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2 years ago

7

In a digital multimeter, what changes inside the meter when you push the button to change from 200v range to the 20v range? by w

hat factor does it change? does it increase or decrease?

1 answer:

charle [14.2K]2 years ago

6 0

So we want to know what changes inside the multimeter when we change the voltage range from 200 V to 20 V, by what factor and does it increase or decrease. What we want when trying to measure the voltage with a multimeter is that a minimal current passes trough the mulitmeter so when we change the voltage range, we decrease the resistance by a factor of 10 because the voltage is decreased by a factor of 10.

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

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Α $=2.4 \frac{m}{s}$

β $=1.2 \frac{m}{s^2}$

$x(t)=at$

$y(t)=3-$ β $t^2$

$Vx(t)=$ α

$Vy(t)=-2$ β $t$

$vectorV=[$ α $;-2$ β $t]$

$ax(t)=0$

$ay(t)=-2$ β $t$

$vector a [0;-2$ β $t]$

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2 years ago

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

kati45 [8]

Answer:

velocity = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = $\frac{\pi }{4}$ × (4.50× $10^{-2}$ )² × velocity

solve it we get

velocity = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = $\frac{\pi }{4}$ × 3 × (4.50× $10^{-2}$ )² × velocity

velocity = 52.4 m/s

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2 years ago

A battery that produces a potential difference V is connected to a 10-W light bulb. Later, the 10-W bulb is replaced with a 20-W

uysha [10]

Answer:

The battery supply the greatest current with the 20-W bulb

Explanation:

The electrical power at a certain time by a two-terminal device is the product of the potential difference between the terminals and the current intensity that passes through the device. For this reason the power is proportional to the current and voltage. Mathematically can be written as:

$P=VI$

Let:

$P_1=Electric\hspace{3}power\hspace{3}dissipated\hspace{3}by\hspace{3}the\hspace{3}10-W\hspace{3}bulb\\P_2=Electric\hspace{3}power\hspace{3}dissipated\hspace{3}by\hspace{3}the\hspace{3}20-W\hspace{3}bulb$

The voltage remains constant for every case so:

$I_1=Current\hspace{3}supply\hspace{3}by\hspace{3}the\hspace{3}battery\hspace{3}with\hspace{3}the\hspace{3}\hspace{3}10-W\hspace{3}bulb\\I_2=Current\hspace{3}supply\hspace{3}by\hspace{3}the\hspace{3}battery\hspace{3}with\hspace{3}the\hspace{3}\hspace{3}20-W\hspace{3}bulb$

Then:

$P_1=V*I_1=10W\\P_2=V*I_2=20W$

Analyzing the equations it makes sense that the battery supplies the highest current with the 20 W bulb, because the power only depends on the voltage and the current, if the voltage remains constant, the only reason for the power to increase is due to current.

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2 years ago

When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th

earnstyle [38]

Answer: The spring of the spring is 25 N/m.

Explanation:

Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)

Oscillation is 4 sec = 20

Oscillation in 1 sec = $\frac{20}{4}=5$

Frequency of the vibration of the spring = $5 s^{-1}=5 Hz$

Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring of the spring is 25 N/m.

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