How much power is generated if 375 J of work is done in 15 s?

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago

As a pendulum moves toward the equilibrium position, velocity and acceleration . As the pendulum moves away from the equilibrium

STatiana [176]

Answer:

As a pendulum moves toward the equilibrium position, velocity increases and acceleration decreases. As the pendulum moves away from the equilibrium position, velocity decreases and acceleration increases.

Explanation:

Using the law of conservation of energy, we know that Em1=Em2.

Em1 (at the highest point) = Eg + Ek, where Ek is 0

Em2 (at the equilibrium point) = Eg +Ek, where Eg is 0

This makes sense. At the highest point, the pendulum is at its maximum height. At this point, however, it stops moving, so its velocity is 0. At the equilibrium point, the pendulum is at its lowest height (i.e. h=0). At this point, however, its moving at its maximum velocity. This velocity is constant, which means that acceleration is 0.

4 0

2 years ago

A satellite is put in a circular orbit about Earth with a radius equal to 35% of the radius of the Moon's orbit. What is its per

Debora [2.8K]

Answer:

0.21 lunar month

Explanation:

the radius of moon = r₁

time period of the moon = T₁ = 1 lunar month

The radius of the satellite = 0.35 r₁

Time period of satellite

The relation between time period and radius

$T\ \alpha\ \sqrt{r^3}$

now,

$\dfrac{T_2}{T_1}=\dfrac{\sqrt{r_2^3}}{\sqrt{r_1^3}}$

$\dfrac{T_2}{T_1}=\dfrac{\sqrt{0.35^3r_1^3}}{\sqrt{r_1^3}}$

$\dfrac{T_2}{1}=\sqrt{0.35^3}$

T₂ = 0.21 lunar month

hence, the time period of revolution of satellite is equal to 0.21 lunar month

6 0

3 years ago

an electron accelerated from rest through a voltage of 770 v enters a region of constant magnetic field. part a if the electron

finlep [7]

The magnitude of the magnetic field on which an electron accelerated from rest through a voltage of 770v enters a region of constant magnetic field is 3.744 * $10^{-4}$ T