Answer:
(D) 30.6 cm
Explanation:
Given the following data;
Measurement A = 18.425 cm
Measurement B = 7.21 cm
Measurement C = 5.0 cm.
To find the total measurement;
Total measurement = A + B + C
Total measurement = 18.425 + 7.21 + 5
Total measurement = 30.635 cm
However, an additional sum requires that all the numbers have a equal number of decimal points; 1 decimal point for 5.0 would be chosen.
Hence, we would round up the answer to 1 decimal point.
Total measurement = 30.6 cm
Why are the shoes soles of fat people get worn out faster than those of thin people?
The acidity model given by pH = -log10[H+] where acidity (pH) is a measure of the hydrogen ion concentration [H+] (measured in moles of hydrogen per liter) of a solution. Using this model, determine the pH of a solution with a hydrogen ion concentration of 2.3 X 10^-5 (round your answer to the nearest hundredth)
Answer:
Hydrogen ion concentration, [H+] :
[tex]{ \bf{[H {}^{ + } ] = 2.3 \times {10}^{ - 5} \: mol \: l {}^{ - 1} }}[/tex]
Substitute to the negative logarithmic model:
[tex] = { \tt{ - log(2.3 \times {10}^{ - 5} ) }} \\ pH = 4.64[/tex]
If a lawn mower is pushed with a distance of 30 meters and 12N-m of work is exerted, calculate the force.
Answer:
Explanation:
W = FΔx so filling in:
12 = F(30) so
F = .4N
Give a general description of how winds form
Answer:
Wind comes from pressure and temperature differences.
Explanation:
Have you ever opened a window on a hot day to let a breeze in? What is happening is a rush of air to balance out temperatures. This goal for uniformity is why ice melts in a glass of warm coffee and air rushes out of a balloon.
Answer:
Wind forms when the sun heats one part of the atmosphere differently than another part. This causes expansion of warmer air, making less pressure where it is warm than where it is cooler. Air always moves from high pressure to lower pressure, and this movement of air is wind.
Explanation:
what changes must be done to the wire to increase its conductance.
Answer:
- Decreasing the resistance
- Using a shorter length
- Using a smaller area wire
Explanation:
Formula for conductance in wires is;
G = 1/R
Where;
G is conductance
R is resistance
This means that increasing the resistance leads to a larger denominator and thus a smaller conductance but to decrease the denominator means larger conductance.
Thus, to increase the conductance, we have to decrease the resistance.
Resistance here has a formula of;
R = ρL/A
Where;
ρ is resistivity
L is length of wire
A is area
Thus, to decrease the resistance, we will have to use a shorter length and smaller area of wire.
If Cl− is the only anion in the solution, what is the Cl− concentration in milliequivalents per liter?
Answer:
155mEq/L Cl-
Explanation:
A Ringer’s solution contains the following concentrations of cations: 146 mEq/L of Na+, 5 mEq/L of K+, and 4 mEq/L of Ca2+.
As Cl- is the only counterion of those cations:
For Na, the molecule is NaCl and the mEq/L of Cl- = mEq of Na+. The Cl- of the first ion is 146mEq/L Cl-
For K+, The molecule is KCl, mEq Cl- = 5mEq/L Cl-
And, for Ca2+, The molecule is CaCl2 but the equivalents of Ca2+ = Equivalents of Cl- = 4mEq/L Cl-
The total concentration of Cl- are:
146 + 5 + 4 =
155mEq/L Cl-Does anyone know the answer ? I forgot my calculator.
12 x sin50
12 × sin50 = 9.192533317........
How is the force useful to us?
Answer:
Force is useful to us because it changes or tries to change the positionof a body, it helps to change the direction of a moving object , it helps to change the speedof a moving body etc.
I hope this will help you
The answer is D. I can't figure out why? The human range is from 20 Hz to 20 kHz...
