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1 year ago

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Read each scenario and then answer the question. Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in

8 s. Scenario C: 200 J of work is done in 10 s. Which scenario uses the most power?.

2 answers:

Alexeev081 [22]1 year ago

4 0

The power in every scenario is the same. Power is the measurement of the energy transfer over time.

<h3>What is power?</h3>

Power is the measurement of the energy transfer over time. It can be calculated by the formula,

$P = \dfrac Wt$

Where,

$P$ - power

$W$ - work

$t$ - time

In scenario A,

$P = \dfrac {120 }6 = 20 \rm \ W$

In scenario B,

$P = \dfrac {160 }8 = 20 \rm \ W$

In scenario B,

$P = \dfrac {200 }{10} = 20 \rm \ W$

Therefore, the power in every scenario is the same.

Learn more about Power:

brainly.com/question/1618040

Andrej [43]1 year ago

3 0

Answer:

Read each scenario and then answer the question.

Scenario A: 120 J of work is done in 6 s.

Scenario B: 160 J of work is done in 8 s.

Scenario C: 200 J of work is done in 10 s.

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To solve this problem we will use the kinematic formula for the final velocity.

$V_f_x = V_0_x + a_xt$

The final speed is 0 at the moment the player stops.

The time until it stops is 1.3 s

The initial speed is 200 feet / s Note (check the speed units in the problem statement, 200ft / s is very much and 200ft / h is very small)

Then, we clear the formula.

$a_x = \frac{(Vfx-V0x)}{t}\\ a_x = \frac{(0-200)}{1.3}\\ a_x = -153.5 ft / s ^ 2$

Because the player is slowing down, the acceleration goes in the opposite direction to the player's movement, and that is why it is negative.

To answer part b) we use the following formula.

$Vf ^ 2 = Vo ^ 2 + 2a * (x_2 - x_1)\\\\ (x_2 - x_1)= \frac{(V_f ^ 2-V_0 ^ 2)}{2a}\\\\ (x_2 - x_1)= \frac{(0-200 ^ 2)}{- 2 * 153.5}\\\\ (x_2 - x_1)= 130.29 feet$

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2 years ago

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

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2 years ago

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andrew-mc [135]

Answer: Parietal

Explanation: The parietal lobe is where the primary somatosensory cortex is located. This cortex is where all tactile stimulation is processed in the brain and allows to you detect/feel someone scratching your back.

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2 years ago

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bazaltina [42]

Answer:

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Explanation:

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2 years ago

A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

$\epsilon = \frac{NBA}{t}$

Where,

$\epsilon =$ electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

$R = R_c + R_d$

And that the current can be expressed as function of charge and time, then

$I = \frac{q}{t}$

Equation Faraday's law and Ohm's law we have,

$V = \epsilon$

$IR = \frac{NBA}{t}$

$(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}$

Re-arrange for Magnetic Field B, we have

$B = \frac{q(R_c+R_d)}{NA}$

Our values are given as,

$R_c = 58.7\Omega$

$R_d = 45.5\Omega$

$N = 120$

$q = 3.53*10^{-5}C$

$A = 3.21cm^2 = 3.21*10^{-4}m^2$

Replacing,

$B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}$

$B = 0.095T$

Therefore the magnetic field in the System is 0.095T

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2 years ago