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tekilochka [14]

3 years ago

10

Why is the following situation impossible? The object of mass m = 4.00 kg in Figure P6.10 is attached to a vertical rod by two s

trings of length ℓ = 2.00 m. The strings are attached to the rod at points a distance d = 3.00 m apart. The object rotates in a horizontal circle at a constant speed of v = 3.00 m/s, and the strings remain taut. The rod rotates along with the object so that the strings do not wrap onto the rod. What if? Could this situation be possible on another planet?

1 answer:

Crank3 years ago

4 0

The situation is impossible mainly because we can't see Figure P6.10 .
It would undoubtedly be the same story on an another planet, until we
see the figure and understand what's going on.

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Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$

Learn more about gravitational constant:

brainly.com/question/1724648

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#LearnwithBrainly

4 0

3 years ago

The volume of an object as a function of time calculated by V=At^3+B/t,where t is time measured in seconds and V is in cubic met

S_A_V [24]

Answer:

A = L^3 T^-3 , B = L^3 T

Explanation:

Given: volume ( V ) = At^3 + B/t ------ ( 1 )

dimension of volume = L^3

and the Dimension of time = T

back to equation ( 1 )

L^3 = A * T^3 ------- ( 2 )

also L^3 = B/T ------- ( 3 )

from equation ( 2 )

A = L^3 T^-3

from equation ( 3 )

B = L^3 T

<u>The dimensions of the constants A and B </u>

A = L^3 T^-3 , B = L^3 T

8 0

3 years ago

A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

5 0

3 years ago

When water vapor rises into the atmosphere and begins to condense, it releases ________, the energy of a thunderstorm.

zubka84 [21]

The answer is latent heat. The specific latent heat of vaporisation, L_v, of a substance is the energy input required for each kilogram to be converted from liquid to gas by evaporation. The 'specific' means per kilogram, so more generally latent heat of vaporisation is the energy taken in during the process for a given mass.

Here we are not vaporising the substance. We are in fact condensing it, the reverse process. All this means is the latent heat is released as electrostatic potential decreases in the water, as opposed to being absorbed. I hope this helps you :)

5 0

3 years ago

Read 2 more answers

How can electric shock be prevented?

Nina [5.8K]

Wearing rubber or stay away from water or/ and a conductor

7 0

3 years ago