Answer:
(C) 2 kHz
Explanation:
The normal hearing frequency range of a person has a minimum of about 20 Hz, which is the about the frequency of the lowest pipe organ pipe. The maximum frequency that a human can ear comfortably is 20 kHz
Therefore, the student will not be able to hear sounds less than 20 Hz, and sounds above 20 kHz and the student will only be able to hear the 2 kHz sound.
For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along the path described by LaTeX: y=2+\frac{1}{x} y = 2 + 1 x . They can shoot rockets tangent to the direction of flight at targets on the x-axis. Where will a rocket fired from LaTeX: \left(1,3\right) ( 1 , 3 ) hit the target? Where will a rocket fired from LaTeX: \left(2,2.5\right) ( 2 , 2.5 ) hit the target? Where will a rocket fired from LaTeX: \left(2.5,2.4\right) ( 2.5 , 2.4 ) hit the target? Where will a rocket fired from LaTeX: \left(4,\:2.25\right) ( 4 , 2.25 ) hit the target?
Answer:
When fired from (1,3) the rocket will hit the target at (4,0)
When fired from (2, 2.5) the rocket will hit the target at (12,0)
When fired from (2.5, 2.4) the rocket will hit the target at [tex](\frac{35}{2},0)[/tex]
When fired from (4,2.25) the rocket will hit the target at (40,0)
Explanation:
All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).
Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.
In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:
[tex]y=2+\frac{1}{x}[/tex]
[tex]y=2+x^{-1}[/tex]
[tex]y'=-x^{-2}[/tex]
[tex]y'=-\frac{1}{x^{2}}[/tex]
so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:
[tex]y'=-\frac{1}{x^{2}}[/tex]
[tex]y'=-\frac{1}{(1)^{2}}[/tex]
m=y'=-1
so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-3=-1(x-1})[/tex]
[tex]y-3=-1x+1[/tex]
[tex]y=-x+1+3[/tex]
[tex]y=-x+4[/tex]
So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.
[tex]-x+4=0[/tex]
and solve for x
x=4
so, when fired from (1,3) the rocket will hit the target at (4,0)
Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)
so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:
[tex]y'=-\frac{1}{x^{2}}[/tex]
[tex]y'=-\frac{1}{(2)^{2}}[/tex]
[tex]m=y'=-\frac{1}{4}[/tex]
so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-2.5=-\frac{1}{4}(x-2})[/tex]
[tex]y-2.5=-\frac{1}{4}x+\frac{1}{2}[/tex]
[tex]y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}[/tex]
[tex]y=-\frac{1}{4}x+3[/tex]
So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.
[tex]-\frac{1}{4}x+3=0[/tex]
and solve for x
x=12
so, when fired from (2, 2.5) the rocket will hit the target at (12,0)
Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)
so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:
[tex]y'=-\frac{1}{x^{2}}[/tex]
[tex]y'=-\frac{1}{(2.5)^{2}}[/tex]
[tex]m=y'=-\frac{4}{25}[/tex]
so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-2.4=-\frac{4}{25}(x-2.5})[/tex]
[tex]y-2.4=-\frac{4}{25}x+\frac{2}{5}[/tex]
[tex]y=-\frac{4}{25}x+\frac{2}{5}+2.4[/tex]
[tex]y=-\frac{4}{25}x+\frac{14}{5}[/tex]
So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.
[tex]-\frac{4}{25}x+\frac{14}{5}=0[/tex]
and solve for x
[tex]x=\frac{35}{20}[/tex]
so, when fired from (2.5, 2.4) the rocket will hit the target at [tex](\frac{35}{2},0)[/tex]
Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)
so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:
[tex]y'=-\frac{1}{x^{2}}[/tex]
[tex]y'=-\frac{1}{(4)^{2}}[/tex]
[tex]m=y'=-\frac{1}{16}[/tex]
so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-2.25=-\frac{1}{16}(x-4})[/tex]
[tex]y-2.25=-\frac{1}{16}x+\frac{1}{4}[/tex]
[tex]y=-\frac{1}{16}x+\frac{1}{4}+2.25[/tex]
[tex]y=-\frac{1}{16}x+\frac{5}{2}[/tex]
So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.
