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3 years ago

If santa slides 5.00 m before reaching the edge, what is his speed as he leaves the roof?

1 answer:

8 0

Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.

Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is

Square root of (19.6h) meters per second.

It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.

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a powerboat accelerates along a straight path from 0 km/hr to 99.8 km/hr in 10.0 s.Find the average acceleration of the boat in

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(27.72-0)/10
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3 years ago

A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in

sesenic [268]

Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v
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3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

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3 years ago

What is the proportionality between pressure and temperature, and the proportionality between atmospheric pressure and tempratur

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According to Ideal gasTo solve this problem, the fastest relationship allows us to observe the proportionality between the two variables would be the one expressed in the ideal gas equation, which is

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Here

P = Pressure

V = Volume

N = Number of moles

R = Gas constant

T = Temperature

We can see that the pressure is proportional to the temperature, then

$P \propto T$

This relationship can be extrapolated to all the scenarios in which these two variables are related. As the pressure increases the temperature increases. The same goes for the pressure in the atmosphere, for which an increase in this will generate an increase in temperature. This variable can be observed in areas of different altitude. At higher altitude lower atmospheric pressure and lower temperature.

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3 years ago

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3 years ago

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