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3 years ago

Please help I’m so desperate I’m giving about all my points

Physics

1 answer:

alexira [117]3 years ago

7 0

Answer:

1 and 5

Explanation:

potential energy is movement energy, and at 1 and 5 there is a lot of kinetic energy (built up energy) and not a lot of potential energy

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At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago

50 point!! Will mark brainliest!! (Please help)

jok3333 [9.3K]

What is the variable?

~the price is the variable.

What happens to demand?

~It'll go down. Since the price of snow blowers will increase then the quantity demanded will go down.

Hope this helps- have a good day bro cya)

4 0

2 years ago

Read 2 more answers

The bonds of oxygen molecules are broken by sunlight. The minimum energy required to break the oxygen-oxygen bond is 495 kJ/mol.

makvit [3.9K]

Answer:

The wavelength of sunlight that can cause this bond breakage is 242 nm

Explanation:

The minimum energy of the sunlight that'll break Oxygen-oxygen bond must match 495 KJ/mol

But 1 mole of any molecule contains 6.02 × 10²³ molecules/mol

Each molecule of Oxygen will require (495 × 10³)/(6.02 × 10²³) = 8.22 × 10⁻¹⁹ J

E = hf

v = fλ

f = v/λ

f = frequency of the sunlight

λ = wavelength of the sunlight

v = speed of light = 3.0 × 10⁸ m/s

E = hv/λ

λ = hv/E

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(8.22 × 10⁻¹⁹)

λ = 2.42 × 10⁻⁷ m = 242 nm.

5 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago

1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

109.6 cm , - 1.74 , real

1.5

Explanation:

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

3 years ago