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3 years ago

Electrical power companies sell electrical energy

Physics

1 answer:

mixer [17]3 years ago

4 0

Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

Q = 8.82 × 10⁶ J

Cost of electrical energy :

Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

Cost = $ 0.3675

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Calculate the density of the stone given that its mass is 250g

stiks02 [169]

Explanation:

density = mass divide volume

the volume must be there in order to answer the question.

8 0

3 years ago

CAN SOMEONE PLEASE HELP ME

Anna71 [15]

Answer:

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3 0

3 years ago

If the mass of a pendulum is increased (with other factors controller), would the period (time) decrease or stay the same?

Akimi4 [234]

The period of the pendulum depends only on the length from the pivot to the "center of mass". So if the string has no mass, then the amount of mass on the end doesn't make any difference.
But if the pendulum is suspended on, say, a chain with mass, then the more mass on the bottom, the lower the center of mass is, and the longer the period is.

8 0

4 years ago

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

mass of the block, = M
coefficient of static friction in section 1, = $\mu_s_1$
angle of inclination of the plane, = θ

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago

15 points. give me the method.

AveGali [126]

Answer:

$\boxed{{160 \: m(s)}^{ - 1} }$

Explanation:

$if \:the \: frequencies \: are \to \\ f_{1} = 640Hz \\ and \\f_{2} = 480Hz \: \\ but \: \boxed{v = f \gamma }: f = \frac{v}{ \gamma } \\ if \: \gamma_{1} - \gamma _{2} = 1 = \gamma \\ f_{1} - f_{2} = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 = \boxed{{160 \: m(s)}^{ - 1} }$

5 0

3 years ago