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3 years ago

12

While traveling south on an expressway, a car traveling 60 mph (miles per hour) slows down to 30 mph in 12 minutes due to traffi

c conditions. Calculate the acceleration.

1 answer:

Cloud [144]3 years ago

5 0

10mph is the acceleration that u are looking for

You might be interested in

True or false? An object at rest has an instantaneous acceleration of zero.

Tpy6a [65]

True. it’s at rest which means it’s not moving so there’s no acceleration

8 0

3 years ago

Your current exam mean is 97.2. if you receive a 99 on the next exam, this will have the effect of ______.

GarryVolchara [31]

If your current exam mean is 97.2. and you receive a 99 on the next exam, then this will have the effect of increasing the mean.

<h3>What is the mean?</h3>

In statistics, the mean is an average value used to calculate when taking different measurements, which can be fundamental to collecting statistically significant information.

In conclusion, if your current exam mean is 97.2. and you receive a 99 on the next exam, then this will have the effect of increasing the mean.

Learn more about the average mean here:

brainly.com/question/20118982

#SPJ1

8 0

1 year ago

I'm trying to find the potential energy of a planet using formula G M(sun) x m(planet) / r. I have given G - newton's graviation

ludmilkaskok [199]

Answer:

$-5.39\times10^{33} J$

Explanation:

Potential energy =

$-\frac{GMm}{r}\\=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times5.97\times10^{24}}{147/1\times10^{9}}\\=-5.39\times10^{33} J$

7 0

2 years ago

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

3 0

3 years ago

A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At

NikAS [45]

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

$B=\frac{1}{2} \mu \epsilon_0 r$

Where,

$\mu =$ Permeability constant

$\epsilon_0 =$ Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

$B \propto r$

Then the distance relationship would be given by

$\frac{r}{R} = \frac{B}{B_{max}}$

$r =\frac{B}{B_{max}} R$

$r = \frac{0.5B_{max}}{B_{max}}R$

$r = 0.5R$

On the outside, however, it is defined by

$B = \frac{\mu_0 i_d}{2\pi r}$

Here the magnetic field is inversely proportional to the distance, that is

$B \not\propto r$

Then,

$\frac{r}{R} = \frac{B_{max}{B}}$

$r = \frac{B_{max}{B}}R$

$r = \frac{B_{max}{0.5B_{max}}}R$

$r = 2R$

7 0

3 years ago