HELP QUICK!!!

Find the acceleration of a car that can go from rest to 50 km/h in 13 s

ziro4ka [17]

The acceleration of the car is 1.067 m/ $s^{2}$ .

Explanation:

Acceleration is the measure of change in velocity experienced by any object for a given time period. So it is determined as the ratio of difference in the velocity to the time period.

As here the initial velocity is stated as zero, so u = 0. And the final velocity is termed as 50 km/h. Then we have to determine the acceleration in 13 s. So here we have to convert the units as common units. Thus, 50 km/h should be converted to m/s as $\frac{50*1000}{3600}=13.88 m/s$

So now, the initial velocity u = 0 and final velocity v = 13.88 m/s and the time period is given as t = 13 s.

$Acceleration = \frac{v-u}{t}=\frac{13.88-0}{13}=1.067 m/s^{2}$

So the acceleration of the car is 1.067 m/ $s^{2}$ .

8 0

3 years ago

PLEASE HELP!!!!I IM GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (30pts)

maksim [4K]

Kinetic Energy

=>It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

Potential Energy 

=>potential energy is the energy held by an

 object because of its position relative to other

 objects, stresses within itself, its electric

 charge, or other factors.

<h2>Difference:</h2>

=>Potential energy is a stored energy on the other hand kinetic energy is the energy of an object or a system's particle in Motion.

6 0

3 years ago

A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from

Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of incidence beam striking the water is $\theta = 49.63^o$

c the angle of refraction beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is

$h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

$tan \theta = \frac{100}{85}$

$\theta = tan^{-1}(\frac{100}{85} )$

$= 49.63^o$

According to Snell's law

$\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water = 1.333

$i$ is the angle of incidence

$\mu_{air}$ is the refractive index of air = 1

r is the angle of refraction

Substituting values accordingly

$1.33 * sin (40.37) = 1 * sin(r)$

Making r the subject of the formula

$r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

$= 59.7^o$

looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°

i.e $90 -59.7 = 30.3$ °

From the diagram we see that the height target above sea level can be obtained by this relation

$tan \theta = \frac{h}{214}\\$

Where h is the is the height

$tan(30.3) = \frac{h}{214}$

$h = 214 * tan (30.3)$

$=125.05m$

4 0

3 years ago

PHYSICS can somebody please answer this

tankabanditka [31]

Yes the answer is recoiled of the Gaian and the hash of attorney then the distance must be bábale of the original number

8 0

3 years ago

In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea

Vikentia [17]

For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).

fringe = (delta t) / (λ/c)

We can find (delta t) with the equation:

delta t = [v^2(L1+L2)]/c^3

Derivation of this formula can be found in your physics text book. From here we find (delta t):

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.

4 0

3 years ago