This question is incomplete, the complete question is;
An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is 20°C (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)
After the car has been driven at high speeds, the tire temperature increases to 55°C
(a) Assuming that the volume of the tire does not change and that air behaves as an ideal gas, find the gauge pressure of the air in the tire. (kPa)
(b) Calculate the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent. (kPa)
Answer:
a) the gauge pressure of the air in the tire is 236 kPa
b) the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent is 205.33 kPa
Explanation:
Given the data in the question;
Gauge pressure at 20°C = 200 kPa
Absolute pressure at 20°C is; p = 200 + 101.3 = 301.3 kPa
Initial temperature T = 20°C = 20 + 273 = 293 k
Final temperature T" = 55°C = 55 + 273 = 328 K
Now, at constant volume, P"/P = T"/T
P"T = PT"
P" = PT" / T
P" = PT" / T
we substitute
P" = ( 301.3 × 328 ) / 293
P" = 98826.4 / 293
P" = 337.29
so Gauge pressure at 55°C is;
⇒ P" - 101.3
⇒ 337.29 - 101.3 = 235.99 ≈ 236 kPa
Therefore, the gauge pressure of the air in the tire is 236 kPa
b)
Final Volume V'' = V + 10% of V
V" = V + 0.1V
V" = 1.1V
we know that;
P"V"/T" = PV/T
P" = P (T"/T)(V/V")
we substitute
P" = 301.3 (328 /293 )(V/1.1×V)
P" = × 1/1.1 × V/V
P" = 337.29 × 1/1.1
P" = 306.628 kPa
so, absolute pressure at 55°C = 306.628 - 101.3 = 205.33 kPa
Therefore, the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent is 205.33 kPa
Draw a conclusion, based on the solubility curves shown above, of which compound would have the greatest
percentage recovered after cooling a saturated solution of that compound from 90°C to 30°C?
A) KCL
B) NaNO3
C) Nacl
D) KNO3
Answer: The answer is D. KNO3
Explanation:
The graph shows that the KN03 going straight up from the temperature sign so you reversed that so that it will make it to 90°C to 30°C
To solve this we must be knowing each and every concept related to solubility. Therefore, the correct option is option D among all the given options.
What is solubility?The greatest amount of one material that may be dissolved in the other is referred to as its solubility. It is the most solute that may be dissolved into a solvent near equilibrium, resulting in a saturated solution.
When specific circumstances are satisfied, more solute can be dissolved further than the solubility limit point, resulting in a supersaturated solution. Adding extra solute after saturation or supersaturation does not enhance the concentration in the solution. Rather, the excess solute begins to precipitated out of solution. KNO[tex]_3[/tex] is the compound that would have the greatest percentage recovered after cooling a saturated solution of that compound from 90°C to 30°C.
Therefore, the correct option is option D.
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Define personal health.
Answer:
Personal Health is the ability to take charge of your health by making conscious decisions to be healthy.
Answer:Personal Health is the ability to take charge of your health by making conscious decisions to be healthy. It not only refers to the physical well being of an individual but it also comprises the wellness of emotional, intellect, social, economical, spiritual and other areas of life.
Explanation:
The variable ______________ describes how quickly something moves.
it's up in Gogle trust me
A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back up 1.50 m before falling to the ground. If the mass of the coconut is
2.00 kg, calculate the potential energy of the coconut relative to the ground at each of the following sites:
(a) while it is still in the tree,
(b) when it hits the bystander on the head,
(c) when it bounces up to its maximum height,
(d) when it lands on the ground,
(e) when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole.
Answer:
A. 240 J
B. 60 J
C. 90 J
D. 0 J
E. 50 J
Explanation:
A. Determination of the potential energy of the coconut while it is still in the tree
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 12 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 12
PE = 240 J
B. Determination of the potential energy of the coconut when it hits the bystander on the head,
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 3
PE = 60 J
C. Determination of the potential energy of the coconut when it bounces up to its maximum height,
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 + 1.5 = 4.5 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 4.5
PE = 90 J
D. Determination of the potential energy of the coconut when it lands on the ground,
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 0 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 0
PE = 0 J
E. Determination of the potential energy of the coconut when it rolls into a ground hole, and falls 2.50 m to the bottom of the hole.
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 2.50 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 2.50
PE = 50 J
(a) The potential energy of the coconut relative to the ground while it is still in the tree is 235.2 J.
(b) The potential energy of the coconut relative to the ground when it hits the bystander on the head is 58.8 J.
(c) The potential energy of the coconut relative to the ground when it bounces up to its maximum height is 88.2 J.
(d) The potential energy of the coconut relative to the ground when it lands on the ground is 0 J.
(e) The potential energy of the coconut when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole is 49 J.
