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loris [4]
3 years ago
6

A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

is neglected, the horizontal distance R from the point on the lake directly below the point of release to the point where the stone strikes the water is given by which formula? A) R = v(2h/g)^2 B) R = (1/2)gt^2 C) R = 2mv squareroot (2h/g) D) R = v squareroot (2h/g) E) None of these is correct.
Physics
1 answer:
seropon [69]3 years ago
4 0

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

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Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?
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)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app
masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

F \propto \frac{1}{r}

F = \frac{k}{r} (Eq. 1)

Where:

F - Force, measured in newtons.

k - Proportionality ratio, measured in newton-meters.

r - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

F_{A}\cdot r_{A} = F_{B}\cdot r_{B}

F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}(Eq. 2)

Where:

F_{A}, F_{B} - Initial and final forces, measured in newtons.

r_{A}, r_{B} - Initial and final distances, measured in meters.

If we know that F_{A} = 5\,N, r_{A} = 0.9\,m and r_{B} = 0.35\,m, then final force is:

F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)

F_{B} = 12.857\,N

A force of 12.857 newtons must be applied to open the door.

3 0
2 years ago
The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?
pishuonlain [190]
The average dissipated power in a resistor in a ac circuit is:
P=I_{rms}^2 R
where R is the resistance, and I_{rms} is the root mean square current, defined as
I_{rms} =  \frac{I_0}{\sqrt{2}}
where I_0 is the peak value of the current. Substituting the second formula into the first one, we find
P=( \frac{I_0}{\sqrt{2} } )^2 R =  \frac{1}{2} I_0^2 R
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
I_0 =  \sqrt{ \frac{2 P}{R} }=  \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A
4 0
3 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
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