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3 years ago

9

Quanto tempo deve ficar ligado um ferro eletrico de 1000 w para que tenha o mesmo consumo de energia que um chuveiro de 4400 w q

ue fica ligado 10 minutos

1 answer:

Cerrena [4.2K]3 years ago

8 0

Answer:

Thus, the time for the first lamp is 44 minutes.

Explanation:

Power of first lamp, P' = 1000 W

Power of second lamp, P'' = 4400 W

time for second lamp, t'' = 10 minutes

Let the time for first lamp is t'.

As the energy is same, so,

P' x t' = P'' x t''

1000 x t' = 4400 x 10

t' = 44 minutes

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A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

3 0

3 years ago

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of

Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0

3 years ago

Did I do these questions correctly?

Elena L [17]

For number 11, you should say that there is more pollution and fossil fuels being burned.

8 0

3 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effect

Fantom [35]

Answer:

$I_s=5.8A$

Explanation:

Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:

$P_p=P_s$

So:

$V_p*I_p=V_s*I_s$

Where:

$V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil$

Solving for $I_s$

$I_s=\frac{V_p*I_p}{V_s}$

Replacing the data provided:

$I_s=\frac{110*29}{550} =5.8A$

4 0

3 years ago