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gizmo_the_mogwai [7]
3 years ago
11

Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci

ty of car 2?
Physics
2 answers:
Aleksandr-060686 [28]3 years ago
5 0

Answer:

50 mph north

Explanation:

EleoNora [17]3 years ago
4 0
We subtract the velocity of car 1 from the velocity of car 2:
v=(30\ mph\ North)-(20\ mph\ South)
=(30\ mph\ North)+(20\ mph\ North)
=50\ mph\ North
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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
Nitella [24]

Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

7 0
2 years ago
A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Pachacha [2.7K]

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

6 0
3 years ago
A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor
Alex17521 [72]

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0
3 years ago
About 3% of the water on Earth is freshwater. Only about 40% of that freshwater is available for human use. Why is so much fresh
Pie
A) it is frozen i think
7 0
3 years ago
Read 2 more answers
Air pressure may be represented as a function of height above the surface of the Earth as shown below.. P(h) = P_0 e^-.00012h. .
Rufina [12.5K]

Answer. Second Option: .85p_o=p_o e^-.00012h


Solution:

P(h)=Po e^(-0.00012h)

Air pressure: P(h)

Height above the surface of the Earth (in meters): h

Air pressure at the sea level: Po

Height at which air pressure is 85% of the air pressure at sea level:

h=?, P(h)=85% Po

P(h)=(85/100) Po

P(h)=0.85 Po

Replacing P(h) by 0.85 Po in the formula above:

P(h)=Po e^(-0.00012h)

0.85 Po = Po e^(-0.00012h)

5 0
3 years ago
Read 2 more answers
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