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4 years ago

Condition stops when all substances have

Physics

2 answers:

Scorpion4ik [409]4 years ago

8 0

Answer:

equal masses

different masses

equal temperatures

different temperatures

Explanation:

barxatty [35]4 years ago

5 0

Equal temperatures.

Heat energy can be transferred from one place to another by three main processes. In convection, heat energy is carried by the movement of particles of matter. In conduction, heat is transferred by particles vibrating. In radiation, heat is carried directly by electromagnetic waves. During conduction heat travels through solids by conduction. If one end of a metal bar is heated, heat energy moves rapidly along the bar. The hot particles do not move along the bar, but vibrate and pass energy to their neighbors.

Hope it helps! :D

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A ball bearing is projected vertically upwards from the ground with a velocity of 15ms. Calculate the time taken by the ball to

cupoosta [38]

Answer:

t = 3 [s]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}-g*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 15 [m/s]

g = gravity acceleration = 10 [m/s²]

t = time [s]

Now replacing we have:

$0 = 15 -10*t\\10*t=15\\t= 1.5[s]$

Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.

The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:

t = 1.5 + 1.5 = 3 [s]

7 0

3 years ago

You are taking care of two small children, Sarah and Rachel, who are twins. On a nice cool, clear day you decide to take them ro

BlackZzzverrR [31]

Answer:

Explanation:

Rachel and Sarah are standing on horizontal ground . Let T₁ be the tension between Rachel and Sarah and Sarah is in between you.

Force = mass x acceleration

Apply this equation on Rachel

T₁ = 23 x 2 = 46 N

Let T₂ be tension between Sarah and me and it is inclined at 60°

Force = mass x acceleration

Apply this equation on Sarah

T₂ cos 60 - T₁ = 23 X 2 = 46

T₂ cos 60 - 46 =46

T₂ cos 60 = 92

T₂ = 92 / .5

= 184 N

5 0

3 years ago

Read 2 more answers

Two Objects of the same size will always have the same mass” Is this statement correct?

attashe74 [19]

No, they won't, mass coincides with density and objects have different densities a one pound lead ball would be smaller than a one pound copper one.

6 0

3 years ago

An airplane is traveling due east with a velocity of 7.5 × 102 kilometers/hour. There is a tailwind of 30 kilometers/hour. What

Delvig [45]

A tailwind means you add thed velocity of the tailwind to the velovity of the plane (no need for any vectors etc. so you just need (7.5x 102 ) +30 then convert it so it like one of the answers (btw - none of the answers is exactly correct - you need to round to get it) thats what you need to do to get the question

4 0

3 years ago

Read 2 more answers

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$