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Crank
3 years ago
8

What causes objects to move or stay still (support your details with claims and evidence)

Physics
1 answer:
maxonik [38]3 years ago
6 0
Force moves the object but if the same anyone force is applied to both sides then it doesn’t move
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Is the star with the longest total lifetime Also the farthest from earth? Explain.
Stels [109]
E) No. Ollie will shine for 30 Billion years but is only 10,000 LY from Earth.
F) No. Cosmo will shine for 3 Million years but is 10 Billion LY from Earth.
G) Yes. Ollie is only 10.000 LY away but will shine for 30 Billion years.
Ga) No. Stars such as Cosmo shine for 3 Million years.
Gb) If Cosmo was also 3 Million LY away we would see it now.
6 0
3 years ago
018 10.0 points
-BARSIC- [3]

Answer:

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6 0
2 years ago
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
3 years ago
Find the intensity of a 55 dB sound given I 0=10^-12W/m^2
MakcuM [25]

Answer:

3.16 × 10^{-7} W/m^{2}

Explanation:

β(dB)=10 × log_{10}(\frac{I}{I_{0} })

I_{0}=10^{-12} W/m^{2}

β=55 dB

Therefore plugging into the equation the values,

55=10 log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

5.5= log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

10^{5.5}= \frac{I}{10^{-12} }

316227.76×10^{-12}= I

I= 3.16 × 10^{-7} W/m^{2}

5 0
2 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
3 years ago
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