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madreJ [45]
3 years ago
14

Si está probando u motor para un nuevo modelo de coche ; este es capaz de pasar de 0 a 100 km/h en 7,5 segundos . Si el coche ti

ene una masa de 550 kg cuál será la fuerza que realiza el motor ?
Physics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

La fuerza que realiza el motor es 2035 N.

Explanation:

Podemos encontrar la fuerza del motor usando la siguiente ecuación:

F = ma   (1)

En donde:

m: es la masa del coche = 550 kg

a: es la aceleración

Se puede calcular la aceleración mediante la siguiente ecuación cinemática:

v_{f} = v_{0} + at   (2)

En donde:

v_{f}: es la velocidad final del coche = 100 km/h

v_{0}: es la velocidad inicial del coche = 0

t: es el tiempo = 7,5 s

Resolviendo la ecuación (2) para "a" tenemos:

a = \frac{v_{f} - v_{0}}{t} = \frac{100\frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s} - 0}{7,5 s} = 3,70 m/s^{2}

Entonces, la fuerza es:

F = ma = 550 kg*3,70 m/s^{2} = 2035 N

Por lo tanto, la fuerza que realiza el motor es 2035 N.

Espero que te sea de utilidad!

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2. A person lifts 200kg seven times over the course of 11.8s. If they displaced the weight 2.2m up each time, how much power did
Aneli [31]

Answer:

<em>The person delivered a power of 2,558 Watt</em>

Explanation:

<u>Work and Power</u>

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being  the force vector and  the displacement vector, the work is calculated as:

W=\vec F\cdot \vec s

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred per unit of time. In the SI, the unit of power is the watt, equivalent to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

If the person lifts a mass of m=200 Kg, then exerts a force equal to its weight:

F = m.g = 200*9.8 = 1,960

F = 1,960 N

The work done when lifting the weight 7 times by a distance of s=2.2 m is:

W = 7*1,960*2.2=30,184

W = 30,184 J

Finally, the power delivered in t=11.8 seconds is:

\displaystyle P=\frac {30,184}{11.8}

P = 2,558 Watt

The person delivered a power of 2,558 Watt

4 0
3 years ago
neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.
Vikentia [17]

As per Kepler's third law we know that

\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}

now here we know that

T_1 = year of Neptune

T_2 = year of Earth

R_1 = distance of Neptune from Sun

R_2 = Distance of Earth from Sun

so now we will have

\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}

T_1^2 = 27000

T_1 = 164.3 years

so length of year of Neptune is 164.3 years

6 0
3 years ago
Help it’s due tomorrow
Sati [7]

Answer:

A., the variables have a direct relationship.

Explanation:

As K rises, L rises.

It's not B. because one isn't rising as the other is lowering.

It's not C. because undefined would be a vertical line.

4 0
3 years ago
A student claims that any object in motion must experience a force that keeps it in motion. Do you agree or disagree? Explain yo
frosja888 [35]

Answer:

I disagree

Explanation:

I think the students claim is wrong because according to Newton's First Law an object that is in motion stays in motion unless acted upon by an unbalanced force. Which makes the students claim wrong because a object doesn't require another force to keep it moving.

5 0
3 years ago
Read 2 more answers
A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to
Xelga [282]

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

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Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

8 0
3 years ago
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