[tex]-\frac{1}{16}x+\frac{5}{2}=0[/tex]
and solve for x
[tex]x=40[/tex]
so, when fired from (4,2.25) the rocket will hit the target at (40,0)
I uploaded a graph that represents each case.
Which of the following provides the best evidence in support of the big bang
theory?
O A. The way a balloon expands
B. The confirmation of special relativity
C. The uniformity of cosmic background radiation
D. The separation of energy levels in an atom
C. The Uniformity of Cosmic Background Radiation
4. Interference is an example of which aspect of electromagnetic radiation?
A) Particle behavior
B) photon behavior
C) the photoelectric effect
D) wave behavior
Answer:
D is the answer wave behavior
Interference, refraction, diffraction, and dispersion are all aspects of wave behavior. (D). That is, particles don't do these things.
A 0.080-kg egg test dummy is fitted with a helmet and attached to a swing. This egg test dummy is pulled back and released, allowing it to collide with a cement block. The impulse on the egg test dummy is - 0.39N.s is over an interval of 0.050 s. What is the magnitude of the force on the egg test dummy during this time interval?
Answer:
7.8 N
Explanation:
Applying,
I = Ft................. Eqaution 1
Where I = Impulse on the egg test dummy, F = Force on the egg test dummy during the time interval, t = time interval
make F the subject of the equation
F = I/t.................. Equation 2
From the question,
Given: T = -0.39 N.s, t = 0.050 s
Substitute these vales into equation 2
F = -0.39/0.050
F = -7.8 N
Hence the force that act on the egg test dummy is 7.8 N
A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal. (a) determine the time taken by the projectile to hit the point P at ground level. (b) determine the range X of the projectile as measured from the base of the cliff at the instant just before the projectile hits point P. Find (x) the horizontal and vertical components of its velocity and (d) the magnitude of the velocity and (e) the angle made by the velocity vector with the horizontal (f) Find the maximum height above the cliff top reached by the projectile
Answer:
(a) The time it takes the projectile to hit the ground is approximately 10.42 seconds
(b) The range of the projectile from the base of the cliff is approximately 540.92 meters
(c) The horizontal component of the velocity is approximately 51.91 m/s
The vertical component of the velocity is approximately 39.12 m/s
(d) The magnitude of the velocity of the projectile is 65.0 m/s
(e) The angle made by the velocity vector of the projectile with the horizontal is 37°
(f) The maximum height above the cliff top reached by the projectile is approximately 77.99 meters
Explanation:
The height of the cliff above ground level, y₀ = 125 m
The initial speed of the projectile, u = 65.0 m/s
The angle of elevation of the projectile, θ = 37°
(a) The time for the projectile to hit the ground, t, is given as follows;
y = y₀ + u·sin(θ)·t -(1/2)·g·t²
g = 9.81 m/s²
At ground level, y = 0, we get;
0 = 125 + 65·sin(37)·t - 4.905·t²
t = (-65·sin(37) ± √((65·sin(37))² - 4×(-4.905)×125))/(2 × (-4.905))
∴ t ≈ -2.45 or t ≈ 10.42
The time it takes the projectile to hit the ground, t ≈ 10.42 seconds
(b) The range of the projectile X = u·cos(θ) × t
∴ X = 65.0 × cos(37°) × 10.42 ≈ 540.92
The range of the projectile from the base of the cliff, X ≈ 540.92 meters
