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forsale [732]
3 years ago
7

A string is stretched by two equal and opposite forces F newton each then the tension in sting is?

Physics
1 answer:
Arlecino [84]3 years ago
3 0

Answer: The tension in the string is zero

Explanation:

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Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci
ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})] (1)

Donde:

w - Ancho de la placa, en centímetros.

l - Longitud de la placa, en centímetros.

\alpha - Coeficiente de dilatación, en \frac{1}{^{\circ}C}.

T_{o} - Temperatura inicial, en grados Celsius.

T_{f} - Temperatura final, en grados Celsius.

Si sabemos que w = 65\,cm, l = 78\,cm, \alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}, T_{o} = 20\,^{\circ}C and T_{f} = 400\,^{\circ}C, entonces el área de la placa a la temperatura final:

A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]

A_{f} = 5102.752\,cm^{2}

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0
3 years ago
A car travels with an average speed of 22 m/s.what is this speed in km/s
Debora [2.8K]
It is .022 kilometers.
4 0
3 years ago
Read 2 more answers
Which of these is a good example of Newton's First Law of inertia
Eddi Din [679]

Answer: The correct option is A.

Explanation:

Inertia is a state of an object or body to maintain its state. It resists any change in its state.

Newton's first law of inertia: When an objects is in state of motion, it will remain its state of motion or if it is in state of rest then it will remain in rest unless it is acted upon by external motion.

In the given options, a ball sits motionless on the ground is good example of Newton's first law motion. No external force is acting in this case.

In options (B), (C) and (D) , the external force is acting.

Therefore, the correct option is A.

8 0
3 years ago
Read 2 more answers
Please help me with this​
Allushta [10]

Answer:

We define voltage as the amount of potential energy between two points on a circuit. One point has more charge than another. This difference in charge between the two points is called voltage.

3 0
2 years ago
The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =
dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar,  θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

                        ∑ Mₐ = Iα

Where,     Mₐ - the moment about A

                 α  - angular acceleration

                 I - moment of inertia of the rod AB

                       -mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha

                        \alpha=\frac{-3gcos\theta}{2l}

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

                          \vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

The acceleration at the center of the bar

                          \vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

                         (\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]

(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]

Comparing the coefficients of i

=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}

Comparing coefficients of j

(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}

Net force on x direction

F_{x}=(\vec{a_{G}})_{x}

substituting the values

F_{x}=1.5(14.58L+11.96)

Similarly net force on y direction

F_{y}=(\vec{a_{G}})_{y}+mg

               = 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

F=\sqrt{F_{x}^{2}+F_{y}^{2}}

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0
3 years ago
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