Ask question

Ask question

All categories

4 years ago

6

A different car in the next lane accelerates from rest to 30 m/s at a rate of 6 m/s^2 for 5 seconds. How far does this car trave

l during this period?

1 answer:

mylen [45]4 years ago

8 0

You can use equation 4 for this:

$\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=0^2+2\left(6\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=75\,\mathrm m$

You might be interested in

If the ultraviolet photon has a wavelength of 249 nm and one of the photons emitted by the fluorescent material has a wavelength

melamori03 [73]

Answer:

601 nm

Explanation:

Energy of photon having wavelength of λ nm

= $\frac{1244}{\lambda}$ eV

Energy of 249 nm photon

= $\frac{1244}{249}$

=4.996 eV

Similarly energy of 425nm photon

= $\frac{1244}{425}$

=2.927 eV

Difference = 2.069 eV.

This energy will give rise to another photon whose wavelength will be

λ = $\frac{1244}{2.069}$

= 601 nm.

6 0

4 years ago

Please help! Will mark Brainliest :)

zepelin [54]

Answer:

i have to say whether its correct or wrong yep

3 0

3 years ago

Read 2 more answers

A uniform rod 8m long weighing 5kg is supported horizontally by two vertical parallel strings at p and q and at distance 2m and

Gala2k [10]

Answer:

Fp = 36 N

Fq = 58 N

Explanation:

Let the left end be the reference end with string p closest to it.

Let CCW moments be positive

Sum moments about p to zero

1(9.8)[2 - 1] + Fq[6 - 2] - 5(9.8)[8/2 - 2] - 1.5(9.8)[5 - 2] - 2(9.8)[7 - 2] = 0

Fq[4] = 23.5(9.8)

Fq = 57.575 ≈ 58 N

Sum moments about q to zero

1(9.8)[6 - 1] - Fp[6 - 2] + 5(9.8)[6 - 8/2] + 1.5(9.8)[6 - 5] - 2(9.8)[7 - 6] = 0

Fp = 35.525 N

or

Sum vertical forces to zero

Fp + 57.575 - (9.8)(1 + 1.5 + 2 + 5) = 0

Fp = 35.525 ≈ 36 N

4 0

3 years ago

If the truck has a mass of 2,500 kilograms , what is its momentum ? (V=45m/s) Express your amswer

vlabodo [156]

Answer:

The answer is 2,500 * 45 = 112500kgm/s

6 0

2 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago