A spring has an elastic constant of 8.75 x 103 N/m. What is the magnitude of the force required to stretch this spring a distance of 28.6 cm from its equilibrium position?

Answer:

21977.56kg⋅m/s

Explanation:

Step one:

Given data

Weight of car= 9800N

The mass of the car is

m= w/g

m= 9800/9.81

m=998.98kg

Velocity of car= 22 m/s

time of impact= 0.5 seconds

Required

The impulse of the force F

Step two:

Ft= mv

Ft= 998.98*22

Ft=21977.56kg⋅m/s

The impulse is 21977.56kg⋅m/s

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

Answer:

If there is a large difference between the concentration of one fluid and the concentration of the other, then the particles will diffuse faster. In the experiment, the dye particles in the food coloring are very concentrated.

Explanation:

Answer: Charge 3 is located on the x-axis a distance of 0.67 cm from charge 1 and 1.33 cm from charge 2.

Explanation: Electrostatic Force is the force of repulsion or attraction between two charged particles. It's directly proportional to the charge of the particles and inversely proportional to the distance between them:

[tex]F=k\frac{|q||Q|}{r^{2}}[/tex]

k is an electrostatic constant

For the system of 3 particles, suppose distance from 1 to 3 is x meters, so, distance from 2 to 3 is (0.02-x) meters.

Force will be

[tex]F_{13}=F_{23}[/tex]

[tex]k\frac{q_{1}q_{3}}{r_{13}^{2}} =k\frac{q_{2}q_{3}}{r_{23}^{2}}[/tex]

[tex]\frac{q_{1}}{r_{13}^{2}} =\frac{q_{2}}{r^{2}_{{23}}}[/tex]

Substituting:

[tex]\frac{2.10^{-6}}{x^{2}} =\frac{8.10^{-6}}{(0.02-x)^{2}}[/tex]

[tex]8.10^{-6}x^{2}=2.10^{-6}(0.0004-0.04x+x^{2})[/tex]

[tex]4x^{2}=x^{2}-0.04x+0.0004[/tex]

[tex]3x^{2}+0.04x-0.0004=0[/tex]

Solving quadratic equation using Bhaskara:

[tex]x_{1}=\frac{-0.04+\sqrt{(0.04)^{2}+0.048} }{6}[/tex]

[tex]x_{2}=\frac{-0.04-\sqrt{(0.04)^{2}+0.048} }{6}[/tex]

x₂ will give a negative value and since distance can't be negative, use x₁

[tex]x_{1}=\frac{-0.04+\sqrt{0.0064} }{6}[/tex]

x₁ = 0.0067 m

The position of charge 3 is 0.67 cm from charge 1 and 1.33 cm from charge 2.

Answer:

[tex]f=mg*sin\theta[/tex]

The component of weight of the cube parallel to the plane

Explanation:

From the question we are told that

Cube slides at constant speed

Generally the equation for frictional force is given by

   [tex]f=\mu.N[/tex]

where

[tex]f = frictional force\\\mu =coefficient of friction\\N=Normal force[/tex]

Generally the equation fo Normal force is mathematically given as

  [tex]N=mg.cos\theta[/tex]

Therefore

[tex]f=\mu*mg*cos\theta[/tex]

Generally at constant velocity frictional force f will be

[tex]f=mg*sin\theta[/tex]

Therefore

The magnitude of the frictional force that acts on the cube due to the surface is the component of weight of the cube parallel to the plane

The magnitude of the frictional force that acts on the cube due to the surface is determined with [tex]\mu_ k \times mg \times cos(\theta)[/tex].

The normal force on the cube is calculated as;

[tex]F_n = W \times cos (\theta)\\\\F_n = mg \times cos (\theta)[/tex]

The frictional force acting on the cube due to the surface is calculated as follows;

[tex]F_f = \mu_k F_n\\\\F_f = \mu_ k \times mg \times cos(\theta)[/tex]

where;

Thus, the magnitude of the frictional force that acts on the cube due to the surface is determined with [tex]\mu_ k \times mg \times cos(\theta)[/tex].