The given parameters;
height of the tree, h = 12 mheight of the bystander, h' = 3 mheight it bounced back = 1.5 mmass of the coconut, m = 2.0 kgThe potential energy of the coconut relative to the ground while it is still in the tree;
[tex]P.E = mgh\\\\P.E = 2 \times 9.8 \times 12\\\\P.E = 235.2 \ J[/tex]
The potential energy of the coconut relative to the ground when it hits the bystander on the head;
[tex]P.E = 2 \times 9.8 \times 3 \\\\P.E = 58.8 \ J[/tex]
The potential energy of the coconut relative to the ground when it bounces up to its maximum height;
[tex]P.E = 2 \times 9.8 (1.5 + 3)\\\\P.E = 88.2 \ J[/tex]
The potential energy of the coconut relative to the ground when it lands on the ground;
[tex]P.E = 2 \times 9.8 \times 0\\\\P.E = 0 \ J[/tex]
The potential energy of the coconut relative to the ground when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole;
[tex]P.E = 2\times 9.8 \times 2.5 \\\\P.E = 49 \ J[/tex]
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If bullets are fired from an airplane in the forward direction of its motion, the momentum of the airplane will be:_______
Answer:
1
Explanation:
A model shows a machine that works using electrical fields. What would this machine need for the electrical field to function properly?
at least two charged interacting parts
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.20 m/sm/s. Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 52.9 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.50 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?
Answer:
a) v₀ₓ = 9.9 m / s, b) x_woman = 32.7 m
Explanation:
A) In this exercise, the movement of the bagels is parabolic, we find the time it takes to reach the floor.
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ gt²
t = [tex]\sqrt{2y_o/g}[/tex]
let's calculate
t = [tex]\sqrt{2 \ 52.9/9.8}[/tex]
t = 3,286 s
Now we can analyze how long Henrieta has walked, she has a walking time before the bagel movement begins (t₀ = 4.50 s)
t_woman = t₀ + t
t_woman = 4.50 + 3.286
t_woman = 7.786 s
The distance traveled in this time is
x_{woman} = v_woman t_woman
x_{woman} = 4.20 7.786
x_{woman} = 32.7 m
For her to grab the bagel, the two of them must be at this point
x_bagel = x_woman
x_bael = vox t
v₀ₓ = x_bagel / t
v₀ₓ = 32.7 / 3,286
v₀ₓ = 9.9 m / s
b) when catching the bagels this point x_woman = 32.7 m
What should Miguel do first and why? What type of healthcare professional will respond to the call?
Answer:
Miguel should hand the phone over to a medical personnel.A nurse or a physician will respond to the callExplanation:
From chapter 2 of the book: Medical Assisting: Administrative Skills, Miguel Perez is an administrative assistant. The duties of an administrative assistant in an healthcare professional setting range from performing medical clerical services like keeping patient's files organized, scheduling appointments and answering calls.
So, when a patient has medical concerns and calls the doctor's office, Miguel should answer the phone, know the patient's concerns, put the patient on hold and hand the phone over to a medical personnel - either a nurse or a physician. This is because Miguel is not a trained healthcare professional and cannot offer medical advice or assistance.
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)
Answer:
q = 2 10⁻⁸ C
Explanation:
For this exercise we use the translational equilibrium equation
F_e -A =
F_e = W
the electric force is given by Coulomb's law
F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case they indicate that the loads on the tapes are equal
F_e = k q² / r²
we substitute
k q² / r² = m g
q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]
calculate
q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]
q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]
q = 1,999 10⁻⁸ C
q = 2 10⁻⁸ C
what's the dimension symbol for thermodynamic temperature
Answer: Throughout the scientific world where measurements are nearly always made in SI units, thermodynamic temperature is measured in kelvins (symbol: K). The Rankine scale uses the degree Rankine (symbol: °R) as its unit, which is the same magnitude as the degree Fahrenheit (symbol: °F).
Explanation:
Please mark me as the Brainiest if I got it right
what's the dimension symbol for thermodynamic temperature
Answer:
°R
A wave in the ocean has a wavelength of 2 m and a frequency of 0’5 Hz. What is the speed of this wave?
Answer:
the speed of the wave =1m/s
PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!!
Because of the forces acting on the cart, it will
A. not accelerate
B. accelerate upwards
C. accelerate to the right
D. accelerate to the left
Answer:
D.
Explanation:
The First Law of Thermodynamics is the same as ______ with heat and work taken into consideration.
