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3 years ago

If you could visit Pangaea, what animals would you most likely find there?

Physics

2 answers:

HACTEHA [7]3 years ago

6 0

Answer:

As given below.

Explanation:

1. Dinosaurs

2. Melted,

3. The skin,

4. It has to cool,

5. Gradual,

6. Sliding under,

7. The Antarctic plate must have once been closer to the equator,

8. Off the West Coast of North America,

9. Mountains

10. Plates move toward each other at convergent boundaries, and side-to-side at transform boundaries.

katen-ka-za [31]3 years ago

5 0

1B
2D
3C
4A
5B
6D
7B
8C
9A
10C

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Starting from rest, a car travels 18 meters as it accelerates uniformly for 3.0 seconds. What is the magnitude of

omeli [17]

Answer:

$a=4\frac{m}{s^2}$

Explanation:

Hello.

In this case, for this uniformly accelerated motion in which the car starts from rest at 0 m/s and travels 18 m in 3.0 s, we can compute the acceleration by using the following equation:

$x_f=x_0+v_0t+\frac{1}{2}at^2$

Whereas the final distance is 18 m, the initial distance is 0 m, the initial velocity is 0 m/s and the time is 3.0 s, that is why the acceleration turns out:

$a=\frac{2(x_f-v_ot)}{t^2} =\frac{2(18m-0m/s*3.0s)}{(3.0s)^2}\\ \\a=4\frac{m}{s^2}$

Best regards.

8 0

4 years ago

calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared usin

Lynna [10]

Answer:

minimum number of photon is 4.05 × $10^{7}$

Explanation:

given data

energy = 50 keV = 50 × $10^{3}$ eV = 50 × $10^{3}$ × 1.602× $10^{-19}$ J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E = n × h × ν .......................1

put here all these value

50 × $10^{3}$ = n × 6.6× $10^{-34}$ × c/ 1× $10^{-3}$

50 × $10^{3}$ × 1.602× $10^{-19}$ = n × 6.6× $10^{-34}$ ×( 3 × $10^{8}$ / 1× $10^{-3}$ )

solve it and find n

n = 4.05 × $10^{7}$

so here minimum number of photon is 4.05 × $10^{7}$

3 0

3 years ago

Is gravity a force ?

alexandr402 [8]

Answer:

Gravity ( gravity , weight, traction force) is the force acting on any body on or near the surface of the celestial body directed at its center. Gravity, that is, the force with which the earth , moon , planets, and other celestial bodies or their systems act on other bodies, is theoretically exercised at any distance, but is practically examined on the surface of these bodies as well as at short distances.

6 0

3 years ago

Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How

Pachacha [2.7K]

Answer:

The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Explanation:

Given;

speed of the faster car, v₁ = 60 mi/h

speed of the slower car, v₂ = 55 mi/h

Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

$\frac{x}{55} - \frac{x}{60} = \frac{15}{60}$

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

$\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles$

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

4 0

3 years ago

A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m 2, point, 0, start frac

dem82 [27]

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

7 0

3 years ago

Read 2 more answers