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shusha [124]
3 years ago
12

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

You might be interested in
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=301.4Ω

\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?

~<em>the</em><em> </em><em>price</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>variable</em><em>.</em>

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:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

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

2

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
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