g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

Answers

Answer 1

Answer:

F = 10.8N

Explanation:

Given the mass m = 0.4kg, v1 = 25m/s, v2 = 12m/s and t =0.5s

From Newtown's second law of motion the average force can be found. This law states that the product of the force experienced by a body and the time t of the force acting on the body is equal to the change in momentum of the body. Mathematically it can be stated as follows

F×t = m(v2 – v1)

F = m(v2 – v1)/t = 0.4(25 – 12)/0.5 = 10.8N

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Which statement best describes how the first quatrain relates to the second quatrain? The first shows the beloved’s actions; the second describes how she imitates them. Both the first and the second show the actions of the speaker and the beloved. The first shows the speaker’s actions; the second shows the beloved’s opposition to them. The first shows the speaker’s sadness; the second shows the beloved’s anger.
) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region
A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of 2.0 × 106 N/m. What is the maximum distance the spring compresses?
The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which data set represents constant acceleration?

A gate with a circular cross section is held closed by a lever 1 m long attached to a buoyant cylinder. The cylinder is 25 cm in diameter and weighs 200 N. The gate is attached to a horizontal shaft so it can pivot about its center. The liquid is water. The chain and lever attached to the gate have negligible weight. Find the length of the chain such that the gate is just on the verge of opening when the water depth above the gate hinge is 10 m.

Answers

The length of the chain such that the gate is just on the verge of opening is mathematically given as

l=8.58m

What is the length of the chain?

Generally, the equation for the is mathematically given as

$F'=\gamma hA$

Therefore

$9810*10* (\pi)/(4)1^2$

Fh= 77048 N

Where

$ycp-y=((\pi r^4)/(4))/(10*(\pi D^4)/(4))$

ycp-y=0.00625

In conclusion, resultant force

x = F'' - W

x = 9810* 10*( \pi/4 )*0.25^2 *(10-l)-200

x = 4615.5-481.5 l

Therefore

77048* 0.00625 - 1 *(4615.5-481.5 l) = 0

l=8.58m

Answers

I think your answer is speed faster than the speed of sound

Answer:

speeds above 343 m/s

Explanation:

I have taken the test got 100%

The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

Answers

Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth = 6 × 10²⁴ kg

r = Distance between Earth and Sun = $1.5* 10^(11)\ m$

t = Time taken = 3 days

Acceleration is given by

$a=(GM)/(r^2)\n\Rightarrow a=(6.67* 10^(-11)* 6* 10^(24))/((1.5* 10^(11))^2)\n\Rightarrow a=1.77867* 10^(-8)\ m/s^2$

Velocity of the star

$v=u+at\n\Rightarrow v=0+1.77867* 10^(-8)* 3* 24* 60* 60\n\Rightarrow v=0.00461031264\ m/s$

The Sun's speed will be 0.00461031264 m/s

Which one of these are true about scale models? a. A map may have a scale model with the proportion of one centimeter to one kilometer. b. A scale model is always smaller than the object it represents, c. A model may be built to different scales.

Answers

Answer:

Option A is true

Explanation:

For option A, it's true because a map that has a scale model with the proportion of one centimeter to one kilometer is known as verbal scale which is a type of scale.

For option B, it's not true because though a scale model is most times always smaller than the object it represents, there are sometimes when the scale model is an enlarged view/representation of a small object.

For option C, it's true because there is no one scale that is confined to just a model. A model can use different scales to depict an object.

Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.

Answers

Answer:

$\phi = 1.78 *10^(-7) \ Weber$

$L = 1.183 *10^(-7) \ H$

Explanation:

From the question we are told that

The radius is $r = 6 \ cm = (6)/(100) = 0.06 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic flux of the coil is mathematically represented as

$\phi = B * A$

Where B is the magnetic field which is mathematically represented as

$B = (\mu_o * I)/(2 * r)$

Where $\mu_o$ is the magnetic field with a constant value $\mu_o = 4\pi * 10^(-7) N/A^2$

substituting value

$B = (4\pi * 10^(-7) * 1.50 )/(2 * 0.06)$

$B = 1.571 *10^(-5) \ T$

The area A is mathematically evaluated as

$A = \pi r ^2$

substituting values

$A = 3.142 * (0.06)^2$

$A = 0.0113 m^2$

the magnetic flux is mathematically evaluated as

$\phi = 1.571 *10^(-5) * 0.0113$

$\phi = 1.78 *10^(-7) \ Weber$

The self-inductance is evaluated as

$L = (\phi )/(I)$

substituting values

$L = (1.78 *10^(-7) )/(1.50 )$

$L = 1.183 *10^(-7) \ H$

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =(k*q)/(d)$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = (8.99e9N*m2/C2*(4e-6C))/(8m) + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

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