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3 years ago

What is the potential energy of a 2500 g object suspended 5 kg above the earth's surface?

Physics

1 answer:

Alinara [238K]3 years ago

8 0

Answer:

$potential \: energy = mgh \\ = ( \frac{2500}{1000} ) \times 10 \times 5 \\ = 125 \: newtons$

if height is 5 m

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(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

6. Velocity is a ____________. a) vector b) value c) arrow d) none of these

Olenka [21]

<span>the speed of something in a given direction. so i think none of these</span>

4 0

3 years ago

Read 2 more answers

For a given wave IF frequency doubles the wavelength

motikmotik

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

$v=f\lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

$\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}$

7 0

3 years ago

Give two examples of contact and non-contact forces and explain why they are contact and non-contact forces respectively.

Sever21 [200]

The Two examples of contact forces are:

frictional force
Contact force.

The two examples of non contact forces are:

Gravitational force
magnetic force.

Contact forces happens due to the contact between two objects

Non Contact forces happens because there is no contact between two objects. There is no attraction.

3 0

2 years ago

Suppose a car is traveling at +20.3 m/s, and the driver sees a traffic light turn red. After 0.207 s has elapsed (the reaction t

olga nikolaevna [1]

Answer:

33.6371 m

Explanation:

t = Time taken

u = Initial velocity = 20.3 m/s

v = Final velocity

s = Displacement

a = Acceleration = -7 m/s²

Distance traveled in the 0.207 seconds

Distance = Speed × Time

⇒Distance = 20.3×0.207 = 4.2021 m

Equation of motion

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20.3^2}{2\times -7}\\\Rightarrow s=29.435\ m$

Distance traveled by the car while braking is 29.435 m

Total distance measured from the point where the driver first notices the red light is 29.435+4.2021 = 33.6371 m

7 0

3 years ago