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3 years ago

9

The force applied when using a simple machine ??

1 answer:

beks73 [17]3 years ago

8 0

Answer : I hope this helps !

The Effort Force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.

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The first astronomer to show that spiral nebulae (today called spiral galaxies) have large Doppler shifts was

solong [7]

William Parsons was the first astronomer to show that spiral nebulae have large Doppler shifts.

<h3>What are Spiral nebulae?</h3>

In space, a nebula is a huge cloud of gas and dust. Some nebulae, including multiple nebulae, are made of gas and dust that have been released from the explosion of a dead star called a supernova. There are other nebulae that are star-forming regions.
Edwin Hubble began observing Cepheid variables in many spiral nebulae, including the alleged Andromeda Nebula, in 1923, demonstrating that they are, in fact, full-fledged galaxies outside our own. Since then, the phrase spiral nebula has lost favor.
One group of astronomers believed that spiral nebulae were parts of our Milky Way galaxy, whereas the other group believed that these objects were actual galaxies that were outside of the Milky Way galaxy.

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2 years ago

What is the number one and most important rule to remember in soccer?

AlekseyPX

To score in the correct goal

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3 years ago

Find the area under the standard normal curve between z=0.19 and z=2.18. round your answer to four decimal places, if necessary.

LiRa [457]

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

<h3>How can we calculate the area under the curve?</h3>

To calculate the The area under the standard normal curve between z=0.19 and z=2.18, we are using two things,

<u>Step 1</u>: The formula,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

<u>Step 2</u>: The statistical values of the area under the curve we get from the picture.

Now we put the known values from the picture in the above formula, we get,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

Or, P(0.19<z<2.18)= 0.9854-0.5753

Or, P(0.19<z<2.18)= 0.4101

From the above calculation we can easily conclude that,

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

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2 years ago

A block of mass 12.2 kg is sliding at an initial velocity of 6.65 m/s in the positive x-direction. The surface has a coefficient

valentina_108 [34]

Explanation:

Given that,

Mass of the block, m = 12.2 kg

Initial velocity of the block, u = 6.65 m/s

The coefficient of kinetic friction, $\mu_k=0.253$

(a)The force of kinetic friction is given by :

$f=\mu_k mg$

mg is the normal force

So,

$f=0.253\times 12.2\times 9.8\\\\f=30.24\ N$

(b) Net force acting on the block in the horizontal direction,

f = ma

a is the acceleration of the block

$a=\dfrac{f}{m}\\\\a=\dfrac{30.24}{12.2}\\\\a=2.47\ m/s^2$

(c) Let d is the distance covered by the block before coming to the rest. Using third equation of motion as follows :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(6.65)^2}{2\times 2.47}\\\\d=-8.95\ m$

Hence, this is the required solution.

3 0

4 years ago

A certain string on a piano is tuned to produce middle c (f = 261.63 hz) by carefully adjusting the tension in the string. for a

Lemur [1.5K]

Frequency and wavelength are two variables which are indirectly proportional.

They are related in the following equation:

f = c / w

Where,

<span>f = frequency c = speed of light w = wavelength</span>

Since c is constant, we can equate condition 1 and condition 2:

f1 w1 = f2 w2

When w2 = 3 w1, then f2 becomes:

261.63 w1 = f2 (3 w1)

Cancelling w1:

f2 = 261.63 / 3

<span>f2 = 87.21 Hz</span>

7 0

3 years ago