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postnew [5]
2 years ago
6

There is a skier at the top of a ski slope. The skier has potential energy. What gives the skier his potential energy?

Physics
1 answer:
masha68 [24]2 years ago
3 0
A is the answer hope it helps yiu
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In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?
tatuchka [14]

B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

v_{x}^{2} = v_{ox}^{2} +2 a x (x - x₀)

v_{x} = velocity at position "x"

v_{ox} = velocity at position "x = 0 "

x = final position

x_{o} = initial position of the object at the start of the motion

6 0
3 years ago
Read 2 more answers
A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
8090 [49]

Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0
3 years ago
Why is it not suitable to determine the volume of an irregular charcoal using displacement method ​
AleksandrR [38]

Answer:

Displacement method of volume measurement is no suitable

Explanation:

Displacement method of volume measurement is no suitable for the objects that do not get immersed into the water completely because of the hindrance in accuracy of the measurement.

4 0
3 years ago
How far does a car go in 30 seconds at a speed of 29 m/s? (The equation for<br> distance is d = st.)
Harrizon [31]

d = speed x time

distance = 29 x 30

distance = 870m

please mark as BRAINLIEST

8 0
2 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
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