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Viefleur [7K]
3 years ago
14

Imagine you are the teacher of a science class. A student brings in a

Physics
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer: Running out of energy means that the magicians will break

Explanation:

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The siren on an ambulance is emitting a sound whose frequency is 2450 Hz. The speed of sound is 343 m/s. If the ambulance is sta
lyudmila [28]

Answer:

The wavelength is 0.14 m

Explanation:

Given that,

Frequency = 2450 Hz

Speed of sound = 343 m/s

We need to calculate the wavelength

Using formula of wavelength

v=  f\lambda

Where, v = speed of sound

f = frequency

Put the value into the formula

\lambda=\dfrac{v}{f}

\lambda=\dfrac{343}{2450}

\lambda=0.14\ m

Hence,  The wavelength is 0.14 m

5 0
4 years ago
Read 2 more answers
2. A small car has a mass of 890 kg. What is its weight? (Hint: remember the definition of weight).
Dafna1 [17]

What is the weight? The weight of a 890 kg car is:

8727.9185 newtons

or

0.981057067 tons force

or

31393.8261 ounces force

Hope that helps! I apologize if it does not answer your question!

I used this site: <u><em>https://www.sensorsone.com/mass-to-weight-calculator/</em></u>

3 0
3 years ago
#2
sladkih [1.3K]

Answer:

mass and distance

Explanation:

force is mass while motion can also be regard as distance or movement

3 0
3 years ago
How many times Father is Proxima Centauri from the sun then is earth from the sun answer choices about 269,000 times or about 0.
Bas_tet [7]

Answer:

269,000

Explanation:

The distance between Proxima Centauri and the Sun, d₁₁ = 4.246 light years

The distance between Earth and the Sun, d₂ = 1.58 × 10⁻⁵ light years

The ratio of the distances are;

d₁/d₂ = 4.246/(1.58 × 10⁻⁵) = 268734.177215 ≈ 269,000 times

Therefore, Proxima Centauri is approximately 269,000 times further from the Sun than the Earth is from the Sun

4 0
4 years ago
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

6 0
4 years ago
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