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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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Two particles of a gas collide. Why is this considered an elastic collision? (1 point)

dezoksy [38]

I don’t know really really I don’t know

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2 years ago

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A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the

marin [14]

Answer:

0.395 m

2.4 m/s

Explanation:

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$x$ = initial position of spring from equilibrium position = 0.21 m

$v_{i}$ = initial speed of the cart = 2.0 ms⁻¹

$A$ = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

$(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m$

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$A$ = amplitude of the oscillation = 0.395 m

$v_{o}$ = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

$(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}$

5 0

3 years ago

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Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$