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Illusion [34]
3 years ago
7

Car 1 drives 35 mph to the east, and car 2 drives 50 mph to the west. From the frame of reference of car 1, what is the velocity

of car 2?
A. 85 mph west
B. 15 mph east
C. 85 mph east
D. 15 mph west
Physics
2 answers:
Sergio039 [100]3 years ago
4 0
The distance between the two cars is increasing at the rate of 85 mph.

A passenger in Car-1 says that he is at rest in his own frame of reference,
and Car-2 is moving away from him at 85 mph, toward the west.
Radda [10]3 years ago
3 0

Answer:

A. 85 mph west

Explanation:

Velocity of car 1 is given as

v_1 = 35 mph towards East

\vec v_1 = 35(-\hat i) mph

Velocity of car 2 is given as

v_2 = 50 mph towards West

\vec v_2 = 50 (\hat i) mph

Now we need to find the speed of car 2 with respect to the frame of car 1

so it is given as

\vec v_{21} = \vec v_2 - \vec v_1

now we will have

\vec v_{21} = 50 (\hat i) - 35 (-\hat i) = 85 \hat i

so the relative speed will be 85 mph towards West

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Manufacturing Plant Power A manufacturing plant uses 2.36 kW of electric power provided by a 50.0 Hz ac generator with an rms vo
Pie

Answer:

(a) Z = 48.3 Ω

(b) cos ∅ = 0.455

(c) Irms = 10.35 A

(d) C = 74.02 μF

(e) Irms = 4.44 A

Explanation:

Power (P) = 2.36 kW

Frequency (f) = 50 Hz

RMS Voltage (Vrms) = 500 V

Resistance (R) = 22 Ω

Inductive Reactance (XL) = 43 Ω

(a) to calculate the total impedance, use the formula:

Z = √(R² + XL²)

   = √((22)² + (43)²)

   = √2333

Z = 48.3 Ω

(b) To calculate the plant's power factor, we will use the formula:

cos ∅ = R/Z

          = 22/48.3

cos ∅ = 0.455

(c) To calculate the RMS current used by the plant, divide the RMS voltage value by the impedance of the plant.

Irms = Vrms/Z

        = 500/48.3

Irms = 10.35 A

(d) For the power factor to become unity, the inductive reactance must be equal to the capacitive reactance i.e. Xc = XL

Xc = XL

1/(2πfC) = XL

1/(2πfXL) = C

C = 1/(2π*50*43)

   = 7.402 x 10⁻⁵

C = 74.02 μF

(e) P = Vrms*Irms*cos∅

    Irms = P/Vrms*cos∅

             = 2.22 x 10³/500*1

    Irms = 4.44 A

4 0
3 years ago
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
Umnica [9.8K]

Answer:

Explanation:

Given

Charge Q is uniformly spread over large non-conducting Elastic sheet      

Electric field due to non-conducting Elastic sheet

E=\frac{\sigma }{2\epsilon }

where \sigma =surface charge density=\frac{q}{d^2}

E=\frac{\frac{q}{d^2}}{2\epsilon }

for side 2d Electric Field is given by

E'=\frac{\frac{q}{2d^2}}{2\epsilon }

E'=\frac{1}{4}\times \frac{\frac{q}{d^2}}{2\epsilon }

E'=\frac{E}{4}

8 0
3 years ago
Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy
Semenov [28]

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

     Q_lost = - Q_absorbed

     

The latent heat is

      Q_absorbed = m L

The heat given by the body

      Q_lost = M c_{e} ΔT

       

where m is the mass of sweat and M is the mass of the body

       m L = M c_{e} ΔT

        m = M c_{e} ΔT / L

let's replace

        m = 90  3.500  1.8 / 2.42 10⁶

 

        m = 0.2343 kg

reduced to grams

        m = 0.2342 kg (1000g / 1kg)

        m = 23.42 g

 the correct answer is c) 23 g

8 0
3 years ago
The aim of the newton's first law experiment ​
Semmy [17]

Answer:

Application of Newton's first law of motion

A body in motion will continue in motion in a straight line unless acted upon by an outside force.

Explanation:

4 0
2 years ago
Para el siguiente conjunto de medidas, calcule EL ERROR RELATIVO PORCENTUAL: 1.34 m, 1.35 m, 1.37 m y 1.36 m
ivanzaharov [21]

Answer:

Ver explicacion abajo

Explanation:

En este caso para poder calcular el error relativo porcentual, es necesario calcular primero el error absoluto, que se calcula de la siguiente forma:

Error absoluto = Resultado exacto - aproximación

Sin embargo, no tenemos el resultado exacto de las medidas, pero podriamos conocerlo tomando el promedio de estas medidas y este es el que tomaremos como el verdadero resultado de las medidas:

Promedio de medidas = 1.34 + 1.35 + 1.37 + 1.36 / 4

Promedio de medidas = 1.355 m

Ya que tenemos el promedio, podemos calcular el error absoluto de cada medida y luego el error relativo porcentual:

Ea1 = 1.355 - 1.34 = 0.015

Ea2 = 1.355 - 1.35 = 0.005

Ea3 = 1.37 - 1.355 = 0.015

Ea4 = 1.36 - 1.355 = 0.005

Ya que tenemos los 4 errores absolutos, es posible calcular el porcentual:

%error relativo = (Error absoluto / resultado exacto) * 100

Aplicando la expresión con cada uno de los valores tenemos:

%Er1 = (0.015/1.34) * 100 = 1.12%

%Er2 = (0.005/1.35) * 100 = 0.37%

%Er3 = (0.015/1.37) * 100 = 1.09%

%Er4 = (0.005/1.36) * 100 = 0.37%

Espero que te sirva.

4 0
3 years ago
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