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Kamila [148]
3 years ago
13

In 1985 in San Antonio, Texas, an entire hotel building was moved several

Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

The building was moved to a distance of, S = 510.2 m

Explanation:

Given data,

The mass of the building, m = 1.45 x 10⁶ kg

The work done to overcome the force of resistance, E = 1 x 10⁸ J

The force of resistance of the building, 2% of building weight

                                          Fₓ =  2 x 1.45 x 10⁶ x 9.8 / 100

                                               = 284200 N

The work done on the building,

                                    E = Fₓ · S

∴                                   S = E /  Fₓ

                                        = 1.45 x 10⁶ / 284200 N

                                        = 510.2 m

Hence, the building was moved to a distance of, S = 510.2 m

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I think it's D

Explanation:

because I have seen the tracks of the tires

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3 years ago
What is the acceleration of the object?<br> m/s²
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Explanation:

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2 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
3 years ago
8x = -6. What does x equal?
allsm [11]

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4 0
3 years ago
Read 2 more answers
A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a
Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = 2\pi radians

So, 2.73 revolutions = 2.73\times 2\pi=17.153\ radians

Therefore, the angular velocity of the tack is, \omega=17.153\ rad/s

Now, radial acceleration of the tack is given as:

a_r=\omega^2 r

Plug in the given values and solve for a_r. This gives,

a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0
3 years ago
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