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Mamont248 [21]
3 years ago
5

Find the value of x. 6x-1 42 10x+5 a 78 x = [?]

Physics
1 answer:
Arada [10]3 years ago
6 0

Answer: x=6

Explanation:

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A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down slowly (but does not dr
Anna11 [10]

1. lifts it chest high

The force opposing to this action is the force due to gravity. Therefore the work done is:

W1 = m g d

where m is mass of the barbell, g is gravity and d is displacement

 

2. holds it for 30 seconds

Work is a product of force and displacement, since there is no displacement, therefore work done is zero.

W2 = 0

 

3. puts it down slowly

If the barbell was dropped, then it would simply be a free fall. But since it was not, so the work done here is also equal to the weight of the barbell times displacement:

W3 = m g d

 

We can see that W1 = W3, and since W2 = 0, therefore the answer is:

<span>w3 = w1 > w2</span>

7 0
3 years ago
Can you hellp me please
mamaluj [8]

Answer:

i think the answer is B but im not sure

7 0
3 years ago
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Why did earth most likely form into distinct, separate layers? (HELP ASAP PLZ)
g100num [7]

Answer:

C

Explanation:

Rocks with hotter temperatures sank to the bottom of the Earth. The rocks with the hottest temperature became the core and the rocks with the least amount of heat became the crust.

Hope this helps!

8 0
3 years ago
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Is the ratio between the sine of an angle of incidence to the sine of an angle of refraction is called the refractive index?
motikmotik

Answer:

Yes

Explanation:

Yes it is called the refractive index denoted by n

n=sin<i/sin<r

6 0
3 years ago
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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
wlad13 [49]

Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}

where:

T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})

4 0
3 years ago
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