All categories

3 years ago

A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?

2 answers:

6 0

The car's acceleration is (6 m/s) / (12 sec) = 0.5 m/s² .

   Force = (mass) x (acceleration)

   = (2,300 kg) x (0.5 m/s²)

   = 1,150 Newtons

pav-90 [236]3 years ago

4 0

F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs

You might be interested in

When objects are heated they tend to expand because

densk [106]

Hello!
===
When objects are heated, their molecules tend to vibrate fast. As they vibrate, the space between each atom increases. This keeps on happening, and the object expands until it has cooled down.
===
Hope this helps! :)

5 0

4 years ago

If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assu

Andreas93 [3]

Answer:

Explanation:

In such situations we cannot determine which one is more valid as both serves the purpose well.

Two theories are carried out in different environment and circumstance keeping different parameters and one can opt for any number of ways to carry out that experiment but what matter at the end is the accuracy they bring.

Each of the theory is a new discovery and follows all the possible logical rules hence it is not possible to decide which one is more valid.

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

3 years ago

A small piece of Styrofoam packing material is dropped from a height of 2.60 m above the ground. Until it reaches terminal speed

Vadim26 [7]

Answer:

Explanation:

a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m

So terminal speed V = 2 / 4.5

= .444 m /s

When it attains terminal speed , acceleration becomes zero

0 = g - B x .444

B = 22.25 s⁻¹

b ) At t = 0 , v = 0

a = g - B v

a = g at t = 0

c ) When v = .15

a = g - 22.25 x .15

= 9.8 - 3.31

= 6.5 m /s²

7 0

4 years ago

Can an ordinary object, like a motorcycle, be mass-less? Yes or No

Drupady [299]

Answer:

no.

Explanation:

because the mass of an object never changes.

4 0

3 years ago