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Vlad1618 [11]
4 years ago
7

What are two ways the sun supports life

Physics
2 answers:
timama [110]4 years ago
8 0
Maintaining temperature and supporting plant life.
Ronch [10]4 years ago
7 0
The sun supports life as it gives plants the ability to carry out photosynthesis and it allows humans to gain vitamin D
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PLEASE ANSWER USING ALL MY POINTS FOR THIS NEED HELP (please answer all of them)
kipiarov [429]
4-C. ... 5-D. ... 6-A. ... 7-D. ... 8-C. ... 9-B. ... 10-D. It's really a joy and a delight that you've learned so much by posting all of these questions.
5 0
3 years ago
Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generato
Volgvan

Answer:

The coil radius of other generator is 5.15 cm

Explanation:

Consider the equation for induced emf in a generator coil:

EMF = NBAω Sin(ωt)

where,

N = No. of turns in coil

B = magnetic field

A = Cross-sectional area of coil = π r²

ω = angular velocity

t = time

It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.

Therefore, EMF for both coils or generators will be:

EMF₁ = NBπr₁²ω Sin(ωt)

EMF₂ = NBπr₂²ω Sin(ωt)

dividing both the equations:

EMF₁/EMF₂ = (r₁/r₂)²

r₂ = r₁ √(EMF₂/EMF₁)

where,

EMF₁ = 1.8 V

EMF₂ = 3.9 V

r₁ = 3.5 cm

r₂ = ?

Therefore,

r₂ = (3.5 cm)√(3.9 V/1.8 V)

<u>r₂ = 5.15 cm</u>

3 0
3 years ago
If a block is in equilibrium the magnitude of the block's acceleration is
kaheart [24]

zero.

from newton's first law of motion

3 0
3 years ago
Read 2 more answers
A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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