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3 years ago

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1. A base-ball of mass 0.3kg approaches the bat at a speed of 30 miles/hour and when the ball hits the bat for 0.5 s, it started

to move away from the bat at a speed of 60 miles/hour. Find the impulse

1 answer:

andre [41]3 years ago

5 0

Answer:

I = 27kg.mi/h

Explanation:

In order to calculate the impulse of the ball, you use the following formula:

$I=m\Delta v$ $=m(v-v_o)$ (1)

m: mass of the ball = 0.3kg

v: speed of the ball after the bat hit it = 60mi/h

vo: speed of the ball before the bat hit it = 30mi/h

You replace the values of all parameters in the equation (1):

$I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}$

where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.

The imulpse of the ball is 27 kg.miles/hour

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Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla

svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

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3 years ago

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

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2 years ago

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True or false When hydrogen joins with oxygen, a hydrogen bond has been formed.

ladessa [460]

Answer:

hydorgen bond

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2 years ago

Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor

arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

$PV = nRT$

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

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La presión cuando T = 67 °C es:

$P_{2} = \frac{nRT_{2}}{V_{2}}$ (2)

Dado que V₁ = V₂ (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

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Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

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