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SVETLANKA909090 [29]
3 years ago
5

Un automóvil se mueve 80 m en 40 s con una velocidad constante. ¿Cuál es la velocidad del automóvil? Ayuda plissssssssssssssssss

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Physics
1 answer:
Alexxandr [17]3 years ago
5 0

Answer:

Velocidad = 2 m/s

Explanation:

Dados los siguientes datos;

Distancia = 80 m

Tiempo = 40 s

Para encontrar la velocidad del automóvil;

La velocidad se puede definir como la tasa de cambio en el desplazamiento (distancia) con el tiempo.

La velocidad es una cantidad vectorial y, como tal, tiene magnitud y dirección.

Matemáticamente, la velocidad viene dada por la ecuación;

Velocidad = \frac {distancia} {tiempo}

Sustituyendo en la fórmula, tenemos;

Velocidad = 80/40

Velocidad = 2 m/s

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A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w
FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

As the proton is coming at rest after travelling 0.200 m to the right,  vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C

⇒ E = 1.04*10⁻¹ N/C

5 0
3 years ago
39. Draw a complete free body diagram of a 40 kg plastic crate at rest on a wooden table (us=0.7). The applied force to the righ
Leya [2.2K]

In order to draw the free body diagram, first let's calculate the friction force acting on the crate:

\begin{gathered} F_f=N\cdot\mu \\ F_f=40\cdot9.8\cdot0.7 \\ F_f=274.4\text{ N} \end{gathered}

Since the friction force is greater than the force applied, the crate will not move, and the friction force will be equal to the force applied.

The weight force is equal to 40 * 9.8 = 392 N.

So, drawing the diagram, we have:

4 0
1 year ago
What is the heat capacity of an object at 25.5∘C that absorbs 45 kJ of heat and is heated to 28.2∘C?
sergey [27]

Answer:

16.6 kJ/°C

Explanation:

given,

Amount of heat absorbed = 45 kJ

initial temperature, T₁ = 25.5°C

final temperature, T₂ = 28.2°C

change in temperature = T₂ - T₁

                                       = 28.2 - 25.5  = 2.7° C

Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}

Heat\ capacity = \dfrac{45\ kJ}{2.7}

Heat\ capacity = 16.6\ kJ/^0C

Heat capacity of the object is equal to 16.6 kJ/°C

4 0
3 years ago
Read 2 more answers
Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​
Mkey [24]

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F  = (1200)(0) = 0

4 0
2 years ago
A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0
2 years ago
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