(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a force of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=

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Answer 1

Answer:

Answer:

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If 80 joules of work were necessary to move a 5 newton box, how far was the box moved?
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Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rimb. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.

When an external magnetic field is applied, what happens to the protons in a sample?A) All protons align with the field.
B) All protons align opposite to the field.
C) Some protons align with the field and some align opposite to it.
D) All protons assume a random orientation.

Answers

On account of external magnetic field, the protons will align with the magnetic field. Hence, option (a) is correct.

The given problem is based on the concept of magnetic field. The region where the magnetic force is experienced is known as magnetic field. Generally, the protons are the charged entities carrying the positive polarity and are one of the major constituents of modern atomic structure.

The origin of magnetic field occurs due to charged particles present in a specific space. And the magnetic field is due to the flowing of liquid metal in the outer core of the planet generates electric currents.
In the condition when an external field is applied, the majority of protons align to the field because these protons tend to act like small magnets under the effect of this external field.

Thus, we can conclude that on account of external magnetic field, the protons will align with the field.

Learn more about the magnetic field here:

brainly.com/question/14848188

Answer:

Some protons align with the field and some align opposite to it.

Explanation:

Majority align to the field because these protons tend to act like small magnets under the effect of this external field

The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence of a collapse of fishing stocks?A.1965
B.1985
C.1995
D.2005

Answers

The correct answer is C. 1995

Explanation:

The graph shows the changes in the harvest of Atlantic cod. In general, this graph illustrates how the peak occurred in the 1980s but then there was a sudden and sharp decline in 1995. Indeed, 1995 is the year with the lowest number of harvested cod as in this year there were approximately least than 10 thousand metric tonnes of cods. Also, this year shows the collapse of fishing stocks or that the population of this fish collapsed, which made it impossible to harvest as many fish as in previous years. According to this, the year that shows the collapse of fishing stocks is 1995.

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answers

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_(1,2)$ = Mass of each object

$v_(1,2)$ = Initial Velocity of each object

$v_f$ = Final velocity

Replacing we have that,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

$1850*13.8+3100*0 = (1850+3100)v_f$

$v_f = 5.1575m/s$

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

$v_f^2-v_i^2 = 2ax$

Since there is no initial speed, then

$v_f^2 = 2ax$

$5.1575^2 = 2a (1.91)$

$a = 6.9633m/s^2$

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

$F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105$

Therefore the Kinetic friction coefficient is 0.7105

. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.

You should obtain e/m = 2V/(B^2)(r^2)

3. The magnetic field on the axis of a circular current loop a distance z away is given by

B = mu I R^2 / 2(R^2 + z^2)^ (3/2)

where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:

|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/R

where B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Answers

Answer:

Explanation:

Magnetic field creates a force perpendicular to a moving charge in its field which is equal to Bev where B is magnetic field , e is amount of charge on the moving charge and v is the velocity of charge particle .

This force provides centripetal force for creation of circular motion. If r be the radius of the circular path

Bev = mv² / r

r = mv / Be

2 ) If an electron is accelerated by an electric field created by potential difference V then electric field

= V / d where d is distance between two points having potential difference v .

force on charged particle

electric field x charge

= V /d x e

work done by field

= force x distance

= V /d x e x d

V e

This is equal to kinetic energy created

V e = 1/2 mv²

= 1/2 m (r²B²e² / m² )

V = r²B²e/ 2 m

e / m = 2 V/ r²B²

3 )

B = $(\mu* I* R^2)/(2(R^2+Z^2)^(3)/(2) )$

In Helmholtz coils , distance between coil is equal to R so Z = R/2

B = $(\mu* I* R^2)/(2(R^2+(R^2)/(4) )^(3)/(2) )$

For N turns of coil and total field due to two coils

B = $(\mu* I* N)/(R*((5)/(4))^(3)/(2) )$

= $(\mu* I* N)/(R)* ((4)/(5))^(3)/(2)$

= 9.0 x 10^-7 NI/R

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed $v = 2.20*10^7\ m/s$

Acceleration $a=1.90*10^(13)\ m/s^2$

We need to calculate the magnetic field

Using formula of magnetic field

$F=qvB$ ....(I)

Using newton's second law

$F= ma$ ....(II)

From equation (I) and (II)

$ma=qvB$

Put the value into the formula

$1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B$

$3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B$

$B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))$

$B=0.009014\ T$

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

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