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morpeh [17]
3 years ago
8

If the initial energy of the system was 845.54 J and the final energy of the system is -104.99 J , how much work has been done?

Physics
2 answers:
Yuri [45]3 years ago
8 0
Whuwuwuwuwusuwbsnsns
enyata [817]3 years ago
8 0

Answer:

740.55

Explanation:

845.54 + -104.99

=740.55

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The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf
kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy  

V0x = V0 cos(θ) 

V0y = V0 sin(θ) 

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2 

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant  

Vx = Vocos(θ) 

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t 

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t 

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2 

3 0
4 years ago
When a bat hits a ball, the impulse is what?
german
I believe the answer is C


4 0
3 years ago
Read 2 more answers
Plzzz help will mark the brainliest
ANEK [815]
Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle


A satellite....etc
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8 0
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

6 0
3 years ago
Any object that has mass and takes up space is considered
sashaice [31]
Matter. I don't really know how to explain it. Sorry. But anyways, Hope this helps!
7 0
3 years ago
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