The work done to lift a 10kg object to a height of 1m above the ground is

Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What i

frutty [35]

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

$v = \sqrt{\frac{F}{\mu} }$

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ = m / l

where m = mass of the string

l = length of the string

This implies that:

$v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }$

Let us make mass, m, the subject of the formula:

$v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}$

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

$m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg$

5 0

2 years ago

In a parallel circuit the current A stays in one path B splits and goes through two components C makes one circle

ArbitrLikvidat [17]

Answer:

B splits and goes through two components

Explanation:

- A series circuit is a circuit in which the components are all connected along the same branch: as a result, the current flowing through the components is the same, while the sum of the potential differences across each component is equal to the emf of the battery

- A parallel circuit is a circuit consisting of separate branches, so that each branch has a potential difference equal to the emf of the battery. As a result, in such a circuit the current in the circuit splits and goes through the different branches/components.

So, the correct answer is

B splits and goes through two components

8 0

3 years ago

What are the equations of angular acceleration, angular velocity and theta?.

soldier1979 [14.2K]

Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t

3 0

2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago

Questions are in the picture ^ (please answer separately)

Klio2033 [76]

A: The battery is a store of internal energy (shown as chemical energy). The energy is transferred through the wires to the lamp, which then transfers the energy to the surroundings as light. These are the useful energy transfers - we use electric lamps to light up our rooms.

B: In the case of the light bulb the 95J of energy transferred as heat is wasted energy as it is not useful because the purpose of the device is to produce light.

SORRY I ONLY HAVE ANSWERS FOR A AND B

5 0

2 years ago

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