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3 years ago

10

The _____ theory proposes that the moon was a passing asteroid pulled into orbit by Earth's gravity.

2 answers:

dem82 [27]3 years ago

5 0

C). capture theory is the correct answer.

Debora [2.8K]3 years ago

3 0

It's the capture theory.

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Determine the equivalent capacitance between points a and <br> b.

Solnce55 [7]

Where is the figure ?

7 0

3 years ago

g An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of It carries a current of 0.300 A. (a) Wha

valkas [14]

Answer:

Explanation:

Energy of an inductor = 1/2 L i²

L is inductance , i is current .

= 1/2 x 12 x .3²

= .54 J

4 0

3 years ago

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (a) At this instant, wha

zaharov [31]

Answer:

(A) L = 115.3kgm²/s

(B) dL/dt = 94.1kgm²/s²

Explanation:

The magnitude of the angular momentum of the rock is given by the foemula

L = mvrSinθ

We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.

Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =

115.3 kgm²/s

(B) The magnitude of the rate of angular change in momentum is given by

dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²

7 0

3 years ago

A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s

Amiraneli [1.4K]

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

$K.E_{l}=\dfrac{1}{2}mv^2$

$K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2$

$K.E_{l}=1.279\ J$

We need to calculate the rotational kinetic energy

$K.E_{r}=\dfrac{1}{2}\times I\omega^2$

$K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2$

$K.E_{r}=\dfrac{1}{4}\times m\times v^2$

$K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2$

$K.E_{r}=0.639\ J$

The total kinetic energy is given by

$K.E=K.E_{l}+K.E_{r}$

$K.E=1.279+0.639$

$K.E=1.918\ J$

Hence, The cylinder’s total kinetic energy is 1.918 J.

8 0

3 years ago

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

Margarita [4]

Answer:

A) $h = 69.58 m$

D) $v = 58.12 \frac{m}{s}$ (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

$V_{ox} = V_{o}Cos\alpha _{o}$ Formula (1)

$V_{oy} = V_{o}Sin\alpha _{o}$ Formula (2)

Where:

Vo: Initial velocity in m/s

$\alpha_{o}$ : Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

$v_{y}^{2} = v_{oy}^{2} -2gy$ Formula (3)

Where:

$v_{f}^{2}$ : Final speed component in vertical direction in m/s

$v_{oy}^{2}$ : Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

$v_{x} = v_{ox}$ Because the movement is uniform in the x direction (constant speed)

$v_{y} = v_{oy}-g*t$ Formula (4) Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

$v_{o} = 65.2 \frac{m}{s}$

$\alpha _{o}$ = 34.5º above the horizontal

$g=9.8 \frac{m}{s^{2}}$

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

$v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}$

$v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}$

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

$0=(36.93)^2-2*9.8*h$

$h=\frac{ (36.93)^2}{2*9.8} = 69.58 m$

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

$v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}$

$v_{x} = v_{ox} = 53.7 \frac{m}{s}$

$v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}$ (Speed magnitude)

$\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})$

α = 22.49° (Speed direction above the horizontal)

6 0

3 years ago