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Paladinen [302]
4 years ago
10

Which layer of the sun do we normally see?

Physics
1 answer:
Novay_Z [31]4 years ago
6 0

Answer: <u>The photosphere </u>

<u />

Explanation: It is the solar surface we see when we look at the Sun.

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A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
ddd [48]

It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,

b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a)the angular velocity, ω = 2600 rpm = \frac{2600 * 2\pi }{60} rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b)The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c)The angular acceleration, a=\frac{v^{2} }{r}

a = \frac{(47.62)^{2} }{0.175}

⇒a = 12958.08 m/s²

Learn more about angular velocity here:

brainly.com/question/13649539

#SPJ1

6 0
2 years ago
67 points plus brainlest if done correctly.I will report you if you answer 3 or less of the questions, also must post all the an
Alchen [17]

16 is both genes and environmental factors

17 is to trace the inheritance of traits in humans

18 is A, C, F, J, N, and O

19 is all 23 pairs of chromosomes

20 is cystic fibrosis

hopefully  this helps

6 0
4 years ago
Give an example of intense aerobics activity. Prompt must be accurate. ​
Lubov Fominskaja [6]

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

7 0
3 years ago
A kid at the junior high cafeteria wants to propel an empty milk carton along a lunch table by hitting it with a 3.0g spit ball.
Natasha_Volkova [10]

Answer:

4.3 m/s

Explanation:

Using conservation of momentum, assume that the collision is elastic and that the 3.0 g ball stuck with the carton

3.0 g = 3 / 1000 = 0.003 kg = m₁

40 g = 40 / 1000 = 0.04 kg  = m₂, u₂  = 0 since the body is stationary and v = 0.3m/s

m₁u₁ + m₂u₂ = v ( m₁ + m₂)

0.003 u₁  = 0.30 ( 0.003 + 0.04) = 0.0129

u₁ = 0.0129 / 0.003 = 4.3 m/s

5 0
3 years ago
An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe —
sattari [20]

Answer:

  r = 2,026 10⁹ m  and   T = 2.027 10⁴ s

Explanation:

For this exercise let's use Newton's second law

        F = m a

where the force is electric

        F = k \frac{q_1q_2}{r^2}

Acceleration is centripetal

        a = v² / r

we substitute

        k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}

        r = k \frac{q_1q_2}{m \  v^2}          (1)

let's look for the charge in the insulating sphere

          ρ = q₂ / V

          q₂ = ρ V

the volume of the sphere is

         v = 4/3 π r³

we substitute

        q₂ = ρ \frac{4}{3} π r³

        q₂ = 3 10⁻⁹ \frac{4}{3} π 4³

        q₂ = 8.04 10⁻⁷ C

let's calculate the radius with equation 1

        r = 9 10⁹  1.6 10⁻¹⁹  8.04 10⁻⁷ /(9.1  10⁻³¹ 628 10³)

        r = 2,026 10⁹ m

this is the radius of the electron orbit around the charged sphere.

Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio

        v = x / t

the distance traveled in a circle is

        x = 2π r

In this case, time is the period

        v = 2π r /T

        T = 2π r /v

let's calculate

        T = 2π 2,026 10⁹/628 103

        T = 2.027 10⁴ s

4 0
3 years ago
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