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Answer:

Mass of player 2 is 55.044kg

Explanation:

Step one:

given data

Weight of player 1= 900 N

mass= weight/acceleration due to gravity

mass of player= 900/9.81

m1= 91.74kg

v1= 8m/s

mass of player 2

m2= ?

v2= 0m/s

Step two:

from the problem description, after impact, they both moved with a common velocity of V= 5m/s, hence the collision is an inelastic collision

the expression for inelastic collision is

m1v1+m2v2=(m1+m2)V

substitute

91.74*8+m2*0=(91.74+m2)5

733.92=458.7+5m2

275.22=5m2

divide both sides by 5

m2=275.22/5

m2=55.044kg

The weight of player two is

=55.044*9.81

=539.98N

Answer:

Let [tex]A[/tex], [tex]B[/tex],  and [tex]k[/tex] denote three constants (with the requirement that [tex]k > 0[/tex].) The following assumes that the mass of this object is [tex]m[/tex]. Assume that [tex]x(t)[/tex] denotes the position of the object at time [tex]t[/tex].

(a) [tex]\displaystyle x^\prime(t) = -\sqrt{\frac{k}{m}}\, A\, \sin\left(t\, \sqrt{\frac{k}{m}}\right) + \sqrt{\frac{k}{m}}\, B\, \cos\left(t\, \sqrt{\frac{k}{m}}\right)[/tex].

(b) [tex]\displaystyle x^{\prime\prime}(t) = -\left(\frac{k}{m}\right)\, A\, \cos\left(t\, \sqrt{\frac{k}{m}}\right) - \left(\frac{k}{m}\right)^{2}\, B\, \sin\left(t\, \sqrt{\frac{k}{m}}\right)[/tex]

Explanation:

The differential equation for a simple harmonic motion might take the following form:

[tex]\displaystyle \frac{d^{2} x}{d t^{2}} = -\frac{k}{m}\, x[/tex].

The minus sign on the right-hand side highlights the fact that the displacement and acceleration of the object should be in opposite directions.

Notice how this equation is in the form of a homogeneous second-order ODE:

[tex]x^{\prime\prime}(t) + \underbrace{P(t)}_{0}\, x(t) + \underbrace{Q(x)}_{\sqrt{k / m}} = 0[/tex]

Let [tex]r[/tex] be a constant. One possible solution to this homogeneous second-order ODE would be in the form [tex]x(t) = e^{r\, t}[/tex], such that [tex]x^{\prime}(t) = r\, e^{r\, t}[/tex] whereas [tex]x^{\prime\prime}(t) = r^{2}\, e^{r\, t}[/tex].

Substitute into the original ODE to obtain:

[tex]\displaystyle \underbrace{r^{2}\, e^{r t}}_{x^{\prime\prime}(t)} + \frac{k}{m}\, \underbrace{e^{r t}}_{x(t)} = 0[/tex].

Rearrange the equation and solve for [tex]r[/tex].

[tex]\displaystyle e^{r t}\, \left(r^{2} + \frac{k}{m}\right) = 0[/tex]

Notice that [tex]e^{r\, t} > 0[/tex]. Hence, it must be true that [tex]\displaystyle r^{2} + \frac{k}{m} = 0[/tex]. Solve for [tex]r[/tex] given that [tex]k > 0[/tex]:

[tex]\displaystyle r_{1, 2} = \pm i\sqrt{\frac{k}{m}}[/tex], where [tex]i[/tex] is the imaginary unit.

The two particular solutions for the ODE would be:

[tex]x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}[/tex] and [tex]x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}[/tex].

Apply Euler's Formula to rewrite both solutions in terms of trigonometric functions:

[tex]\displaystyle x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}= \sqrt{\frac{k}{m}} \left(\cos\left( \sqrt{\frac{k}{m}} t\right) + + i\, \sin\left( \sqrt{\frac{k}{m}} t\right)\right)[/tex].

[tex]\displaystyle x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}= -\sqrt{\frac{k}{m}} \left(\cos\left( -\sqrt{\frac{k}{m}} t\right) + i\, \sin\left( -\sqrt{\frac{k}{m}} t\right)\right)[/tex].

The general solution would be in the form:

[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \left(i\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)\right)[/tex],

Where [tex]C_1[/tex] and [tex]C_2[/tex] are constants (not necessarily real numbers.)

Since position is supposed to assume a real value for any real [tex]t[/tex], set [tex]B[/tex] to a multiple of [tex]i[/tex] such that the general solution is real-valued:

[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)[/tex].

Differentiate to obtain general expressions for velocity (first derivative) and acceleration (second derivative.)

