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TEA [102]
1 year ago
5

The theory of plate tectonics describes this process as a gradual movement of the tectonic plates across the Earth’s lithosphere

. Using the map of plate motion, predict where the continents would most likely lie 100,000 years in the future. South America and North America would become one continent. South America and Africa would be farther apart. Australia would move toward the South Pole. North America and Europe would be closer together.
Physics
2 answers:
RideAnS [48]1 year ago
8 0
<span>South America and Africa would be farther apart.</span>
sweet-ann [11.9K]1 year ago
3 0

Answer:

South America and Africa would be farther apart.

Explanation:

Continental drift is the displacement of the continental masses with respect to each other. With the development of plate tectonics, this motion could be adequately explained. Currently, America is moving westward from Africa and Europe. Therefore, in 100.000 years South America and Africa would be farther apart.

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10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
2 years ago
Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc
lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

8 0
2 years ago
The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune
Veronika [31]
Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
\frac{r^3}{T^2}= \frac{GM}{4 \pi^2} (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
1 AU = 1.5 \cdot 10^{11} m
so the radius of the orbit is
r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
T=\sqrt{ \frac{4 \pi^2}{GM} r^3} =  \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years
3 0
1 year ago
A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial
Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

               t=\frac{2usin\theta}{g}

           Here θ = 30° and t = 20 s

a) t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s

  Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

                  H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m

      Maximum altitude = 490.5 m

c) Range of projectile is given by

                              R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m

    Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

   Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

   We have equation of motion s = ut + 0.5 at²

   Horizontal motion

                         u = 169.91 m/s

                         a = 0 m/s²

                          t = 15 s

                Substituting

                          s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

      Vertical motion

                         u = 98.1 m/s

                         a = -9.81 m/s²

                          t = 15 s

                Substituting

                          s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

   \texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m

   Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
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