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oksian1 [2.3K]
4 years ago
6

According to research, college graduates are more likely to engage in all of the following EXCEPT:

Physics
2 answers:
zheka24 [161]4 years ago
8 0

The answer is A, this is question is strange but I believe the answer is A

Ksivusya [100]4 years ago
7 0
Strange question. I’d say B)
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Which of the following is one part of a chemical formula?
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Answer:A

Explanation: number that shows the total atomic mass of the substance

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3 years ago
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A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

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8 0
3 years ago
A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a
finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$  after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y $

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

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Therefore, speed of the disk 2 after collision is 4.33 m/s

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3 years ago
How do flower petals help a flowering plant reproduce?
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3 years ago
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Explain how an object moving at a constant speed can have a nonzero acceleration
SOVA2 [1]
Its very simple if a body is moving in circle the magnitude of its velocity remain constant  but its direction changes because velocity is directed towards tangent and at every point in a cirlce its direction will be different (along tangent) so velocity is not uniform .As acceleration is the rate  change of velocity  so it will be non zero because velocity is changing due to its direction.
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3 years ago
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