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3 years ago

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You have two identical beakers A and B. Each beaker is filled with water to the same height. Beaker B has a rock floating at the

surface (like a pumice stone). Which beaker, with all its contents, weighs more. Or are they equal?

1 answer:

gregori [183]3 years ago

4 0

Answer:

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water the weight of the two beakers is the same

Explanation:

The beaker weight is

beaker A

W_total = W_ empty + W_water

Beaker B

W_total = W_ empty + W_water + W_roca

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same

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A car is traveling at a speed of 45 km/h into town. It takes the car 2 hours to get there. How far has the car traveled?

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3 years ago

Read 2 more answers

The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,

omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

$g = \frac{GM}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

$M_p = \frac{1}{100} M_e$

$r_p = \frac{1}{4} r_e$

The relationship of gravity would then be given:

$g_e = \frac{GM_e}{r_e^2}$

The relationship with the new planet, from the gravity of the earth would be given

$g_p = \frac{GM_p}{r_p^2}$

$g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}$

$g_p = \frac{GM_e 16}{100 r_e^2}$

$g_p = 0.16 \frac{GM_e}{r_e^2}$

$g_p = 0.16g_e$

The relationship with the weight of the earth would be given as:

$W_e = m*g_e = 600N$

$W_p = m*g_p = m(0.16g_p)$

$W_p = (m*g_p)(0.16)$

$W_p = 600*0.16$

$W_p = 96N$

Therefore the weigh on this planet would be 96N

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3 years ago

Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00

lions [1.4K]

Answer:

b. $\Delta KE = 390 eV$

Explanation:

As we know that the electric field due to infinite line charge is given as

$E =\frac{\lambda}{2\pi \epsilon_0 r}$

here we can find potential difference between two points using the relation

$\Delta V = \int E.dr$

now we have

$\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr$

now we have

$\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})$

now plug in all values in it

$\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})$

$\Delta V = 216ln6 = 387 V$

now we know by energy conservation

$\Delta KE = q\Delta V$

$\Delta KE = (e)(387V) = 387 eV$

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3 years ago