Answer:

44.5s ; 22.64 m/s

Explanation:

The motion of the truck could be separated into 3 different phases :

First :

Time of motion :

Initial Velocity, u = 0 ; final velocity, v = 29 m/s

Acceleration, a = 2 m/s²

Recall: acceleration = change in velocity / time

Time = change in velocity / acceleration

Time = (29 - 0) / 2 = 14.5 second

Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m

Second :

Time = 25 seconds at constant speed

29 m/s for 25 seconds

v*t = 29 * 25 = 725 m

Third:

5 seconds before coming to rest

((29 + 0) /2) * 5

14.5 * 5 = 72.5 m

A.)

Length of journey = (14.5+ 25 + 5) = 44.5 seconds

B.)

Average velocity = total distance / total time taken

Average velocity = (210.25 + 725 + 72.5) / 44.5

= 1007.75 / 44.5

= 22.646067

= 22.64 m/s

Answer:

Collection. or Precipitation.

Explanation:

b or d

The part of the water cycle when ocean water may end up in the river is to be considered as the condensation.

It is the change with regard to the state of matter from the gas phase into the liquid phase. Also, it should be considered as the reverse of vaporization.

So based on this, we can say that The part of the water cycle when ocean water may end up in the river is to be considered as the condensation.

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Complete Question:

When a 40.0-g nugget of pure gold is heated from 49°C to 60.0°C, it absorbed 5339.01 of energy. What is

the specific heat of gold?

Answer:

Specific heat capacity, c = 12.134J/KgC

Explanation:

Given the following data;

Mass = 40g

Initial temperature, T1 = 49°C

Final temperature, T2 = 60.0°C

Quantity of heat = 5339.01J

To find the specific heat capacity of gold;

Heat capacity is given by the formula;

[tex] Q = mcdt[/tex]

Where;

dt = T2 - T1

dt = 60 - 49

dt = 11°C

Substituting the values into the equation, we have;

[tex] 5339.01 = 40*c*11 [/tex]

[tex] 5339.01 = 440c [/tex]

[tex] c = \frac {5339.01}{440} [/tex]

Specific heat capacity, c = 12.134J/KgC

Answer:

F • t= m•∆v

Explanation:

the impulse experienced by the object equals the change in momentum of the object. in equation form: F • t= m•∆v

Answer:

The answer is point C because as the planet gets closer to the edge, then it starts to accelerate.

Explanation:

Answer:

Please see list of answers below

Explanation:

If the FORCE applied to the car doubles, then following the equation F = m * a , the acceleration will DOUBLE as well.

The mass of the car stays the SAME.

And the velocity of the car will INCREASE with time based on the formula of kinematics of an accelerated object:

v(t) = vi + a * t

Answer:

Consider the relationship of the variables in Newton’s second law. In a drag car race, the force applied to the car is doubled by the driver stepping on the gas pedal.

The acceleration of the car will

✔ double

.

The mass of the car will

✔ remain unchanged

.

The velocity of the car will

✔ increase

.

Explanation:

Answer:

the speed of electron is 4.42 x 10⁶ m/s

the speed of proton is 2406.7 m/s

Explanation:

Given;

electric field strength, E = 478 N/C

charge of the particles, Q = 1.6 x 10⁻¹⁹ C

mass of proton, Mp = 1.673 x 10⁻²⁷ kg

mass of electron Me = 9.11 x 10⁻³¹ kg

time of motion, t = 54.2 ns = 54.2 x 10⁻⁹ s

The magnitude of charge experienced by the particles is calculated as;

F = EQ

F = 478 x 1.6 x 10⁻¹⁹

F = 7.648 x 10⁻¹⁷ N

The speed of the particles is calculated as;

[tex]F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{9.11\times 10^{-31}} \\\\v_e = 4.42 \ \times \ 10^6 \ m/s[/tex]

[tex]v_p = \frac{Ft}{m_p} \\\\v_p = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{1.673\times 10^{-27}} \\\\v_p = 2406.7 \ m/s[/tex]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 [tex]\pi[/tex] [tex]r^{2}[/tex]  = [tex]\frac{Q1}{E_{0} }[/tex]

So,

Rearranging the above equation to get Electric field, we will get:

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex]

Multiply and divide by [tex]r1^{2}[/tex]

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] x [tex]\frac{r1^{2} }{r1^{2} }[/tex]

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

c) r > r2 :

Electric Field = ?

