Can someone help me with this?

Why is the unit of work is a derived unit<br>

masha68 [24]

Answer:

Since energy can be measured as work, we can write energy = force x distance. Thus SI derived unit of energy has the units of newtons x meter or kg m2/s2.

Explanation:

5 0

2 years ago

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Perhaps to confuse a predator, some tropical gyrinid beetles(whirligig beetles) are colored by optical interference that is duet

gulaghasi [49]

Answer:

The grating spacing of the beetle is $1.078*10^{-6}m$

Explanation:

The concept to solve this problem is relate to interference effect given in the Young's Slits. Here was demonstrated that the length of the side labelled \lambda is known as the path difference. The equation is given by,

$n\lambda = dsin\theta$

Where,

$\lambda$ = wavelenght of light

N = a positive integer: 1,2,3...

$\theta$ = Angle from the center of the wall to the dark spot

d= width of the slit

Replacing our values we have that for n=1,

$dsin\theta = n\lambda$

$dsin(30)=(1)(539*10^{-9})$

$d=1.078*10^{-6}m$

Therefore the grating spacing of the beetle is $1.078*10^{-6}m$

6 0

3 years ago

A chemical reaction occurs. Which of the following would indicate that energy is transformed during the reaction?

weqwewe [10]

The answer would be 4. All of the above

4 0

3 years ago

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The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t

slamgirl [31]

Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length $L=5.0\times5.0\times10^{-4}\ m$

We need to calculate the linear charge density

Using formula of linear charge density

$E=\dfrac{2k\lambda}{r}$

$\lambda=\dfrac{Er}{2k}$

Put the value into the formula

$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

$\lambda=1.1\times10^{-10}\ C/m$

We need to calculate the charge

Using formula of charge

$Q=\lambda\timesL$

Put the value into the formula

$Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})$

$Q=2.75\times10^{-13}\ C$

Hence, The charge is $2.75\times10^{-13}\ C$

5 0

3 years ago

"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat

vagabundo [1.1K]

Answer:

(a) g = 8.82158145 $m/s^2$ .

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

$g = \frac{M_{(earth)} }{r^2} G$ , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145 $m/s^2$ .

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

$a_{centripetal}=\frac{V^2}{r} =g$ here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c) S = vT, here T is time period or time required to complete one full revolution.

S = earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0

3 years ago