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worty [1.4K]
3 years ago
11

Which tool can be used to measure the volume of a liquid to one decimal place

Physics
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

a beaker

Explanation:

:)

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How many orbitals are allowed in a subshell if the angular momentum quantum number for electrons in that subshell is 3?
svp [43]

Answer:

7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.

Explanation:

We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.

    Range for given l lies between -l to +l .

The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .

    Total number of orbitals are 7.

4 0
3 years ago
A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally
dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

4 0
3 years ago
All of the following are important reasons to take notes except.
jeka57 [31]

Its a waste of time, you have to not only write it down, but study it after too . other than that notes are great.

4 0
3 years ago
Read 2 more answers
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
3 years ago
Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. The best type of gr
lara31 [8.8K]

Answer:

B line graph

Explanation:

sorry if I am wrong

7 0
3 years ago
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