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Lera25 [3.4K]
3 years ago
8

Convection currents in air and water occur because _

Physics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0

warm fluids are less dense than cold fluids

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Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (
pentagon [3]

Answer:

q=2.997\times 10^{-4}C

Sign-Negative

Explanation:

We are given that

Electric field =E=100NC^{-1} (Radially downward)

Acceleration=0.19 ms^{-2}(Upward)

Mass of charge=3 g=3\times 10^{-3}kg

1kg=1000g

We have to find the magnitude and sign of  charge would have to be placed on a penny .

By newton's second law

\sum F_y=ma

\sum F_y=qE-mg

Substitute the values then we get

qE-mg=ma

Substitute the values then we get

q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)

100q-29.4\times 10^{-3}=0.57\times 10^{-3}

100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}

q=\frac{29.97\times 10^{-3}}{100}

q=2.997\times 10^{-4}C

Sign of charge =Negative

Because electric force acting  in opposite direction of electric field therefore,charge on penny will be negative.

4 0
3 years ago
In a game of tug of war, team one pulls to the right with a force of 500 newtons and team two pulls to the left with a force of
seraphim [82]

Answer:

Explanation:

There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.

That is completely arbitrary. It could be the other way around. It does not matter.

Left is minus so: - 600 N   is the force going left.

Right plus so: + 500 N

Now just add.

Net Force = +500 - 600

Net Force = - 100 N

So the Net Force is - 100 N going to the left.

8 0
2 years ago
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the
pantera1 [17]

Answer:

Explanation:

F = k*n

F = 30N

k = ???

N = 0.73 M

F = k* N

30N = k * 0.73

k = 30N / 0.73

k = 41.1

5 0
2 years ago
The density of gold is 19 300kg/m cube. what is the mass of gold cube with the length 0.2015m?
Sergeeva-Olga [200]

Answer:

157.9 kg

Explanation:

Density: This can be defined as the ratio of the mass of a body and it's volume.

The S.I unit of density is kg/m³.

From the question,

Density = Mass/volume

D = m/v............................ Equation 1

Where D = Density of gold, m = mass of gold, v = volume of gold.

make m the subject of the equation

m = Dv.................... Equation 2

Since the gold is a cube,

v = l³................... Equation 3

Where l = length of the gold cube.

Substitute equation 3 into equation 2

m = Dl³............... Equation 4

Given: D = 19300 kg/m³, l = 0.2015 m

Substitute into equation 4

m = 19300(0.2015)³

m = 157.9 kg.

4 0
3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
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