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3 years ago

Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan m

Physics

1 answer:

lara31 [8.8K]3 years ago

6 0

Degree Celsius to Kelvin

<u>Explanation</u>:

Two units of measuring temperature are Degree Celsius and Kelvin.
A measured value in degree Celsius can be converted to Kelvin by adding 273 to it. Here, Juan is doing that conversion - degree Celsius to Kelvin.
°C + 273 = K

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Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

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For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
$r_1=0.915\cdot 10^6$km$

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4 years ago

Which of the following phases of the Moon produces the highest and lowest tides on Earth?

Nezavi [6.7K]

I believe the answer is D. Full Moon
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4 years ago

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A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp

dusya [7]

from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

Answer:

(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A) What is its speed after falling for 2.00s?

from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

from the equation of motion $s=ut+0.5at^{2}$ we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

$v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}$

v = 15.4 m/s

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3 years ago

For what absolute value of the phase angle does a source deliver 71 % of the maximum possible power to an RLC circuit

kherson [118]

Answer: the absolute value of the phase angle is 28°

Explanation:

taking a look at expression for the instantaneous electric power in an AC circuit;

P = VI -------let this be equation 1

p is power, v is voltage and I is current;

for maximum power

P_max = V_rms × I_rms --------let this be equ 2

where P_max is the maximum power, V_rms is the rms value of voltage and I_rms is the rms value of current.

Also for average electric power in an AC circuit

P_avg = V_rms × I_rms × cos²∅ -------let this be equ 3

where P_avg is the average power and cos∅ is the power factor

now from equation 2; P_max = V_rms × I_rms

so p_max replaces V_rms × I_rms in equation 3

we now have

P_avg = P_max × cos²∅

so we substitute

expression for the given value of the average power is

P_avg = P_max × 75%

p_avg = P_max.78/100

for the expression of the average electricity in an AC circuit

P_max.78/100 = P_max × cos²∅

78/100 = cos²∅

to get the absolute value of phase angle

∅ = cos⁻¹ ( √(78/100))

∅ = cos⁻¹ ( 0.8832)

∅ = 27.969 ≈ 28°

Therefore the absolute value of the phase angle is 28°

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3 years ago

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3 years ago

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