²/π²

But T = 2.1

Pi= 3.14

8(2.1)²/3.14²

35.28/9.85

= 3.6feet

5 0
2 years ago
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow
allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

l = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

l = l1 + l2 + 2√l1l2cos(∅)

where

l1 = 2.25 mW

l2 = 2.025 mW

∅ = π

so we substitute

l = 2.25 + 2.025 - 2√(2.25 × 2.025)

l = 4.275 - 4.26907

l = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer  

5 0
2 years ago
On earth which of the following are being investigated to contain fusion reactions
Vera_Pavlovna [14]

Answer:

A. Inertial Confinement and B. Magnetic Confinement

3 0
2 years ago
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