8. Una fuerza de 100 N actúa sobre un cuerpo que se desplaza a lo largo de

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

scoray [572]

Jim walks 7 miles from the starting point to the end point
i guess so

3 0

4 years ago

Read 2 more answers

What does the first law of thermodynamics state?

Paraphin [41]

Answer:

b) There is and always will be the exact same amount of energy in the universe.

Explanation:

The first law of thermodynamics establishes the principle of energy conservation, that is to say that in any process the energy is not destroyed, it only transforms into another type of energy, so when analyzing thermodynamic systems it is established that the energy that enters A system is the same that comes out. For the above reason, it can be inferred that if the energy is not destroyed or created there will always be the same amount of energy in the universe.

8 0

3 years ago

Give reasons:

Nata [24]

Explanation:

1) 1 : to Prevent a burning we use tongs

2:it can burn our skin hoa it's help

2) 1: because pulley helps in water draw from a well

3) 1: second class lever has mechanical advantage more than one as load is in between fulcrum an effort making the effort arm longer than the load arm

6 0

4 years ago

Ca+ is an example of a(n)

umka21 [38]

Answer:

Calcium

hope this helped mah dude :>

3 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

4 years ago