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Anastaziya [24]
3 years ago
12

How much work in joules must be done to stop a 920-kg car traveling at 90 km/h?

Physics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

Mass of the car, m = 920 kg

The car is travelling at a speed of, u = 90 km/h = 25 m/s

We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

W=\Delta K

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

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Cual es la fuerza electrica sobre el electrón (-1.6 x 10¹⁹c) de un atomo de hidrógeno ejercida por el protón (1.6 x 10¹⁹c)? Supó
kkurt [141]

Answer:

La  fuerza eléctrica es -8.2*10⁻⁸ N

Explanation:

El enunciado correcto es: <em>¿Cuál es la fuerza eléctrica sobre el electrón (-1.6 x 10⁻¹⁹c) de un átomo de hidrógeno ejercida por el protón (1.6 x 10⁻¹⁹c)? Supóngase que la distancia entre el electrón y el protón es de 5.3 x 10⁻¹¹ m</em>

Entre dos o más cargas aparece una fuerza denominada fuerza eléctrica. Su valor depende del valor de las cargas y de la distancia que las separa, mientras que su signo depende del signo de cada carga. Las cargas del mismo signo se repelen entre sí, mientras que las de distinto signo se atraen.

La fuerza eléctrica con la que se atraen o repelen dos cargas puntuales en reposo es directamente proporcional al producto de las mismas e inversamente proporcional al cuadrado de la distancia que las separa:

F=K*\frac{q1*q2}{d^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. En el Sistema Internacional (S.I.) se mide en Newtons (N).
  • q1 y q2 son lo valores de las dos cargas puntuales. En el S.I. se miden en Culombios (C).
  • d es el valor de la distancia que las separa. En el S.I. se mide en metros (m).
  • K es una constante de proporcionalidad llamada constante de la ley de Coulomb. Depende del medio en el que se encuentren las cargas. Para el vacío K tiene un valor aproximadamente de 9*10⁹ \frac{N*m^{2} }{C^{2} }.

En este caso:

  • F=?
  • K= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q1= -1.6*10⁻¹⁹ C
  • q2= 1.6*10⁻¹⁹ C
  • d= 5.3*10⁻¹¹ m

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{(-1.6*10^{19} C)*(1.6*10^{19} C)}{(5.3*10^{-11} )^{2} }

Resolviendo:

F= -8.2*10⁻⁸ N

<u><em>La  fuerza eléctrica es -8.2*10⁻⁸ N</em></u>

6 0
3 years ago
The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6
Leona [35]

Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

Velocity peak to peak is 6.28 cm/s

Explanation:

Speed of 2 particles = 5.3 cm/s

Speed of 4 particles = 1.4 cm/s

Speed of 6 particles = 7.14 cm/s

Speed of 8 particles = 1.52 cm/s

Speed of 2 particles = 7.68 cm/s

v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s

Average velocity is 3.93 cm/s

v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s

Root mean square velocity is 4.79 cm/s

v_p=7.68-1.4=6.28\ cm/s

Velocity peak to peak is 6.28 cm/s

8 0
3 years ago
Calculate the force necessary to accelerate a 10 kg table from<br> O m/s to 4 m/s in 2 seconds.
yanalaym [24]

Answer:

a= v/t

a = 4/2

a = 2 m/s^2

And F = M a

F = 10 × 2

F = 20 N

6 0
3 years ago
What state of matter has a define volume but no definite shape ?
mariarad [96]

Liquid has no definite volume or shape.

6 0
3 years ago
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Find the work w1 done on the block by the force of magnitude f1 = 65.0 n as the block moves from xi = -3.00 cm to xf = 3.00 cm .
inn [45]
W = F•dx
F = 65 N,
dx = xf - xi = 0.03 m - (-0.03 m) = 0.06 m
W = 65 N × 0.06 m = 3.9 J
5 0
3 years ago
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