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3 years ago

What conditions are typcial of the soil the deeper we dig into it?

1 answer:

8 0

When we first dig into soil, it's nice and soft.( Depending on what soil it is; usually topsoil)

But when we go deeper into it, there are rocks and the soil is harder. There is bedrock at the bottom, subsoil in the middle, and topsoil at the top, which is where we walk.

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A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. A) cm/h B) cm/min C) m/h

Crazy boy [7]

A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h

7 0

3 years ago

In an emergency situation, firemen need to respond as quickly as possible. If a fireman is responding from the second floor, how

rjkz [21]

Answer:

4m/s

Explanation:

May be different considering how long the pole is and how heavy the firefighter is.

8 0

3 years ago

A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.A

kenny6666 [7]

Answer: 1.76 Nm

Explanation:

If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.

In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:

τ = F * r = 16 N. (0.11) m = 1.76 Nm

7 0

3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

3 years ago