All categories

3 years ago

Phineas and Ferb are riding their bikes on Main Street. They begin at 11 Mile Rd at

Physics

1 answer:

Lady bird [3.3K]3 years ago

7 0

Answer:

1 hour and 25 minutes

Explanation:

hopefully this helps ,, but like WHY phineas and ferb hshjdshjdhdhdbd im so childish

You might be interested in

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

Yes or no

Tju [1.3M]

Answer:

1. No, it is a colloid

2. Yes

Explanation:

8 0

2 years ago

A monatomic ideal gas undergoes an adiabatic expansion to double its volume. the same final state can be reached by an isobaric

Vlada [557]

(2^(1-γ)-1)/(1-γ) where γ is the heat capacity ratio, Cp/Cv. See attached image for the working.

http://prntscr.com/htqqte

5 0

3 years ago

A punted football hits the ground with a force of 50 N . According to Newton’s 3rd law of motion what also happens at the same e

Nadya [2.5K]

As per Newton's III law every action has equal and opposite reaction

So here we can say that

every body which apply force on other body must have a reaction force of same magnitude in opposite direction

So here if ball hits the ground by 50 N force then the ball must have a reaction force on itself with same magnitude and opposite direction

the magnitude of the force will be 50 N

and its direction is opposite to the force that ball apply on the floor

4 0

3 years ago

An echo is heard from a cliff 3.71 s after a rifle is fired. How many feet away is the cliff

LenKa [72]

The cliff is 2042 ft away.

We know that the speed of sound in air is directly proportional to the absolute temperature.

First convert the Fahrenheit temperature to Celsius;

°C = 5/9(44.5 - 32)

°C = 6.9 °C

Applying the formula;

V1/V2 = √T1/T2

Where; V1 = velocity of sound in air at 0°C

V2 = Velocity of sound in air at 6.9 °C

1087/V2 = √273/279.9

V2= 1101 ft/s

Given that; V = 2s/t

Where s is the distance of the cliff

t is the time taken

1101 ft/s = 2s/3.71 s

s = 1101 ft/s × 3.71 s/2

s = 2042 ft

Learn more:brainly.com/question/15381147

3 0

3 years ago