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3 years ago

Calculate the electric current that flows through a electric heater of power 1000watt when collected with 220 volt ac supply

Physics

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

4.545 A

Explanation:

From the question,

Using,

P = VI....................... Equation 1

Where P = power of the heater, V = Voltage supply, I = Electric current that flows through the heater

make I the subject of the equation above

I = P/V.................. Equation 2

Given: P = 1000 watt, V = 220 volt.

Substitute these values into equation 2

I = 1000/220

I = 4.545 A

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The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately 0.199 rad/s

Second question:

The telescope should accelerates away by approximately F = 0.0005·m

Third question:

The pulling force between the Earth and the satellite

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ 0.199 rad/s

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

The force of attraction between the Earth and the telescope which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

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3 years ago

What type of objects cannot pull to magmets

boyakko [2]

Answer:

brass, copper, zinc and aluminum, wood, glass, and plastic

Explanation:

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3 years ago

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A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

JulsSmile [24]

$m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}$

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3 years ago

Harry and Cassie wanted to conduct an experiment to see how many licks it takes to get to the center of a tootsie roll pop. Belo

umka2103 [35]

Hey there!

The answer is that The data was not reliable because there was not any repetition.

An experiment should be completed multiple times to be sure of accurate results.

A screenshot is attached.

Hope this helps! :)

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4 years ago

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While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for

Nimfa-mama [501]

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

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3 years ago