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2 years ago

10

What is the result of seafloor spreading?

1 answer:

Over [174]2 years ago

6 0

Answer:

Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

Explanation:

eventually the crust cracks.

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A lamp draws a current of 0.1 A when it is connected to a 122-V source. (

Setler79 [48]

The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W

4 0

3 years ago

An incompressible fluid (water) is flowing through a pipe of diameter 20 cm with

sergey [27]

Answer:

115 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Plugging in values:

(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²

P = 115,000 Pa

P = 115 kPa

3 0

3 years ago

A mover does 422 J of work pushing a crate 8.39 m. How much force did he exert?

kicyunya [14]

Answer:

50.3N

Explanation:

Work done = force x distance

422J. = force x 8.39m

÷8.39 both side to get force

Force is 50.3N to 1 d.p.

Check:

50.3 x 8.39=422.017J

Same as 422J to 1 d.p

4 0

2 years ago

Read 2 more answers

Ai là người phát hiện trái đất hình cầu đầu tiên ?

Scorpion4ik [409]

Answer:

Can't understand the language

5 0

2 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago