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2 years ago

what does the presence of the mosesaurus fossils tell the initial location and positioning of soyth america africa and antartica

1 answer:

3 0

The remains of Mesosaurus- a crocodile-like reptilian tells United States that South America, Africa and Antarctica were once joined along which implies they were once an enormous continent. However, because of the idea called the geological phenomenon, this one massive continent began to break so making individual continents.

You might be interested in

What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.

Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler

7 0

2 years ago

Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm

Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]

here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]

= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0

2 years ago

A 120.0 kg crate is placed on a 15.00°

Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0

2 years ago

A 0.5 kg air-track car is attached to the end of a horizontal spring of constant k = 20 N/m. The car is displaced 15 cm from its

Dimas [21]

Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.

1/2kx^2 = 1/2mv^2

Solve to find velocity. Remember, the spring is displaced .15 m, not 15!

To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.

For problem C I don't know, haven't done that yet in my class. Sorry!

3 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your

Nataly [62]

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

$1.66 = 0 + 0t + \frac{1}{2}9.8t^2$

$1.66 = 4.9t^2$

$\frac{1.66}{4.9} = t^2$

$\sqrt{0.339} = t\\ t = 0.582s$

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

5 0

3 years ago