(c) The horizontal component of the velocity, uₓ = u×cos(θ)
∴ uₓ = 65.0 ×cos(37°) ≈ 51.91
The horizontal component of the velocity, uₓ ≈ 51.91 m/s
The vertical component of the velocity, [tex]u_y[/tex] = u×sin(θ)
∴ [tex]u_y[/tex] = 65.0 ×sin(37°) ≈ 39.12
The vertical component of the velocity, [tex]u_y[/tex] ≈ 39.12 m/s
(d) The magnitude of the velocity, [tex]\left | u \right |[/tex], is the given speed = 65.0 m/s
[tex]\left | u \right |[/tex] = √([tex]u_y[/tex]² + uₓ²)
∴ [tex]\left | u \right |[/tex] = √(39.12² + 51.91²) = 65
[tex]\left | u \right |[/tex] = 65.0 m/s
(e) The angle made by the velocity vector with the horizontal = The angle of elevation with which the projectile is launched = 37°
(f) The maximum height above the cliff top reached by the projectile, [tex]y_{max \ of \ cliff}[/tex], is given as follows;
[tex]y_{max \ of \ cliff}= \dfrac{ \left(u \times sin(\theta)\right)^2}{2 \cdot g}[/tex]
∴ [tex]y_{max \ of \ cliff}[/tex] = (65.0 × sin(37°))²/(2 × 9.81) ≈ 77.99
[tex]y_{max \ of \ cliff}[/tex] ≈ 77.99 meters.
Hai vận động viên chạy trên cùng 1 đoạn đường, vận động
viên A chạy với vận tốc 12 m/s vận động viên B chạy với vận tốc 36
km/h. Hỏi vận động viên nào chạy nhanh hơn?
Answer:
Pemain A
Explanation:
Mengingat data berikut;
Kecepatan pemain A = 12 m/s
Kecepatan pemain B = 36 km/h
Untuk menentukan siapa pelari tercepat di antara dua pemain;
Pertama-tama, kita harus mengubah kecepatan menjadi satuan standar pengukuran yang sama.
Jadi, mari kita gunakan pengukuran umum meter per detik.
Konversi:
36 km/h = (36 * 1000)/(60 * 60)
36 km/h = 36000/3600
36 km/h = 10 m/s
Kecepatan pemain B = 10 m/s
Oleh karena itu, dibandingkan dengan kecepatan pemain A; pemain A lebih cepat.
On the graph of voltage versus current, which line represents a 3.0 resistor
Answer:
C. Line A
Explanation:
V = A*R
12V = 4A*R
R = 12V/4A
R = 3ohms
research paper on tsunamis (4-5 paragraphs minimum) 3) Tsunamis how they work….evacuation routes, structures built in oceans to protect, early detection, etc
Expectations:
A. Must be at least 4 full paragraphs. (probably will need to be more)
B. Must have an introductory paragraph that explains what the problem is, and the general overview of the paper.
C. Must Define Problem, including details and how it relates to Irvine
D. Main Body of Paper (probably couple of paragraphs) This is where you will show what you found out in your research. Make sure to include things like current use/techniques, as well as what technological advancements are being researched to improve your specific topic/issue.
E. Conclusion: Short summary of main points. Your Opinion on what is working and why (or what is not working and why)
F. Citations: Must use at LEAST 3 sources. Must site works at end, in either MLA or APA format, AND must site them where used in the paper (also MLA or APA format).
Make sure all work is yours and original. Copying is cheating. Using information from a source without citing is plagiarism.
can u guys help me with this question.
State one example of the cause of a short circuit. Explain your answer
Answer:
the positive and negative terminals of a battery are connected with a low resistance conductor.
Can the magnitude of a vector ever (a) be equal to one of its components, or (b) be less than one of its components? 9. Can a particle with constant speed be accelerating? What if it has constant velocity?
