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Phantasy [73]
3 years ago
9

How do molecules at warm temperatures differ from molecules at cool temperatures? Question 1 options: At warm temperatures, mole

cules move around more. At warm temperatures, molecules grow in size. At warm temperatures, molecules exchange electrons more easily. At warm temperatures, moleucles combine with each other more easily.
Physics
2 answers:
yaroslaw [1]3 years ago
7 0

Answer:

A. At warm tempetures, molecules move around more.

Explanation:

I'm at k12 and I just took the test got it right. Physical Science: Unit 2 Test

lutik1710 [3]3 years ago
5 0

Answer:

At warm temperatures, molecules move around more.

plz mark me as brainliest.

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Solve the equation 3x^2+8x+5=0
Pachacha [2.7K]
3x^2+8x+5=0\\
3x^2+3x+5x+5=0\\
3x(x+1)+5(x+1)=0\\
(3x+5)(x+1)=0\\
x=-\dfrac{5}{3} \vee x=-1
4 0
3 years ago
A student on her way to school walks eastward in a straight line 20.0 meters towards the bus stop, but realizes she dropped her
larisa [96]

Answer:

Total displacement will be 47 meter

Total distance will be 83 meters

Explanation:

We have given that first the student go eastward towards bus stop 20 meters

But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters

So displacement = 20-18 = 2 meters

And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters

Total distance traveled by the student = 20+18+45 = 83 meters  

3 0
3 years ago
The unit of electric potential or electromotive force is the
Natalka [10]
That would be called VOLT
4 0
3 years ago
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
RoseWind [281]

Answer:

9 Brainly hahaha ............huh

4 0
2 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
3 years ago
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