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2 years ago

When do sunspots disappear?

Physics

1 answer:

dedylja [7]2 years ago

8 0

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

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What does the slope of a velocity versus time graph represent?

liberstina [14]

The slope represents the acceleration

7 0

3 years ago

How fast (in rpm) must a centrifuge rotate if a particle 7.3 cm from the axis of rotation is to experience an acceleration of 1.

guajiro [1.7K]

The formula we use here is:

radial acceleration = ω^2 * R 

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s

and since: ω = 2pi*f --> f = ω/(2pi)
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
= 36,714.77 rpm

5 0

3 years ago

4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2

bazaltina [42]

Answer:

$1030.83\ \text{N}$

Explanation:

$\Delta L$ = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = $\dfrac{\pi}{4}d^2$

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

$\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}$

La tensión en la cuerda es $1030.84\ \text{N}$ .

8 0

2 years ago

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

$W= F*d= m*g*d$

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

$W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]$

Converting from Joules to Kcals:

$\frac{10620.75}{4186} =2.537 Kcal$

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

x ---> 100%

$x=\frac{2.537*100}{20} =12.685$

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

$\frac{9000Kcals}{12.685Kcals} =709.5 times$

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

$Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\$

$P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]$

*****

4 0

3 years ago

Braddy connected the loose wire to the battery and created an electromagnet. He picked up 45 thumb tacks with his electromagnet,

Katarina [22]

Answer:

C) Use two batteries instead of one.

Explanation:

-Strength of an electromagnet depends on the electrical current which flows through the wires.

-Two batteries have a higher current than one and thus higher strength in the electromagnet.

3 0

3 years ago