Answer:
a) the other components are zero, in the direction of one of the coordinate axes
b) the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction
9) constant velocity the acceleration must necessarily be zero,
constant speed can be accelerated since it may be changing the direction of the velocity vector
Explanation:
Vectors are quantities that have modulo (scalar) direction and sense.
a) If in a vector its magnitude is equal to one d its components implies that the other components are zero, therefore the vector must be in the direction of one of the coordinate axes
b) if the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction, other than the direction of the axes, that is
R² = x² + y²
where R is the magnitude of the vector e x, and are the components
9) When a particle has a constant velocity, the acceleration must necessarily be zero,
v = vo + a t
The bold letters indicate vectors If a = 0 implies that v = vo
If a particle has constant speed it can be accelerated since it may be changing the direction of the velocity vector, this type of acceleration has the name of centripetal acceleration
Si la experiencia de Torricelli se llevara a cabo al nivel del mar usando un aceite de 890
kg/m3
de densidad en lugar de mercurio, ¿cuál sería la altura que alcanzaría la columna de
aceite?
Answer:
La columna de aceite tendría una altura de 11,613 metros.
Explanation:
El modelo de Torricelli basa el cálculo de la presión atmosférica mediante el concepto de presión hidrostática, en donde la densidad del fluido ([tex]\rho[/tex]), en kilogramos por metro cúbico, es inversamente proporcional a la altura de la columna ([tex]h[/tex]), en metros, para una igual presión atmosférica ([tex]P[/tex]), en atmósferas. Entonces, podemos estimar la altura de la columna mediante el siguiente modelo:
[tex]\rho_{Hg}\cdot h_{Hg} = \rho_{o}\cdot h_{o}[/tex] (1)
Donde:
[tex]\rho_{Hg}[/tex] - Densidad del mercurio.
[tex]h_{Hg}[/tex] - Altura de la columna de mercurio.
[tex]\rho_{o}[/tex] - Densidad del aceite.
[tex]h_{o}[/tex] - Altura de la columna del aceite.
Si sabemos que [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], [tex]h_{Hg} = 0,760\,m[/tex] y [tex]\rho_{o} = 890\,\frac{kg}{m^{3}}[/tex], entonces la altura de la columna del aceite es:
[tex]h_{o} = \left(\frac{\rho_{Hg}}{rho_{o}} \right)\cdot h_{Hg}[/tex]
[tex]h_{o} = \frac{13600\,\frac{kg}{m^{3}} }{890\,\frac{kg}{m^{3}} }\times (0,760\,m)[/tex]
[tex]h_{o} = 11,613\,m[/tex]
La columna de aceite tendría una altura de 11,613 metros.
List out the fundamental and derived units
why our eyes donot feel cold ?
Answer:
The corneal tissue at the front of the eye has high thermal conductivity so it doesn't stray too far from body temperature even when the ambient temperature is extreme. For example, if the environment is at -11°C, the surface temperature of a human cornea only falls to 28.6°C.
Calculate the maximum absolute uncertainty for R if:
R = 9A / B
A = 32 +/- 2 seconds
B = 11 +/- 3 seconds
1 second
0.33 seconds
9 seconds
2 seconds
6 seconds
Answer:
ΔR = 9 s
Explanation:
To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value
The given expression is R = 2A / B
the uncertainty is ΔR = | [tex]\frac{dR}{dA}[/tex] | ΔA + | [tex]\frac{ dR}{dB}[/tex] | ΔB
we look for the derivatives
[tex]\frac{dR}{dA}[/tex] = 9 / B
[tex]\frac{dR}{dB}[/tex] = 9A ( [tex]- \frac{1}{B^2 }[/tex] )
we substitute
ΔR = [tex]\frac{9}{B}[/tex] ΔA + [tex]\frac{9A}{B^2}[/tex] ΔB
the values are
ΔA = 2 s
ΔB = 3 s
ΔR = [tex]\frac{9}{11}[/tex] 2 + [tex]\frac{9 \ 32}{11^2 }[/tex] 3
ΔR = 1.636 + 7.14
ΔR = 8,776 s
the absolute error must be given with a significant figure
ΔR = 9 s
Impulse can be defined as
The frequency at which something occurs
The change in velocity
The change in momentum from an force within the system
The change in momentum from a force outside of the system
Explanation:
The change in momentum from an force within the system.
