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Softa [21]
3 years ago
9

I need help please ASAP

Physics
2 answers:
mash [69]3 years ago
6 0

Answer:

Have a great day..

Explanation:

01)275 miles=c

02)42 miles=c

03)15mph=d

04)7 hours=c

05)5.7mph=b

den301095 [7]3 years ago
5 0

Answer:

1st. C            

2nd. C

3rd. D

4th. C

Explanation:

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Yes

Explanation:

Kinetic energy is K.E1/2mv2 so that means it is directly proportional to mass and velocity.

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The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular mo
emmainna [20.7K]

Answer:

b. 84 minutes

Explanation:

a_c=g = Centripetal acceleration = 9.81 m/s²

r = Radius of Earth = 6378.1 km

v = Velocity

Centripetal acceleration is given by

a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s

Time period is given by

T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes

The time period of Earth’s rotation would be 84.43835 minutes

7 0
3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a
Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

d_x = 5 m

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d_y = 6 m

The rock is thrown horizontally with velocity v_x, so we can rewrite the horizontal distance as

d_x = v_x t

where t is the time of flight. Re-arranging the equation,

t=\frac{d_x}{v_x} (1)

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d_y = \frac{1}{2}gt^2

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t=\sqrt{\frac{2d_y}{g}} (2)

Equating (1) and (2), we can solve the equation to find v_x:

\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s

6 0
3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

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- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

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- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

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The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

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