All categories

2 years ago

StA

Physics

1 answer:

sweet-ann [11.9K]2 years ago

3 0

the Answer is 91 meters

You might be interested in

Which of the following leads to new discoveries and new theories?

Fynjy0 [20]

)scientific knowledge

3 0

3 years ago

Read 2 more answers

What is the relationship between wavelength of light and the quantity of energy per photon?

slamgirl [31]

Answer:

The quantity of energy per photon is inversely proportional to the wavelength of the light.

Explanation:

Energy of light is given as

E = hf

where E = energy of the photons,

f = frequency of the light

If the number of photons = n

(E/n) = (h/n) f

Let (E/n) = E'

(h/n) = h'

But the frequency of light is related to wavelength through the relation

v = fλ

where v = speed of light = c

λ = wavelength of light

f = (c/λ)

E' = h' f

Substituting for f

E' = h' (c/λ)

h' and c are both constants, h'×c = K

E' = (K/λ)

So, the quantity of energy per photon is inversely proportional to the wavelength of the light.

Hope this Helps!!!

3 0

3 years ago

Read 2 more answers

An electromagnetic wave with high frequency and high energy is A. Safe for humans. B. Is helpful to humans. C. Is helpful to hum

Vilka [71]

B and C are both correct choices.

4 0

3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago

A cheetah starts from rest and accelerated at 8.7 m/s^2 for 3s. How far did the cheetah go in that time

nika2105 [10]

Answer:

cheetah goes 52.2 m in that time.

Explanation:

4 0

3 years ago