Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Answers

Answer 1

Answer:

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

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true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields

Answers

Answer:

True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the positive x axis; the second charge,q2= 6.00 nC, is placed a distance 9.00 mfrom the originalong the negative x axis.[Give the x and y components of the electric fieldas an ordered pair. Express your answer innewtons per coulomb to three significant figures.Keep in mind that an x component that points tothe right is positive and a y component thatpoints upward is positive.]

Answers

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ (12)/(20)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(20)\ \right )\n\Rightarrow \theta_1\ =\ 36.87^o\n\n\therefore sin\theta_1\ =\ (12)/(9)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(9)\ \right )\n\Rightarrow \theta_1\ =\ 53.13^o\n$

Electric field by charge $q_1$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 8* 10^(-9))/(20^2)\n\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 6.0* 10^(-9))/(16^2)\n\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_(1x)\ +\ F_(2x)\n\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\n\Rightarrow F_x\ =\ 0.18*( -cos36.87^o)\ +\ 0.24* cos53.13^o\n\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\n\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_(1y)\ +\ F_(2y)\n\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\n\Rightarrow F_y\ =\ 0.18* sin36.87^o\ +\ 0.24* sin53.13^o\n\Rightarrow F_y\ =\ 0.180\ +\ 0.192\n\Rightarrow F_y\ =\ 0.299\ N/C$

Hence, the net electric field at point A,

$F\ =\ ( 0, 0.299 )\ N/C.$

What is the average kinetic energy of hydrogen atoms on the 5500°C surface of the sun?

Answers

Answer: The average kinetic energy of hydrogen atoms is $1.19562* 10^(-19)J$

Explanation:

To calculate the average kinetic energy of the atom, we use the equation:

$K=(3)/(2)kT$

where,

K = average kinetic energy

k = Boltzmann constant = $1.3807* 10^(-23)J/K$

T = temperature = $5500^oC=[5500+273]K=5773K$

Putting values in above equation, we get:

$K=(3)/(2)* 1.3807* 10^(-23)J/K* 5773K\n\nK=1.19562* 10^(-19)J$

Hence, the average kinetic energy of hydrogen atoms is $1.19562* 10^(-19)J$

The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Answers

Part a)

Equation of position with time is given as

$x = (2.0 m/s)t + (-3.0 m/s2)t^2$

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

$x = 2* 0.45 - 3 * 0.45^2$

$x_1 = 0.2925 m$

at t = 0.55 s

$x = 2* 0.55 - 3*0.55^2$

$x_2 = 0.1925$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.1925 - 0.2925 = -0.1 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.1}{0.1) = -1 m/s$

part c)

at t = 0.49 s

$x = 2* 0.49 - 3 * 0.49^2$

$x_1 = 0.2597 m$

at t = 0.51 s

$x = 2* 0.51 - 3*0.51^2$

$x_2 = 0.2397 m$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.2397 - 0.2597 = -0.02 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.02}{0.02) = -1 m/s$

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in- dependently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular

Answers

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

F = m a₀

a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

x = v₀ t + ½ a t²

x = ½ a₀ t²

t² = 2x / a₀

28² = 2x /a₀ (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

in the direction of axis a, there is a motor so its force is F/2

the acceleration on this axis is

a = F/2m

a = a₀ / 2

so if we use the distance equation

x = v₀ t + ½ a t²

as part of rest v₀ = 0

x = ½ (a₀ / 2) t²

let's clear the time

t² = (2x / a₀) 2

we substitute the let of equation 1

t² = 28² 2

t = 28 √2

t = 39.60 s

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A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5.42 x 10^4 rad/s. In the process, the bit turns through 1.72 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.42 x 10^4 rad/s, starting from rest?