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Slav-nsk [51]
3 years ago
8

The simple sugar made during photosynthesis is called _____.

Physics
2 answers:
xz_007 [3.2K]3 years ago
6 0
The answer is glucose
Andre45 [30]3 years ago
3 0

Answer:

glucose

Explanation:

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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
ASAP ONLY CORRECT ANSWERS PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEE I RLLY NEED HELPPPPPPP
nadya68 [22]

Answer:

attached below

Explanation:

5 0
3 years ago
In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/
Hunter-Best [27]

Answer:

n_{T} = 31.68\,rev

Explanation:

The angular acceleration is:

\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}

\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}

And the angular deceleration is:

\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}

\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}

The total number of revolutions is:

n_{T} = n_{1} + n_{2}

n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}

n_{T} = 31.68\,rev

4 0
3 years ago
A model airplane of mass 0.6 kg is attached to a horizontal string and flies in a horizontal circle of radius 6 m, making 1.6 re
nikitadnepr [17]

Answer:

speed= 12.15\frac{m}{s}

Explanation:

In this question we have given

mass of airplane=.6 Kg

Radius of horizontal circle,r=6m

Time taken to complete 1.6 revolution=5s

Therefore time taken to complete 1 revolution, t=\frac{5}{1.6}

t=3.1 s

We have to find the speed of airplane

We will first find the distance covered by airplane in tracing one circle which is equal to circumference of circle

We know that

Circumference,d = 2 \pi \times r

or d=2\times 3.14 \times 6

d=37.68 m

We know that speed=\frac{d}{t}............(1)

Put value of d and t in equation 1

speed=\frac{37.68}{3.1}

speed= 12.15\frac{m}{s}

5 0
3 years ago
.<br> The spot on Jupiter is a _______?<br><br> Volcano<br><br> Ocean<br><br> Storm<br><br> Unknown
MaRussiya [10]
Storm!! Hope I could help
8 0
3 years ago
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