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10 days ago

What happens to the particles of a liquid when energy is removed from them?

Physics

1 answer:

Softa [913]10 days ago

3 0

Response:

D: The distance among the particles diminishes

Clarification:

Removing energy reduces the activity of molecules, similar to how one slows down in cold temperatures (I believe).

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A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

ValentinkaMS [1144]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is $\frac{4}{3} \pi (\frac{D}{2})^3$

Thus $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

The radius of the disk is $1.15(\frac{ D}{2} )$

Applying angular momentum conservation;

The $M_i$ of the sphere = $\frac{2}{5} m \frac{D}{2}^2$

$M_i$ of the disk = $m*\frac{ \frac{1.15*D}{2}^2 }{2}$

$\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }$

= 0.6

5 0

4 days ago

A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the

Yuliya22 [1153]

The answer is letter d, 8.3.

Here’s a solution for the given problem:

We have:

B = 6i - 8j

Let A be unknown; we'll denote A as = mi + nj

The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

and we also know the magnitude of A+B is equivalent to the magnitude of A,

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2.

Using vector addition, A+B becomes (m+6)i + (n-8)j.

Since we know A+B = Ki + 0j, we can establish that:

m + 6 = K

n - 8 = 0, which gives n=8.

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2

= m^2 + 12m + 36 = m^2 + 64

which gives us 12m = 28

m = 2.33333...

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.

5 0

3 days ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Sav [1095]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

The wire’s diameter is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

Aluminum's resistivity is $2.75*10^{-8} \ ohm-meters.$

The electric field variation is described as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

The charge is effectively given by the equation

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area expressed as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

Thus,

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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10 days ago

Ariel dropped a golf ball from her second story window. The ball starts from rest and hits the sidewalk 3.5 s later with a veloc

inna [987]

Utilizing the acceleration formula,

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7 days ago

A tennis player who is recovering from an ankle injury and is not allowed to change directions can maintain her cardio fitness l

Softa [913]

A tennis player recovering from an ankle injury, restricted from pivoting, can keep her cardiovascular fitness by using the rowing machine, pedaling on a stationary bike with one leg, or swimming. These exercises do not necessitate directional changes and are safe for her injury.

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4 days ago

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