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2 years ago

The process of wind blowing sand from one location to another is called

Physics

2 answers:

erastovalidia [21]2 years ago

6 0

Answer:

deposition

Explanation:

you are depositing something over

lesya [120]2 years ago

3 0

Answer:

weathering

Explanation:

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A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a

Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged (x)= 0.13 m

now,

weight of the mass attached = Kx

m g = k x

0.35 x 9.8 = k x 0.13

k = 26.38 N/m

now, using conservation of energy

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2$

$v = \sqrt{\dfrac{kx'^2}{m}}$

$v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}$

$v = \sqrt{1.6958}$

v = 1.30 m/s

6 0

3 years ago

A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5

Paul [167]

Answer:

Energy lost due to friction is 22 J

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball $KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}\times 4\times 6^2=72J$

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

$KE=\frac{1}{2}\times 4\times 5^2=50J$

Therefore energy lost due to friction = 72 -50 = 22 J

8 0

3 years ago

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

2 years ago

Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of

sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .

I₁ / I₂ = ω₂ / ω₁

I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3 x 2.3 x 10¹⁹ / M

= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0

2 years ago

Magnetic pole reversal worksheet

Likurg_2 [28]

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8 0

3 years ago