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sveta [45]
3 years ago
9

A pair of fuzzy dice is hanging by a string from your rearview mirror. while you are accelerating from a stoplight to 28 m/s in

6.0 s, what angle does the string make with the vertical? (assume your acceleration is constant.)
Physics
1 answer:
Lerok [7]3 years ago
4 0
We can use the formula of motion in physics (2nd law od newton) in this problem:
x direction: Fsin ∅ = ma 
y direction: Fcos ∅ -mg = 0
∅ is equal to sin ∅ / cos ∅  or x/y
tan ∅ = ma / mg = a /g

Applying acceleration formula:
v = vo + at ; 28 = 0 + 6a ; a = 4.67 m/s^2
∅ = tan-1 (a/g) = tan-1 (4.67/9.81) = <span>25.4 degrees.</span>
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Suppose your vehicle is moving 12m/s as it crosses the 7m line. It took 6 seconds to get to the 7m line. What was its accelerati
Inga [223]

Answer:

3.61 m/s²

Explanation:

Given:

v = 12 m/s

x = 7 m

x₀ = 0 m

t = 6 s

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a

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7 m = 0 m + (12 m/s) (6 s) - ½ a (6 s)²

a = 3.61 m/s²

8 0
3 years ago
A point source of light is 1.24 m below the surface of a pool. What is the diameter of the circle of light that a person above t
Ksivusya [100]

Answer:

D = 2.828 m

Explanation:

given,

distance of source of light = 1.24 m below surface of pool

refractive index of the water  = n₁ = 1.33

refractive index of air = n₂ = 1

refraction angle be = 90°

let C be the critical angle

Radius = d tan C

d is the depth of the source

Using Snell's law

n₁ sin C = n₂ sin R

1.33 x sin C = 1 x  sin 90°

sin C = \dfrac{1}{1.33}

C = sin^{-1}(0.752)

C = 48.75°

hence,

R = 1.24 x tan 48.75°

R = 1.414 m

Diameter = 2 x R

D = 2 x 1.414

D = 2.828 m

5 0
3 years ago
Satellites can focus on specific latitudes using:
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B- east west orbits
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3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
3 years ago
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