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coldgirl [10]
3 years ago
7

Which of the following represents an element?

Physics
1 answer:
Digiron [165]3 years ago
7 0
H2 is the correct answer
You might be interested in
PLEASE HELP ME WITH THIS PROBLEM
valentinak56 [21]

1) The mass of the continent is 2.13\cdot 10^{21} kg

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

1)

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3

The mass of the continent is given by

m=\rho V

where:

\rho = 2790 kg/m^3 is its density

V=7.62\cdot 10^{17} m^3 is its volume

Substituting, we find the mass:

m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg

2)

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is

v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s

Now we can find the kinetic energy of the continent, which is given by

K=\frac{1}{2}mv^2

where

m=2.13\cdot 10^{21} kg is the mass

v=5.08\cdot 10^{-10}m/s is the speed

Substituting,

K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J

3)

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

K=\frac{1}{2}mv^2

where

v is the speed of the man

And solving the equation for v, we find his speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo
luda_lava [24]

Answer:

?

Explanation:

8 0
2 years ago
Explain why the top of the loop cannot be the same height as (or higher than) the top of the first hill. Assume the roller coast
Ivahew [28]

Answer:

By conservation of energy, it can climb up to a height equal to that it went down before. However, due to the friction in the machines, the total mechanical energy of the roller coaster will decrease. As a result, the first "hill" of many roller coasters are the highest, but the followings will have decreasing heights.

Explanation:

7 0
3 years ago
Read 2 more answers
A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

5 0
3 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
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