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3 years ago

In your opinion, how can we increase a student’s understanding and respect for science?

Physics

2 answers:

nordsb [41]3 years ago

7 0

Answer:

In my opinion make it fun or take time to think or make it something that kids wanna do but they are still learning from it.

astraxan [27]3 years ago

6 0

Answer: i would do something that is really easy and i will do test 2 where there aint a word problems and i am going 2 put the work 2 where it is 2 there standard i aint going 2 do sum over there standards of science.

Explanation:

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the average power of the Sun is 3.79x1026 Watts, answer the following questions: 13. What is the average intensity of light at E

Lemur [1.5K]

Explanation:

Given that,

Average power of sun $P=3.79\times10^{26}\ Watt$

We need to calculate the intensity of light at Earth's position

Using formula of intensity

$I=\dfrac{P_{avg}}{4\pi r^2}$

Where, I = intensity

P = power

Put the value into the formula

$I=\dfrac{3.79\times10^{26}}{4\pi\times(1.496\times10^{11})^2}$

$I=1347.616\ W/m^2$

So, The intensity is 1347.616 W/m².

(A). We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully reflective

$P = \dfrac{2I}{c}$

Put the value into the formula

$P=\dfrac{2\times1347.616}{3\times10^{8}}$

$P=8.984\times10^{-6}\ N/m$

(B). We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully absorptive

$P=\dfrac{I}{c}$

$P=\dfrac{1347.616}{3\times10^{8}}$

$P=4.492\times10^{-6}\ N/m$

Hence, This is the required solution.

6 0

3 years ago

In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an

natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

$A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}$

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

$=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa$

σ=5517.25 Pa

Strain in x direction

ε=σ/E

$=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}$

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0

3 years ago

what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground

love history [14]

Answer:

I will say that the the potential energy will be at its maximum.

Explanation:

potential energy deals with gravity and gravity deals with height, so when a object is in its maximum height it will have the maximum potential energy.

6 0

3 years ago

A running back with a mass of 70 kg travels down the field with a velocity of 5.0 m s . Calculate the kinetic energy of the foot

jarptica [38.1K]

K.E = 1/2 mv²

= 1/2 (70kg) (5.0ms)²

= 1 ˣ ( 1750) / 2

= 875 kgms²

3 0

4 years ago

Read 2 more answers

Which statement is true of stellar evolution?

noname [10]

There are no true statements at all on the list of choices
that you included with the question.

8 0

3 years ago