A. The First Law of Robotics
B. The Law of Conservation of Energy
C. Newton's First Law of Motion
D. The Law of Conservation of Momentum
Answer:
the law of conservation of energy
A fuel tank for a rocket in space under a zero-g environment is rotated to keep the fuel in one end of the tank. The system is rotated at 3 rev/min. The end of the tank (point A) is 1.5 m from the axis of rotation, and the fuel level is 1 m from the rotation axis. The pressure in the nonliquid end of the tank is 0.1 kPa, and the density of the fuel is 800 kg/m3 . What is the pressure at the exit (point A)
Answer:
P₂ = 4098.96 Pa
Explanation:
For this exercise let's use Bernoulli's equation
Let's use the subscript 1 for the point of the liquid surface and the subscript 2 for the ends (point A)
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
the velocity at the end of the tank
v₂ = w r₂
the velocity at the surface of the liquid is
v₁ - w r₁
where r₂ = 1.5 m and r₁ = 1 m
the tank pressure is P₁ = P₀ = 0.1 10³ Pa
P₂ = P₁ + ½ ρ [w² (r₁² - r₂²)] + ρ g (y₁ -y₂)
We must remember that the pressure measurements the distances are measured from the lowest part to the surface that has zero height
let's reduce the magnitudes to the SI system
w = 3 rev / min (2π rad / 1rev) (1 min / 60 s) = 0.314159 rad / s
let's calculate
P₂ = 0.1 10³ + ½ 800 0.314159² |(1² -1.5²)| + 800 9.8 |(1-1.5)|
P₂ = 0.1 103 +78.96 + 3920
P₂ = 4098.96 Pa
A 120 W lightbulb and a 90 W lightbulb each operate at a voltage of 120 V. Part A Which bulb carries more current? Which bulb carries more current? 120 W lightbulb 90 W lightbulb The currents are equal. It is impossible to determine.
Answer:
120 W lightbulb
Explanation:
Let the two lightbulb be A and B respectively.
Given the following data;
Power A = 120W
Power B = 90W
Voltage = 120V
To find the current flowing through each lightbulb;
a. For lightbulb A
Power = current * voltage
120 = current * 120
Current = 120/120
Current = 1 Ampere.
b. For lightbulb B
Current = power/voltage
Current = 90/120
Current = 0.75 Amperes
Therefore, the lightbulb that carries more current is A with 1 Ampere.
The bulb that carries more current is :
- A with 1 Ampere.
Let the two lightbulb be A and B respectively.
Given :Power A = 120WPower B = 90WVoltage = 120VTo find the current flowing through each lightbulb;
a. For lightbulb APower = current * voltage120 = current * 120Current = 120/120Current = 1 Ampere.b. For lightbulb BCurrent = power/voltageCurrent = 90/120Current = 0.75 AmperesThus, the lightbulb that carries more current is A with 1 Ampere.
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Help plsssssssssss I write it 100 time no one answer
Answer:
1.93×10²⁸ s
Explanation:
From the question given above, the following data were obtained:
Number of electron (e) = 2×10²⁴
Current (I) = 10 A
Time (t) =?
Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:
1 e = 96500 C
Therefore,
2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e
2×10²⁴ e = 1.93×10²⁹ C
Thus, 1.93×10²⁹ C of electricity is passing through the point.
Finally, we shall determine the time. This can be obtained as follow:
Current (I) = 10 A
Quantity of electricity = 1.93×10²⁹ C
Time (t) =?
Q = it
1.93×10²⁹ = 10 × t
Divide both side by 10
t = 1.93×10²⁹ / 10
t = 1.93×10²⁸ s
Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point
A car with mass m travels over a hill with a radius of curvature of r at a speed of 15 m/s. What is the normal force on the car when the car is at the top of the hill?
Answer:
zero
Explanation:
The computation of the normal force is shown below:
As we know that
F_c = mg - N
F_c = mv^2 ÷ r
N = mg - mv^2 ÷ r
N = m(g - v^2 ÷ r)
Assume that
The mass of the car is 1200 kg
And, r = 10 m
So,
= 1200 (9.8 - 15^2 ÷ 10)
= -15240 N
Since it comes in negative so the normal force would be zero
Standing at a crosswalk, you hear a frequency of 530 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 424 Hz. Determine the ambulance's speed from these observations.
Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀ [tex]\frac{v}{v - v_s}[/tex]
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]
the speed of sound is v = 340 m / s
let's calculate
v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]
v_s = 340 [tex](\frac{106}{954}[/tex])
v_s = 37.77 m / s
what is the average velocity of a van that moves from 0 to 60 m east and 20 seconds
Explanation:
I have a lot to say it was very nice to meet my parents are u doing well I dont want too its been so much I love you so I was like u know I am not a man but you are the auditions I have been in a long long long life is a triangle and a chair for me and my parents think about the way I
Arun runs 9 meters across Mr. Scharff's classroom in 7.1 seconds. How fast did Arun run
Answer:
The answer would be 180 meters.
Explanation:
help please i will mark brainlist!!!
