All categories

2 years ago

Is erosion a constructive or destructive force and why?

Physics

1 answer:

Andrei [34K]2 years ago

5 0

Destructive force
It’s a force that slowly wears away mountains and other features on earths surface, such as erosion.

You might be interested in

Neglecting friction, of a pendulum bob has 100 joules of kinetic energy at the bottom of its swing, how much potential energy do

galben [10]

100 joule of kinetic energy

7 0

3 years ago

1.)how is climate different from weather

Zolol [24]

(sorry I can only answer no. 1)
the climate is the weather over a long period of time (could be days, weeks, months, or years) while weather is only how the day has been on one day (e.g. sunny)

6 0

2 years ago

If you lived on the Moon, would you see Earth go through phases? If so, would the sequence of phases be the same as those of the

marissa [1.9K]

Answer:

yes; yes

Explanation:

Phases of the moon refers to the shapes of the moon due to the lit part of it visible from the Earth. On a new moon day, the moon comes between the sun and the earth such that the lit portion is not visible from the Earth. On a full moon day, the earth comes between the sun and the moon and the whole lit part is visible.

When one would view the earth from the moon, the earth would also be visible as going through the phases. The order would be reversed. Understand this with the following example, On a new moon day, the Earth would be visible completely lit from the moon. So it will be full Earth day on the moon. On a full moon day, the lit side of the Earth would be completely away and hence, from the moon, new earth would be there.

8 0

2 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$