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3 years ago

Help pleasee i give brainliest

Physics

1 answer:

MA_775_DIABLO [31]3 years ago

5 0

Answer:

A is a solid. C is a gas. In solid an liquid the particals are touching. In C, the particals have less affect on each other because they are so far apart.

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A heat engine uses thermal energy that flows from a heat source with a temperature of 325 K to a cold sink, which has a temperat

Ilya [14]

The answer is "58.5"

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3 years ago

Read 2 more answers

Which of these best describes heat? total thermal energy degree of warmth transfer of thermal energy average kinetic energy

Papessa [141]

Total thermal energy is the answer to your question.

3 0

3 years ago

Help with this physics task pls

cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

7 0

3 years ago

A 500 W hair dyer is used to dry hair for 6 minutes a day for 3 days. How much energy is used? A laptop uses 75 W. The cot of el

zzz [600]

Answer:

See the answers below.

Explanation:

We will solve this problem by calculating each part separately.

A 500 W hair dyer is used to dry hair for 6 minutes a day for 3 days.

Energy can be calculated by multiplying the value of the power of the equipment by the amount of time of use.

$500 [W]*[\frac{6min}{1day} ]*[\frac{1day}{24hr} ]*[\frac{1hr}{60min} ]=2.083 [W]$

The cots of electricity is 5.6 cents per kWh. How much would it cost to operate the laptop for 24 hours a day for one week?

We know that the power of the latop is 75 [W], then we can calculate the cost, multiplying the value of the power by the value of the cost by the time of use of the computer.

$0.075[kW]*5.6[\frac{cents}{kw*h}}]*[\frac{24hr}{1day}]*[1week]*[\frac{7days}{1week} ]=70.56 [cents]$

A toaster oven is 85% efficient. It uses 1200 J of energy. How much thermal energy is it producing?

Efficiency is defined as the relationship between the energy obtained on the energy delivered. Almost always the energy delivered is greater than the energy obtained (first law of thermodynamics).

Therefore.

$Effic = E_{obtained}/E_{delivered}\\0.85=E_{obtained}/1200\\E_{obtained}=1020[J]$

4 0

3 years ago

A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m

Alex73 [517]

Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

6 0

3 years ago