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Alekssandra [29.7K]
3 years ago
13

si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p

uede alcanzar el carrito?
Physics
1 answer:
forsale [732]3 years ago
6 0

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

E_{kin}=E_{pot}

Donde:

E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h

Ahora reemplazando:

\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} }   \\\\v=5.24[m/s]

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3 years ago
A 57-kg woman holds a 6-kg package as she stands within an elevator which briefly accelerates upward at a rate of 0.15g. Determi
Temka [501]

Answer:

R = 710.7N

L = 67.689 N

During gravity fall L = R = 0 N

Explanation:

So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is

g + 0.15g = 1.15g

The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):

a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N

The force L that she exerts on the package would be:

am = 1.15g *6 = 1.15*9.81*6 = 67.689N

When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.

7 0
3 years ago
What does this circle graph tell you about water on Earth? (2 points)
vekshin1

Answer:

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8 0
3 years ago
An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _
Viefleur [7K]
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
5 0
3 years ago
(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t
Alex73 [517]

Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

Q = m * Lf

Lf = 333,000 J/kg

Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

8 0
3 years ago
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