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2 years ago

Which method of measurement would be accurate but lack precision?

Physics

1 answer:

kodGreya [7K]2 years ago

7 0

Answer:

Explanation:

reading the volume of water in a graduated cylinder which can be read to the nearest mL is accurate, it lacks precision due to the bottom meniscus formed.

the bottom meniscus may cause a wrong reading due to refraction of light

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If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re

-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = $\frac{1}{2}kx^2$ ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = $\frac{1}{2}k\times0.021^2$

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = $\frac{1}{2}\times13605.44\times (0.062)^2$

U = 26.149 J

Also,

Work done = change in energy

W = 26.149 - 3.0 = 23.149 J

4 0

3 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

k = 3.5 N/m

4 0

3 years ago

A 2.6 kg mass attached to a light string rotates on a horizontal,

Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

maximum mass the string can support before breaking, m = 17.9 kg
radius of the circle, r = 0.525 m

The maximum speed the mass can have before it breaks is calculated as follows;

$T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s$

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

8 0

3 years ago

A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in. part a what is the car's sp

Talja [164]

Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s

8 0

3 years ago

Which statement below best describes the relationship between the voltage, resistance, and current?

aliina [53]

Answer:

As voltage increases, current increases and resistance stays the same .

Explanation:

Ohm's law gives the relationship between the voltage, resistance, and current. The mathematical form of Ohm's law is given by :

$V\propto I\\\\V=IR$

R is resistance

I is current

V is voltage

So, as voltage increases, current increases and resistance stays the same. The correct option is (A).

4 0

3 years ago