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3 years ago

What is the power p supplied to a resistor whose resistance is r when it is known that it has a voltage Î´v across it?

1 answer:

4 0

... power p supplied to a resistor whose resistance is r when it is known that it has a voltage Î´v across ... supplied to a resistor whose resistance

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A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu

tekilochka [14]

Answer: $0.43\ rad/s$

Explanation:

Given

Mass of child $m=34\ kg$

speed of child is $v=2.8\ m/s$

Moment of inertia of merry go round is $I=510\ kg.m^2$

radius $r=2.31\ m$

Conserving the angular momentum

$\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s$

7 0

3 years ago

Multiply the following numbers, using scientific notation and the correct amount of significant digits. 1.003 m⋅3.09 = _____ 3.0

lozanna [386]

Answer

correct answer is 3.10

Explanation:

in this question we have to multiply two numbers 1.003 and 3.09.

1.003 has 4 significant digits and 3.09 has 3 significant digits so answer must have 3 significant digits.

$1.003\times 3.09=3.09927$

$1.003\times 3.09=3.10$

Hope it will help you

3 0

3 years ago

Read 2 more answers

A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.

Ivenika [448]

Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

8 0

3 years ago

Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no

Cloud [144]

Answer:

Propels in the opposite direction

Explanation:

7 0

3 years ago

A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by

Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0

3 years ago