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2 years ago

Why does oil not dissolve in water?

Physics

1 answer:

Fudgin [204]2 years ago

4 0

Answer:

C. Oil is made of non-polar molecules, and water is made of polar molecules.

Explanation:

Non-polar molecules = hydrophobic (water-fearing, doesn’t readily dissolve in water)

Ions & Polar molecules: hydrophilic (water-loving, does readily dissolve in water)

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How can an object overcome static friction?

larisa86 [58]

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

4 0

3 years ago

A battery with an internal resistance ofrand an emf of 10.00 V is connected to a loadresistorR=r. As the battery ages, the inter

yanalaym [24]

Answer:

The current is reduced to half of its original value.

Explanation:

Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

$I_{1} = \frac{V}{R_{int} +r_{L} }$

where Rint = r and RL = r
Replacing these values in I₁, we have:

$I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

$I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

$\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.

7 0

3 years ago

high school physics, no need detail explain, just give the answer, but you have to make sure thank you

Goshia [24]

Answer:many questions add point

6 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

1. When in the past have you pushed your personal limits? Give at least one

Yuri [45]

Answer:

Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.

Explanation:

4 0

3 years ago