All categories

3 years ago

What is the degree of precision of the graduated cylinder 0.01 mL 0.1 mL 1 mL 10 mL

Physics

2 answers:

Elodia [21]3 years ago

5 0

Answer:

0.01 ml

Explanation:

Graduated cylinder is used to measurement of volume of liquid.Generally it gives precise result up to 2 digit or up to 3 digit.

Graduated cylinder are of different types .10 mL graduated cylinder measure up to 2 digit while 100 mL graduated cylinder measure up to 1 digit .

So here it is not size of cylinder is not specified so for most precise result the degree of precision of graduated cylinder will be 0.01 mL (up to 2 didgit).

enot [183]3 years ago

4 0

I believe it is 0.1mL or letter B on Edgenuity.

You might be interested in

A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circuit with two lightbulbs rated at 3.0

Blababa [14]

Answer:

$R = 24 ohm$

Explanation:

As we know that terminal voltage and EMF of cell is given as

$V = EMF - ir$

$ir = 12 - 11.8$

$ir = 0.2 Volts$

resistance of two bulbs is given as

$R = \frac{V^2}{P}$

$R = \frac{12^2}{3}$

$R = 48 ohm$

now these two bulbs are connected in parallel

so equivalent resistance is given as

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{48} + \frac{1}{48}$

so we have

$R = 24 ohm$

3 0

3 years ago

A fanis rotating at a constant 360.0 rev/min. What is the magnitude of the acceleration of a point on one of its blades 10.0 cm

Mariana [72]

Answer:

142m/s2

Explanation:

The solution is in the attached file below

4 0

3 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

4 years ago

Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm

Sphinxa [80]

Answer:

The speed is $s = 5.857 m/s$

The distance is $D = 22.4 \ m$

Explanation:

From the question we are told that

The speed of the banana is $v = 16 \ m/s$

The distance from my location is $d = 8.2 \ m$

The time taken is $t = 1.4 \ s$

The speed of the ice cream is

$s = \frac{d}{t}$

substituting values

$s = \frac{8.4}{1.4}$

$s = 5.857 m/s$

The distance of separation between i and Valdimir is the same as the distance covered by the banana

$D = v * t$

substituting values

$D = 16 * 1.4$

$D = 22.4 \ m$

3 0

3 years ago

Derive an expression for the de Broglie wavelength of a thermal neutron and calculate its value at the temperature T = 290 K. Th

Vladimir [108]

Answer:

1.479 A

Explanation:

Expression for de Broglie wave length

λ = $\frac{h}{\left ( 2mKE \right )^{.5}}$

Where m is mass of neutron ,h is plank constant and KE is kinetic energy of

proton.

Kinetic Energy ( KE) = 3 / 2 kT

=1.5 X 1.38 X 10⁻²³ X 290 = 600.3 X 10⁻²³ J

$\sqrt{2mKE}$ = $\sqrt{2\times1.67\times10^{-27}\times600.3\times10^{-23}}$

44.777 x 10⁻²⁵

λ = $\frac{6.626\times10^{-34}}{44.777\times10^{-25}}$

=1.479 A.

7 0

3 years ago