Using hookes law find the distance a spring with am elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it

Which statement is correct about the car?

JulijaS [17]

Answer:

B

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

6 0

3 years ago

Read 2 more answers

You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric

marysya [2.9K]

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\$

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is $0.42\times10^6\ \rm N.m^2/C$

8 0

3 years ago

Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart

Romashka [77]

Answer:

E = 5291.00 N/C

Explanation:

Expression for capacitance is

$C = \frac{\epsilon A}{d}$

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

$C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}$

$C = 24.06 \epsilon$

$C = 24.06\times 8.854\times 10^{-12} F$

$C =2.1\times 10^{-10} F$

We know that capacitrnce and charge is related as

$V = \frac{Q}{C}$

$= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}$

v = 9.523 V

Electric field is given as

$E = \frac{V}{d}$

= $\frac{9.52}{1.8*10^{-3}}$

E = 5291.00 V/m

E = 5291.00 N/C

5 0

3 years ago

Why is it important that we learn about our solar system, the sun, and planets in our solar system other than Earth?

tangare [24]

Answer:

To understand all the different kind of elements and how they work. Nature is a beautiful thing and it is important we understand how they survive and how we can help. The solar system is a mandatory nature. Learning about other planets also helps us feel safe and more knowledgable.

8 0

3 years ago

A block of mass m attached to a horizontally mounted spring with spring constant ???? undergoes simple harmonic motion on a fric

Firdavs [7]

Answer:

(E) It would increase by a factor of 2.

Explanation:

Maximum speed (v) is given by:

$v = \omega *A$

Where ω is the angular frequency and A is the amplitude.

The angular frequency of a spring-mass system is given by:

$\omega = \sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the block attached to the spring.

Therefore, maximum speed can be written as:

$v= A\sqrt{\frac{k}{m}}\\$

Since mass and amplitude are constant values, a relationship between maximum velocity and spring constant can be defined as:

$v =C \sqrt{k}$

There is a linear relationship between maximum velocity and the square root of the spring constant. Therefore if the spring constant is increased by a factor of 4, the velocity is increased by a factor of the square root of 4, which is 2.

The answer is (E) It would increase by a factor of 2.

8 0

3 years ago