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Leviafan [203]
3 years ago
5

A student would like to determine how heating a liquid changes its volume. The student hypothesizes that the liquid will increas

e in volume. The following list shows the steps taken by the student in order to test the hypothesis.
Select the liquid to test.
Place the liquid in a sealed container.
Use a Bunsen burner to heat the liquid by 10°C.
Measure the volume of the liquid.
Record the results.
What is wrong with how the student conducted the investigation?
Physics
2 answers:
Verizon [17]3 years ago
4 0
10 degrees c f is not hot
Sholpan [36]3 years ago
4 0

Answer:

The volume of the liquid should be measured before it is heated.

Explanation:

During an experiment to test how a variable changes a substance, it is important to first observe and record the characteristics of the substance before the variable is introduced. In this case, the variable is heat energy. The student must first record the liquid[?]s volume, before attempting to change the volume by introducing heat energy.

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At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is per
geniusboy [140]

Answer:

\tau = 7.63 Nm

Explanation:

As we know that moment of force is given as

\tau = \vec r \times \vec F

now we have

\vec r = 1.2 m

\vec F = 14 N

now from above formula we have

\tau = r F sin\theta

here we know that

\theta = 27 degree

so we have

\tau = (1.2)(14) sin27

\tau = 7.63 Nm

3 0
3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant
Alika [10]

Answer:

Isolated or Closed system, both are correct

4 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
A car accelerates from rest to 19 m/s in 6 seconds. What is the acceleration of the car?
choli [55]

Answer:

the car is moving so that how it gos so fast

Explanation:

7 0
3 years ago
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