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3 years ago

7

Match each fossil fuel with its common use.

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

gasoline and natural gas

electricity and coal

heating homes and gas

Explanation:

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In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p

Delicious77 [7]

<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
m1 * v1 + m2 * v2 = m * v
1 * 1 + 5 * 0 = (1 + 5) * v
1 = 6 * v
v = 1/6 m/s
So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>

6 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then th

nlexa [21]

I think that this is false but I am not sure

5 0

3 years ago

Suppose a car approaches a hill and has an initial speed of

kvv77 [185]

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

5 0

3 years ago

PLZZZ HELP 100 POINTS WILL MARK BRAINELST

Kitty [74]

Explanation:

Crust...molten

a. Oceanic, iron

b. Continental, silicates

c. less

3. Mantle, Denser

a. Lithosphere

b. Asthenosphere

4. Core

a. elements, rocks

b. liquid, magnetic

(I guess the liquid should come after the is)

Couldn't answer all but wanted to help

3 0

3 years ago

Read 2 more answers