E x 4 [tex]\pi[/tex] [tex]r^{2}[/tex]  = [tex]\frac{Q1 + Q2}{E_{0} }[/tex]

Rearranging the above equation for E:

E = [tex]\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }[/tex]

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] + [tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex]

As we know from above, that:

[tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] =  (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

Then, Similarly,

[tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex] = (σ2 x [tex]r2^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

So,

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] + [tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex]

Replacing the above equations to get E:

E = (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex]) + (σ2 x [tex]r2^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x [tex]r1^{2}[/tex] = - σ2 x [tex]r2^{2}[/tex]

Answer:

5

Explanation:

Answer:

117.8 m/s

Explanation:

Given that:

String length, L = 0.28

pipe length, L' = 0.82

Speed of sound in air, v = 345 m/s

n = 3 (3rd harmonic)

Frequency, f of 3rd harmonic ;

f = (v*n) / 2L - - - - (1)

for the pipe: ; 3rd harmonic

f = (v*n) / 2L' - - - (2)

Equating (1) and (2)

(v*n) / 2L = (v*n) / 2L'

2L' * v * n = v* n * 2L

v = vL / L'

v = (345 * 0.28) / 0.82

v = 96.6 / 0.82

v = 117.80487

v = 117.8 m/s

Answer:

because 5 6 7

Explanation:

Answer: Because 6/8 is greater than 5/8, it has 1/8 more and it is less than 7/8 because it is one less.

some positions are quarterback, offence, defense, linebacker, and wider receiver hoped it helped :)

Answer:

D

Explanation:

took the quiz!!! I picked B and got it wrong!

An example of convection currents is a radiator emitting warm air and drawing in cool air.

This is a method of heat transfer which is characterized by movement of a heated fluid such as air or water.

A radiator emitting warm air and drawing in cool air involves a fluid which is why option D was chosen as the most appropriate choice.

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The work done by the given force should be -45 joules

Since force acts in the negative y direction

So,

F = -15j

here the displacement vector is d = 3i+3j-1k

Now

work done is W = dot product of force and displacement

                        = [-15j ] . [ 3i+3j-1k]

                        =-45 joules

Therefore, The work done by the given force should be -45 joules

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Answer:

a ball sitting motionless on the ground

Explanation:

If a circler shape is balanced its going to have to be motionless

Answer: I think this is the answer, In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Explanation: I had a question similar to this, Hope this helps!

Answer:

392 Nm or J

Explanation:

Work is equal to force times distance. The force required is the mass times acceleration - in this case F = 5kg x 9.81 ( gravity) = 49N. So work is 49N x 8m = 392 Nm or Joules.

Answer:

the net electric charge of the iron atom is 2.24 × 10⁻¹⁸ C

Explanation:

Given the data in the question;

we know that neural iron atoms should have same number of protons and electrons i.e 26 and 26 but in the sun's atmosphere, many of their electrons stripped away.

so; we have 26 protons and 12 electrons, meaning ( 26-12=14) 14 electrons have been stripped away.

when we loss electron, we gain positive charge

so, the Net charge of the iron atom of with 26 protons and 12 electrons will be;

q = 26 ( proton charge) + 12( electron charge )

we know that; a singular proton has a charge of 1.6×10⁻¹⁹ C

protons carry negative charge while electrons carry negative

we substitute

q = 26 ( 1.6×10⁻¹⁹ ) + 12( -1.6×10⁻¹⁹  )

q = 2.24 × 10⁻¹⁸ C

Therefore, the net electric charge of the iron atom is 2.24 × 10⁻¹⁸ C

Answer:

.

Explanation:

Since Saturn's volume is equivalent to 763.59 earths and Mercury is only 0.055 of one earth, then Saturn's volume is over 10,000 times larger than Mercury's.

Answer: TRUE

Answer:

12 miles per hour

Explanation:

time is equal to speed times distance

Answer:

The correct answer is - true.

Explanation:

Changes in the higher authorities or managers or coworkers can be a warning sign before a layoff, however, a company or organization needs to give an official notice before a particular time period before the layoff.

The layoff is a temporary suspension or permanent termination from the employment of an employee due to reasons related to business such as reducing manpower or staff and less work in the organization. These reasons can lead to a change in the behavior of the managers, coworkers, and subordinates.