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Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

Answers

Answer 1

Answer:

Answer:

The speed of the arrow immediately after it leaves the bow is 38.73 m/s

Explanation:

given information:

force, F = 150 N

x = 50 cm = 0.5 m

mass of arrow, m = 50 g = 0.05 kg

We start from the force of the spring

F = kx

k = $(F)/(x)$

= $(150)/(0.5)$

= 300 N/m

The potential energy, EP of the spring is

EP = $(1)/(2) kx^(2)$

the kinetic energy, EK of the spring

EK = $(1)/(2) mv^(2)$

According to conservative energy,

EP = EK

$(1)/(2) kx^(2)$ = $(1)/(2) mv^(2)$

$kx^(2)$ = $mv^(2)$

$v^(2)$ = $(kx^(2) )/(m)$

v = $x\sqrt{(k)/(m) }$

= $0.5\sqrt{(300)/(0.05) }$

= 38.73 m/s

Answer 2

Answer:

Final answer:

Using Hooke's Law, we can determine the speed of the arrow. The speed of the arrow immediately after it leaves the bow is approximately 38.7 m/s.

Explanation:

In this problem, we can use Hooke's Law to determine the speed of the arrow. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring:

F = -kx

Where F is the force, x is the displacement, and k is the spring constant.

In this case, the force exerted by the bow on the arrow is acting like a spring force. The force of the bow is 150N, and the displacement is 50cm (which is equivalent to 0.5m). So we can set up the equation as:

150N = -k * 0.5m

Now we can solve for k:

k = -150N / 0.5m = -300 N/m

Now that we have the spring constant, we can use it to find the potential energy stored in the bow:

PE = 0.5kx^2 = 0.5*(-300N/m)*(0.5m)^2 = 37.5 J

Next, we can use the conservation of energy to find the kinetic energy of the arrow right after it leaves the bow. The potential energy stored in the bow is converted into kinetic energy:

KE = PE = 37.5 J

The kinetic energy is given by the equation:

KE = 0.5mv^2

Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for v:

v = sqrt(2KE/m) = sqrt(2*37.5 J / 0.05 kg) = sqrt(1500) m/s ≈ 38.7 m/s.

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Much of our knowledge of the interior of the Earth comes from the study of planetary vibrations, which is the science of

Answers

Answer:

Seismology.

Explanation:

Seismology is the beach of physical science that deals with the study of vibrations that comes out from the interior of the earth onto the surface and these vibrations are in the form of seismic waves that are primary, secondary, and surface waves.

The science of seismology tells about the magnitude and intensity of these waves that lead to planetary vibrations. These waves trigger earthquakes, floods, and even landslides.

An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:i. The current density ii. The drift velocity

Answers

Answer:

The current density is $J = 2.04 * 10^(6) A /m^2$

The drift velocity is $v_d = 1.5 * 10^(-4) m/s$

Explanation:

From the question we are told that

The nominal diameter of the wire is $d = 1.02 mm= (1.02)/(1000) = 0.00102 \ m$

The current carried by the wire is $I = 1.67 A$

The power rating of the lamp is $P = 200 W$

The density of electron is $n = 8.5 * 10^(28) \ e/m^3$

The current density is mathematically represented as

$J = (I)/(A)$

Where A is the area which is mathematically evaluated as

$A = \pi (d^2)/(4)$

Substituting values

$A = 3.142 * ((1.02 * 10^(-3))^2 )/(4)$

$A = 8.0*10^(-4)m^2$

$J = (1.67)/(8.0*10^(-4))$

$J = 2.04 * 10^(6) A /m^2$

The drift velocity is mathematically represented as

$v_d = (J)/(ne)$

Where e is the charge on one electron which has a value $e = 1.602 *10^(-19) C$

$v_d =(2.04 * 10^6 )/(8.5 *10^(28) * 1.6 * 10^(-19))$

$v_d = 1.5 * 10^(-4) m/s$

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

$E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J$

So, the change in its gravitational potential energy is 1.76 J.

A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?

Answers

Answer:

(A). The time is 5.47 sec.

(B). The speed of the rock just before it strikes the ground is 39.59 m/s.

Explanation:

Given that,

Initial velocity = 14.0 m/s

Height = 70.0 m

(A). We need to calculate the time

Using second equation of motion

$s=ut+(1)/(2)gt^2$

Put the value into the formula

$70=-14* t+(1)/(2)*9.8* t^2$

$4.9t^2-14t-70=0$

$t =5.47\ sec$

(B). We need to calculate the speed of the rock just before it strikes the ground

Using third equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v^2=(14)^2+2*9.8*70$

$v^2=1568$

$v=√(1568)$

$v=39.59\ m/s$

Hence, (A). The time is 5.47 sec.

(B). The speed of the rock just before it strikes the ground is 39.59 m/s.

A physicist is creating a computational model of a falling person before and after opening a parachute. What boundary conditions would be important here?the air resistance encountered as the person falls

the speed at which the person falls

the change in kinetic and potential energy

the location where potential energy is zero

Answers

Answer:

the location where potential energy is zero

Explanation:

Answer:

Air resistance

Explanation:

Air resistance encountered as the person falls

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. Part A How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

Answers

Final answer:

The total work done in moving an alpha particle from the center to the side of a square with electrons at its corners involves finding the potential energy change, which can be calculated using the charges, distances, and Coulomb's constant.

Explanation:

The question deals with the fundamental concepts of electrostatics and the energy associated with moving charges in an electric field. Given the aforementioned question, we are required to find the work done moving an alpha particle (a helium nucleus, having a charge of +2e) from the center of a square to one of its sides, with electrons (each having a charge of -e) being situated at its corners.

To determine the work done, we must consider the potential energy changes resulting from moving the alpha particle. The potential energy associated with two point charges is given by the formula: U = k*q1*q2/r, where q1 and q2 are charges, r is the distance between them, and k is Coulomb's constant.

First, we calculate the potential energy at the center due to all four electrons then find the potential energy at the midpoint of the side. The work done is the difference between these two potential energies. As the electrons are all at an equal distance from the alpha particle (in the center and on the side), the calculations would involve plugging in the values for the charge of an electron, the charge of an alpha particle, the given distance values, and Coulomb's constant into the aforementioned formula.

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Final answer:

The work required to move the alpha particle from the midpoint to the midpoint of one of the side of the square with four electrons at its corners would be zero as the net electric field at the midpoint due to the electrons is zero.

Explanation:

The subject of this question pertains to the concept of electrostatics and potential energy in physics. In this scenario, the alpha particle is initially at the midpoint of a square with four electrons at its corners. As per Coulomb's Law, the electrostatic force between two charges is inversely proportional to the square of the distance between them.

Since the alpha particle placed in the center of the square and four electrons at the corners form a symmetrical system, the net force and hence the net electric field at the midpoint due to the electrons is zero. Thus, no work would be required to move the alpha particle to the midpoint of one of the sides of the square as work done is calculated by the formula W = F x d x cos(θ), where F is force, d is the displacement, and θ is the angle between the force and displacement. Since F is equal to zero, the work done will also be zero.

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