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nadezda [96]
3 years ago
15

¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?

Physics
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

s=0+\frac{1}{2}(5)(10)^2=250 m

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The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

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t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0
3 years ago
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]

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El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0
3 years ago
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Answer:

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Explanation:

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7 0
3 years ago
An 8.5-l tire is filled with 0.55 mol of gas at a temperature of 305 k. what is the pressure of the gas in the tire?
const2013 [10]

Answer:

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Explanation:

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n = 0.55 mol is the number of moles of the gas

R = 0.0821 atm L / K mol is the gas constant

T = 305 K is the temperature of the gas

By re-arranging the equation and substituting the numbers in, we find:

p=\frac{nRT}{V}=\frac{(0.55 mol)(0.0821 Latm/Kmol)(305 K)}{8.5 L}=1.62 atm

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