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Ilia_Sergeevich [38]
3 years ago
12

A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. What is the vertical component

of the ball’s velocity? What is the horizontal component of the ball’s velocity?
0 m/s; 70 m/s
61.8 m/s; 32.9 m/s
32.9 m/s; 61.8 m/s
70 m/s; 0 m/s
Physics
1 answer:
icang [17]3 years ago
3 0

Answer:

answer is 61.8 m/s; 32.9

l am not sure

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Which of these is an advantage of using models to study tectonic plate movements?
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Actual plate movements can be made les frequent.
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What is speed of wave in terms of frequency and wavelength??​
ratelena [41]

Answer:

In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form, If the crest of an ocean wave moves a distance of 20 meters in 10 seconds, then the speed of the ocean wave is 2.0 m/s.

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Explanation:

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3 years ago
What is newton's second law of motion?
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3 years ago
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

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