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mrs_skeptik [129]
3 years ago
6

What are the two main processes carried out by the excretory system?​

Physics
1 answer:
Paha777 [63]3 years ago
5 0
Water and carbon dioxide
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A light-year is the distance that light travels in one year. The speed of light is 3.00 × 108 m/s. How many miles are there in o
Alinara [238K]

Answer:

(a) 5.88\times 10^{12}mi

Explanation:

We have given speed of light c=3\times 10^8m/sec

Given time = 1 year =365 days

We know that in 1 minute = 60 sec

1 hour = 60×60 = 3600 sec

In one day = 24 hour = 24×60×60=86400 sec

So in 365 days = 365×86400=3.1536\times 10^7sec

We know that distance = speed ×time=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m

We have given that 1 mi = 1609 m

So 9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}miSo option (a) is correct

8 0
3 years ago
After a while, the car started to go around a long bend, still maintaining its constant speed of 55 miles per hour. Is there a n
sukhopar [10]

Answer:

4) True. The change of direction needs an unbalanced force

Explanation:

Let us propose the resolution of the problem using Newton's second law.

    F = m a

As the car is spinning the acceleration is centripetal

    a = v2.r

   

    F = m v2 / r

We can see that as the velocity of a vector even if its module does not change, the change of direction requires an external force.

Now we can analyze the statement if they are true or false

1) and 3) False, even when the speed changes, the direction changes

2) False with the speed change can be determined

4) True. The change of direction needs an unbalanced force

5) False are different things. the direction is where it is going and the speed is the magnitude of the vector

6 0
3 years ago
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively
MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

7 0
3 years ago
A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin
erastovalidia [21]

Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's  = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫   dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5  / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

8 0
3 years ago
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
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