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3 years ago

Why, on a sunny day, it is normally hot inside a greenhouse

1 answer:

5 0

Because of the greenhouse gases inside the greenhouse, and the gases trap the heat from the sun so the plant don't freeze, hope this helps

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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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2 years ago

Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to

wlad13 [49]

Answer:

m,lkj,mkn,njn

Explanation:because she is telling you to do a project

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2 years ago

What are three benefits of being assertive and what are three tips to help you develop an assertive style of communication? (Sit

uysha [10]

1. The benefits of been assertive include the following:
a. It improve one self image: When you choose to be assertive, you adopt a self realistic image.
b. It enhance how we understand others: Been assertive helps you to see others in a more realistic context.
c. Promote self awareness and self confidence: Been assertive helps you to develop a greater respect for your own points of view when dealing with issues.
d. Been assertive is a less stressful way of communicating.

2. The three steps needed to develop assertive communication, include:
a. Learn to 'no' more often: refuse to please everyone and do not behave according to people expectations of you. Be yourself.
b. Your tone must be properly turned when talking: Do not raise your voice or rush a communication. Be calm and cool when talking.
c. Be open to communication: Be willing to discuss issues until a suitable solution is found.
d. Pay attention to non-verbal communication of others: Keep eye contact when communicating and pay attention to other body gestures.

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3 years ago

What kind of medium does ocean waves have

Anika [276]

Ocean waves propagate through the medium called 'sea water'.

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3 years ago

Students in a lab are doing experiments involving the motion of objects. They pull an object across a table and use

Svetlanka [38]

Answer:

Explanation:

Did you ever end up getting an answer? Or like did you find out which segment it was?

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2 years ago