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ahrayia [7]
2 years ago
11

How do i do this is hard

Physics
1 answer:
Zolol [24]2 years ago
7 0

Answer:

I believe it would be E)none of the above.

Explanation:

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suppose that during any period of 1/4 second there is one instant at which the crests or troughs of component waves are exactly
ivanzaharov [21]
The correct answer for the question that is being presented above is this one: "(a)4." Suppose that during any period of 1/4 second there is one instant at which the crests or troughs of component waves are exactly in phase and maximum <span>reinforcement occurs, in 1 second, there will be 4 beats.</span>
8 0
3 years ago
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(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
(a) Calculate the acceleration due to gravity on the surface of the Sun.
Ira Lisetskai [31]
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

                      g=\frac{GM}{r^2}

 Here we need to find acceleration due to gravity of Sun,

                G = 6.67259 x 10⁻¹¹ N m²/kg²

    Mass of sun, M = 1.989 × 10³⁰ kg

    Radius of sun, r = 6.957 x 10⁸ m

Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

   Ratio of gravity = 274.21/9.81 = 27.95

   Weight = mg

  Factor of increase in weight = 27.95

8 0
3 years ago
N2O is the chemical formula of a covalent compound used in the production of whipping cream. The elements that make up this comp
vlabodo [156]
 N2O is nitrous oxide becuase of the nitrogen and oxygen

7 0
3 years ago
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If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the
hoa [83]

Answer:

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

4 0
3 years ago
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