Answer:
.50 M
Explanation:
5*.50=2.5 + 2*.25=.5 = 3n
6*.50= 3N
Final answer is .50M
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25 cm and the point-like object with charge q2 = −2.14 µC is located at x2 = −1.80 cm.
A) Determine the total electric potential (in V) at the origin.
B) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
a) The total electric potential is 2282000 V
b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
What is electric potential?The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
Electric potential at p in diagram 1 below is;
[tex]V_P=V_1+V_2[/tex]
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we know that; the Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
[tex]r_1^2=0.015^2+0.0125^2[/tex]
[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]
[tex]r_1 = \sqrt{0.00038125}[/tex]
[tex]r_1 = 0.0195[/tex]
Also
[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]
[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]
[tex]r_2 = \sqrt{0.000549[/tex]
[tex]r_2 = 0.0234[/tex]
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
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Two objects are electrically charged. The net charge on one object is doubled.
Therefore, the electric force _____.
reverses
doubles
quadruples
divides
the product of 2.03 and 0.05
Answer:
2.03 x 0.05= 0.1015
.........
You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$ at Harbor Freight. Using the meter in voltage mode, you measure 10 V across the battery. You then connect the 2 resistors in series across the battery and use the meter in current mode to find the current flowing through the circuit. It reads 0.111A. You then connect the 2 resistors in parallel across the battery and use the meter again to measure the current now coming from the battery to be 0.5A. With this information you have gathered, you find the value of the 2 resistors.
Value of smallest resistance in ohms.
a. 60
b. 90
c. 20
d. 30
Answer:
the answers, the correct one is D, Rb₂ = 29.97 ohm
Explanation:
For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.
Serial circuit
(Ra + Rb) is = V
(Ra + Rb) 0.111 = 10
(Ra + Rb) = 10 / 0.111 = 90.09
parallel circuit
[tex]\frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}[/tex]
R = [tex]\frac{Ra \ Rb}{Ra + Rb}[/tex]
\frac{Ra \ Rb}{Ra + Rb} i_p = V
\frac{Ra \ Rb}{Ra + Rb} 0.5 = 10
\frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20
we write and solve our system of equations
Ra + Rb = 90.09
\frac{Ra \ Rb}{Ra + Rb} = 20
we solve for Ra in the first equation
Ra = 90.09 - Rb
RaRb = 20 (Ra + Rb)
we substitute Ra in the second equation
(90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]
90.09 Rb - Rb² = 20 90.09
Rb² - 90.09 Rb + 1801.8 = 0
we solve the quadratic equation
Rb = [90.09 ±[tex]\sqrt{90.09^2 - 4 \ 1801.8}[/tex] ] / 2
Rb = [90.09 ± 30.15] / 2
Rb₁ = 60.12 ohm
Rb₂ = 29.97 ohm
the smallest value is Rb = 30 ohm
When checking the answers, the correct one is D
A potter’s wheel moves from rest to an angular speed of 0.10 rev/s in 36.5 s.
Assuming constant angular acceleration,
what is its angular acceleration in rad/s2?
Answer in units of rad/s2
.
Answer:
please find attached pdf
Explanation:
The relationship between frequency and period is...
[tex] \\ [/tex]
Frequency, f, is how many cycles of an oscillation occur per second and is measured in cycles per second or hertz (Hz). The period of a wave, T, is the amount of time it takes a wave to vibrate one full cycle. These two terms are inversely proportional to each other: f = 1/T and T = 1/f.
[tex] \\ [/tex]
Hope It Helps!
Answer:
Inverse
Explanation:
Frequency is the number of cycles in a second. Frequency is the inverse of a period
frequency = 1 / period
Find the specific heat of a substance that requires 8000 J of energy to heat up 400g by 20 C?
Answer:
[tex]c=1\ J/g^\circ C[/tex]
Explanation:
Given that,
Heat required, Q = 8000 J
Mass, m = 400 g
The change in temperature, [tex]\Delta T = 20^{\circ}[/tex]
The heat required due to change in temperature is given by :
[tex]Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{8000 }{400\times 20}\\\\c=1\ J/g^\circ C[/tex]
So, the specific heat of the substance is [tex]1\ J/g^\circ C[/tex]
The monkey experiment is an example of what?
A. top down processing
B. bottom up processing
C. inattentional blindness
D. sensory adaption
Answer:
D.) Sensory adaptation
Explanation:
Assuming you are talking about the cloth and metal monkey experiment performed in the field of psychology (not physics), the monkey formed an attachment to the cloth mother because it felt closer to it, as it was more appealing to its senses.
Determine the resultant force exerted on an object if these three forces are exerted on it: F1=3.0N upwards,F2=6.0N at 45° to the horizontal and F3=5.0 at 120° from the positive x-axis
I couldn't upload the complete pic because I'm browsing from phone
pls give me brainliest