Raquel establece que la rapidez del sonido en el aire en un día es de 346 m/s . Días después hace la misma medición obteniendo una rapidez de 340 m/ s . ¿Cuál será la temperatura del aire en cada día?
Answer:
La ecuación para la velodiad del sonido en aire esta dada por:
[tex]v = \sqrt{ \frac{\gamma*R*T}{M} }[/tex]
Resolviendo esto para T, obtenemos:
[tex]T = v^2*(\frac{M}{\gamma*R} )[/tex]
donde:
T = temperatura del aire en grados Kelvin
γ = constante adiabatica = 1.4
R = constante del gas ideal =
M = masa molar del aire = 29*10^(-3) kg/mol
Si sabemos que el primer día la velocidad es 346 m/s, tenemos:
v = 346 m/s
Ahora podemos reemplazar todos esos valores en la ecuación para T, y asi obtener:
[tex]T = (346m/s)^2*(\frac{29*10^{-3}kg/mol}{1.4* 8.31 J/mol*K} ) = 298.42 K[/tex]
Para el segundo día la velocidad es 340 m/s, entonces ese día la temperatura va a ser:
[tex]T = (340m/s)^2*(\frac{29*10^{-3}kg/mol}{1.4* 8.31 J/mol*K} ) = 288.16 K[/tex]
Answer the question based on this waveform.
A wave has a velocity of 24 m/s and a period of 3.0 s.
Choose the correct amplitude.
1/2 m
1/72 m
3 m
Cannot be determined from the given information
Answer:
Cannot be determined from the given information
Explanation:
Given the following data;
Velocity = 24 m/s
Period = 3 seconds
To find the amplitude of the wave;
Mathematically, the amplitude of a wave is given by the formula;
x = Asin(ωt + ϕ)
Where;
x is displacement of the wave measured in meters.
A is the amplitude.
ω is the angular frequency measured in rad/s.
t is the time period measured in seconds.
ϕ is the phase angle.
Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.
However, the given parameters can be used to calculate the frequency and wavelength of the wave.
A car drives 215km east and then 45km north. What is the magnitude of the cars displacement? Round you answer to nearest whole number.
Answer: [tex]219.65\ \text{km}[/tex]
Explanation:
Given
Car drives 215 km east and then 45 km North
Displacement is East direction is
[tex]\Rightarrow \vec{r_1}=215\hat{i}[/tex]
Now, the displacement from that to 45 km North is given by
[tex]\Rightarrow \vec{r_{21}}=45\hat{j}[/tex]
Net displacement is [tex]\vec{r_2}[/tex]
[tex]\Rightarrow \vec{r_2}=\vec{r_1}+\vec{r_{21}}\\\Rightarrow \vec{r_2}=215\hat{i}+45\hat{j}[/tex]
Magnitude of the displacement is
[tex]\Rightarrow \left | r_{2} \right |=\sqrt{\left ( 215 \right )^2+\left ( 45 \right )^2}\\\Rightarrow \left | r_{2} \right |=\sqrt{48250}\\\Rightarrow \left | r_{2} \right |=219.65\ \text{km}[/tex]
If we go on the top of the Mountain we will our weight increase or decrease?
Answer:
increase
Explanation:
guesshejwoowsnnsnakalalapqjhdjd
Answer:
our weight would decrease.
Explanation:
because of you burning all those calories and fat from hiking up the mountain.
A 105kg astronaut lands on the moon, with a weight of 170N. What is the acceleration due to gravity
Explanation:
given: m= 105kg,f=170N,a=?
F = m×a
a = f/m
a=170/105
therefore, a=1.62m/s^2
A student increased the resister of the voltmeter year not the circuit
Answer:
Say the full question I can't understand what it is
