Una rueda gira con una frecuencia de 530 rpm. Determina la velocidad angular, el periodo y la frecuencia.

. The earth revolves around the sun in the counterclockwise direction, completing one full revolution about every 365 days. In r

LuckyWell [14K]

Answer:

Explanation:

The earth makes in 365 day 1 revolution

The earth makes in 1 day 1 / 365 revolutions

1 / 365 revolution per day

n = 1 / 365 per day .

n is called frequency of revolutions .

Angular velocity = 2π n

= 2 π x 1 / 365

= .0172 radian / day

6 0

3 years ago

Exposure to what type of radiant energy is sensed by human skin as warmth?

tia_tia [17]

Answer:Another name for heat energy is... Infrared waves. These cannot be seen by your eyes but can be felt by your skin. ... Gamma Rays. ... Which type of electromagnetic radiation can result in sunburns and skin cancer? Ultraviolet Radiation ... X Rays. What is the distance between one point of a wave to the same point in the ...

4 0

3 years ago

Read 2 more answers

How do you calculate elastic potential energy

Nimfa-mama [501]

U=1/2kx2

This image sums it up

5 0

3 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

$\frac{v^{2} }{r}$ = a

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

F = $\frac{mv^{2} }{r}$

F = ma

equating the above equations and making a subject of formula, we get:

a = $\frac{v^{2} }{r}$

a = 25/10 = 2.5 $ms^{-2}$

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0

3 years ago