A skateboarder starting from rest accelerates down a ramp at 2 m/s for 2 s. What is the final speed of the skateboarder?

A pendulum has 711 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

Anika [276]

According to law of conservation of energy,
Energy can neither be constructed nor be destroyed but can be transformed from one form to another.

At the highest point of the pendulum(point b), pendulum is associated with potential energy only and no kinetic energy.
Therefore total energy at point b = potential energy = 711 J.... i

At the bottom most point(point a), pendulum is associated only with kinetic energy and no potential energy.
Therefore total energy at point a = kinetic energy ---- ii
From i and ii,
Kinetic energy = potential energy = 711 J.(Conserving energy)

Hence kinetic energy at the bottom most point is 711 J.
Hope this helps!!

7 0

3 years ago

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

8 0

3 years ago

For work to be done on an object,

djverab [1.8K]

Answer:B

Explanation:

For work to be done, the object must move some distance as a result of a force

3 0

3 years ago

A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended

Alik [6]

Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
$k= \frac{F}{\Delta x}$
where F is the force applied, and $\Delta x$ is the stretch of the spring with respect to its equilibrium position. Using the data, we find
$k= \frac{15 N}{3.0 cm}=5.0 N/cm$

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
$\Delta x = \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm$

4 0

3 years ago

Read 2 more answers

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0