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3 years ago

For a solar eclipse to occur which of the following alignments must be necessary

1 answer:

6 0

Very specific alignment of the Sun, Earth, and Moon. If the Moon is lined up precisely with the Sun from the Earth's point of view, the Moon will block Sunlight from reaching the Earth, causing a solar eclipse.

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You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?

Murljashka [212]

Answer:

20 J

Explanation:

Given:

Weight of the book is, $W=10\ N$

Height or displacement of the book is, $d=2\ m$

The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.

So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.

Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.

Therefore, work done is given as:

$Work=W\times d\\Work=10\times 2=20\ J$

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.

3 0

2 years ago

Read 2 more answers

A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts

galina1969 [7]

Answer:

$\mu_s = 0.62$

$\mu_k = 0.415$

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

$x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2$

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

$F = ma\\F = 1.7m$

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

$F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415$

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

$mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62$

5 0

3 years ago

As the average kinetic energy of a substance increases, the tempature of the sample

taurus [48]

I have no clue. Sorry bro

6 0

2 years ago

Read 2 more answers

Which is correct? I need to know!!

Mnenie [13.5K]

What website are you using?

6 0

3 years ago

A diagram of a closed circuit with a power source on the left labeled 120 V. There are 3 resistors in parallel, separate paths,

Zina [86]

1) The equivalent resistance is $2.73\Omega$

2) The voltage in the circuit is 120 V

3) The total current in the circuit is 44.0 A

Explanation:

Resistors are said to be in parallel when they have the same potential difference at their terminals.

The formula to calculate the equivalent resistance for three resistors in parallel is:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

where $R_1, R_2, R_3$ are the resistances of the three resistors.

In this problem, the resistance of each resistor is:

$R_1 = 5.0 \Omega$

$R_2 = 10.0 \Omega$

$R_3 = 15.0 \Omega$

Substituting,

$\frac{1}{R_T}=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}=\frac{11}{30}$

So the equivalent resistance is

$R_T = \frac{30}{11}=2.73 \Omega$

In this problem we are told that the three resistors are connected in parallel. This means that their terminals are connected to the same point of the circuit: therefore, this means that they also have the same potential difference across them.

Therefore, the voltage in each branch containing each resistor is the same, and it is equal to the voltage of the battery, which is

$V=120 V$