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AfilCa [17]
3 years ago
5

What is the velocity of the girl in Figure b? Please help me!!!! Is the answer A,B,C,D,E, or F?

Physics
1 answer:
attashe74 [19]3 years ago
6 0
There is a Blank screen

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Plz help will give 26 points !and be mark as brainliest !
Dimas [21]

<em>Answer: </em>

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4 0
3 years ago
Read 2 more answers
What acceleration of a 45 kg mass pushed with 15 N of force?
Hitman42 [59]

Explanation:

Fnet = ma.

Assume that there is no resistance acting on the mass. Therefore a = 15N / 45kg = 0.33N/kg or 0.33m/s^2.

3 0
4 years ago
When light passes from a faster medium into a slower medium, which of the following explains what will occur?
beks73 [17]

When light passes from a faster medium into a slower medium, light will be refracted toward a line drawn perpendicular to the point of refraction. <em>(B)</em>

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4 years ago
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Froghopper insects have a typical mass of around 11.1 mg 11.1 mg and can jump to a height of 55.3 cm. 55.3 cm. The takeoff veloc
Molodets [167]

Answer:

a = 2710m/s²

Explanation:

See attachment below.

4 0
3 years ago
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Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
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