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iragen [17]
3 years ago
14

If a melon has a a mass of 1 kg, how much does the melon weigh?

Physics
1 answer:
My name is Ann [436]3 years ago
6 0

Explanation:

1kg = approximately 10N

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What is the formula for silver nitrate
serious [3.7K]

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Explanation:

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4 years ago
Where is the highest-energy electron found in an atom of hydrogen?
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In the basic hydrogen atom, shown below left, the cloud is densest in the center and thins out with distance from the nucleus, which means the electron is most likely to be found near the nucleus, in a region about 1/20 nm in size.

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3 years ago
Magnesium is a shiny, flexible metal that can be burned in the presence of air.
alukav5142 [94]

Answer:

A new substance was formed

Explanation:

According to this question, a shiny and flexible metal called Magnesium (Mg) is burnt in air to produce a white powder that has no shiny or flexible properties, however, has more weight than the magnesium metal itself.

This is possible because a CHEMICAL CHANGE has occured, hence, a new substance has been formed. The formation of a new substance during the burning process (chemical reaction), induced the increase in mass.

8 0
3 years ago
An electric dipole consisting of charges of magnitude 2.00 nC separated by 8.40 μm is in an electric field of strength 1390 N/C.
amid [387]

Answer:

(a) The magnitude of the electric dipole moment is 1.68 x 10⁻¹⁴ C.m

(b) The difference between the potential energies ΔU, is 4.6704 x 10⁻¹¹ J

Explanation:

Given;

magnitude of charge, q = 2 nC = 2 x 10⁻⁹ C

distance of separation, d = 8.4 μm = 8.4 x 10⁻⁶ m

strength of electric field, E = 1390 N/C

(a) the magnitude of the electric dipole moment

p = qd

p = (2 x 10⁻⁹ C)(8.4 x 10⁻⁶ m)

p = 1.68 x 10⁻¹⁴ C.m

(b) the difference between the potential energies for dipole orientations parallel and anti-parallel to E

ΔU = U(180) - U(0)

ΔU = 2pE

ΔU = 2(1.68 x 10⁻¹⁴ )(1390)

ΔU = 4.6704 x 10⁻¹¹ J

6 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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