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Svetach [21]
3 years ago
14

A rock falls off the edge of a building and falls for 10 seconds. What was

Physics
2 answers:
solmaris [256]3 years ago
4 0
Ricks velocity would be zooomin out because it would fall off so strongly so it’d change and it’s weight too
In-s [12.5K]3 years ago
3 0
Assuming the rock falls straight down, we can determine that this rock is under the influence of solely gravity (the rock is in free fall). therefore, the rock will accelerate at 9.8 m/s^2 (acceleration due to gravity). we can utilize this to solve for vf by plugging our known variables into the kinematics equation vf=v0+at. this will lead us to vf=0+9.8(10), or vf=98 m/s^2. this answer, while correct, seems a tad bit unreasonable, so you should mention to your teacher to consider the feasibility of his physics problems.
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A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum
qwelly [4]

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

π

2α

≤x≤+

π

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(c) Find the particle’s average position.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

6 0
3 years ago
A 1210 kg rollercoaster car is
ratelena [41]

Answer: 4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

3 0
3 years ago
As a planet's semimajor axis gets smaller, its speed will _______ and its period will _________ .
Rufina [12.5K]
Well first of all, a planet doesn't have a semimajor axis, although it's orbit does.

In an orbit with a smaller semimajor axis, the planet moves faster, and its orbital period is shorter.

That's why the International Space Station circles the Earth in less time than the Moon does.
4 0
3 years ago
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

6 0
3 years ago
What describes the movement of a fluid during convection?
damaskus [11]
Convection is the movement<span> of groups of molecules within </span>fluids<span> such as gases and liquids, including molten rock (rheid).</span>
5 0
3 years ago
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