I WILL GIVE BRAINLIEST

A 80 kg bungee jumper is on a bridge that is 100 meters above a river. Attached to the jumper is a bungee cord that is 50 meters

Usimov [2.4K]

Answer:

a) 70,560 J

b) 88.2 N/m

Explanation:

The spring potential will equal the change in gravity potential

PS = PE = mgh = 80(9.8)(100 - 10) = 70,560 J

PS = ½kx²

k = 2PS/x² = 2(70560)/(100 - 50 - 10)² = 88.2 N/m

7 0

3 years ago

A force of 333 N is applied 33 cm from the pivot point. What is the maximum torque of this situation

exis [7]

Answer: 109.89 Nm

Explanation:

The maximum torque will be calculated as the force multiplied by the perpendicular distance. This will be:

Torque = force × perpendicular distance

torque = 333 × 0.33

= 109.89 Nm

5 0

3 years ago

A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis

GarryVolchara [31]

Resultant displacement is 29.2 km at $83.1^{\circ}$ north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

$A_x = 25.5 km\\A_y = 0 km$

- Second displacement is 41.0 km northwest, so its components are

$B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km$

So, the components of the resultant displacement are

$R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km$

And so, the magnitude is calculated using Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km$

And the direction is given by

$\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}$

Where the angle is measured from the west direction, since Rx is negative.

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

3 0

3 years ago

In an RL series circuit, an inductor of 3.54 H and a resistor of 7.76 Ω are connected to a 26.6 V battery. The switch of the cir

podryga [215]

Answer:

Energy stored in inductor will be 20.797 J

Explanation:

We have given inductance L = 3.54 H

And resistance R = 7.76 ohm

Battery voltage V = 26.6 VOLT

After very long time means at steady state inductor behaves as short circuit

So current $i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp$

Now energy stored in inductor $E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J$

So energy stored in inductor will be 20.797 J

4 0

4 years ago

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

Lorico [155]

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

8 0

4 years ago