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Blizzard [7]
2 years ago
10

A superman cyclist rode a bike uphill at 20 miles/hour for two hours. To sustain this constant speed the cyclist was exerting 50

pounds. (a) How much work has the cyclist done during the trip?
Physics
1 answer:
NeX [460]2 years ago
5 0

Answer:

1.056 x 10⁷ lb-ft

Explanation:

v = Speed of the bike = 20 mph

t = time of travel = 2 h

d = distance traveled by cyclist

Distance traveled by cyclist is given as

d = v t

d = (20) (2)

d = 40 miles

We know that, 1 mile = 5280 ft

d = 40 (5280) ft

d = 211200 ft

F = force applied by cyclist = 50 lb

W = work done by cyclist

Work done by cyclist is given as

W = F d

W = (50) (211200)

W = 1.056 x 10⁷ lb-ft

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A car is driving east at 120. km/h from Toronto to Ottawa. The distance between the two cities is 425.5 km, how long will it tak
Ket [755]

Answer:

The time it will take for the driver to reach Ottawa is 3 hours 32 minutes and 45 seconds

Explanation:

The given parameters are;

Speed of the car = 120 km/h

Distance from Toronto to Ottawa = 425.5 km

The formula for speed is given as follows;

Speed = Distance/Time

Therefore, to find the time duration it takes from Toronto to Ottawa, we have;

Time duration = Distance from Toronto to Ottawa/(Speed of the car)

The time duration = 425.5/120 = 3.54583 hours = 212.75 min = 12765 seconds

The time it takes from Toronto to Ottawa while driving at 425.5 km/h = 12765 seconds.

4 0
2 years ago
Consider a vacuum-filled parallel plate capacitor (no dielectric material between the plates). What is the ratio of conduction c
Lorico [155]

Answer: 1798

Explanation:

Given that there is no dielectric material between the plates,

the permittivity of free space = 9 × 10^-12 f/m

The frequency F = 10 MHz

The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.

Please find the attached file for the solution

8 0
2 years ago
A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in
blsea [12.9K]

Answer:

b) It is impossible to tell without knowing the masses.

Explanation:

The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by

\Delta T= \frac{Q}{m C_s}

where

Q is the amount of heat

m is the mass of the substance

Cs is the specific heat capacity of the substance

In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.

4 0
2 years ago
Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard
podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0
2 years ago
. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
vichka [17]
Transverse wave as the wave is going up and down no compressions
3 0
2 years ago
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