A cable raises a mass of 198.0 kg with an acceleration of 1.2 m/s2. What force (in N) of tension is in the cable

AlekseyPX

Friction is caused by the uneven surfaces of touching objects

3 0

2 years ago

Calculate the pressure exerted on the floor when an elephant who weighs 6000 N stands on one foot which has an area of 20 m2

kondor19780726 [428]

Answer:

300 Pascal

Explanation:

Given

weight or force (F) = 6000 N

area (A) = 20 m²

pressure (p) = ?

we know

the force acting normally per unit area is pressure. So

P = F / A

= 6000 / 20

= 300 Pascal

Hope it will help :)

3 0

3 years ago

How do i solve?<br><br>a = bc<br>(b)

stira [4]

I don’t even know I’m so dumb.

7 0

3 years ago

An atom of arsenic has how many electron-containing orbitals

irina [24]

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

$1s^{2}$

$2s^{2}$

$2p^{6}$

$3s^{2}$

$3p^{6}$

$3d^{10}$

$4s^{2}$

$4p^{3}$

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

8 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago