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mixer [17]
3 years ago
8

HELP!! Worth a lot of points :) (Physics question)

Physics
2 answers:
Temka [501]3 years ago
8 0
My answer would be B.
Source:
i had this question before.
Hope that this helps you! =)
nadezda [96]3 years ago
6 0
It's B if that helps at all :) (I'm like, 98% sure that it's totally B!
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A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} ​F ​⃗ ​​ so that the swing ropes
Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

 T cos θ = m g

 T cos 30° = 30 x 9.8

 T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

                                           = 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

7 0
3 years ago
In regions where few species existed before or where species were wiped out what occurs?
marysya [2.9K]
I cant really say but i believe it is called extinction
5 0
3 years ago
A(n) ________ microscope keeps an object in focus when the objective lens is changed.
Anvisha [2.4K]

Parfocal is the term used to describe a microscope that maintains focus when the objective lenses are replaced.

<h3>What is the name of the objective lens ?</h3>

For observing minute features within a specimen sample, a high-powered objective lens, often known as a "high dry" lens, is perfect. You can see a very detailed image of the specimen on your slide thanks to the 400x total magnification that a high-power objective lens and a 10x eyepiece provide.

The four objective lenses on your microscope are for scanning (4x), low (10x), high (40x), and oil immersion (100x).

The first-stage lens used to create a picture from electrons leaving the specimen is referred to as the "objective lens." The objective lens is the most crucial component of the imaging system since the quality of the images is determined by how well it performs (resolution, contrast, etc.,).

To learn more than objective lens , visit

brainly.com/question/17307577

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6 0
2 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the
Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

(n1)*(sin(theta1))=(n2)*(sin(theta2))

Then we will find the refraction angle relative to the vertical this way:

(n1/n2)*(sin(theta1))=sin(theta2)

(1/1.33)*(sin(45))=sin(theta2)

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

tan(32.12)=(X/30cm)

X=(tan(32.12))*(30cm)

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

tan(32.12)=(24cm/Y)

Y=(24cm/tan(32.12))

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0
3 years ago
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