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leonid [27]
3 years ago
5

Is it true or not true

Physics
1 answer:
maria [59]3 years ago
6 0

Answer:

true

Explanation:

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Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe
solniwko [45]
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr 
</span><span>Θ = arctan(v0² / gr) </span>

<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>

<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
4 0
3 years ago
If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in
Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, N_P = 50 turns

number of turns in the secondary winding, N_S = 10 turns

the secondary load resistance, R_S = 250 Ω

Determine the turns ratio;

K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5

Now, determine the reflected resistance in the primary winding;

\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0
3 years ago
El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es
Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión (P), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

P = \frac{F}{A}

Donde:

F - Fuerza motriz, medida en newtons.

A - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón (A), medido en metros cuadrados, es determinado por:

A=\frac{\pi}{4}\cdot D^{2}

Donde D es el diámetro del pistón, medido en metros.

Si D = 0.32\,m y F =17,640\,N, entonces la presión neumática es:

A = \frac{\pi}{4}\cdot (0.32\,m)^{2}

A \approx 0.080\,m^{2}

P = \frac{17,640\,N}{0.080\,m^{2}}

P = 220,500\,Pa

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

8 0
3 years ago
A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l
hram777 [196]

Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

v_i = 20.86 m/s

now the final speed of the train when it crossed the crossing

v_f = 15 km/h

v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

Now the time to cross that junction is given as

v_f - v_i = at

4.17 - 20.86 = a(-0.44)

t = 37.6 s

4 0
3 years ago
What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

5 0
3 years ago
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