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kobusy [5.1K]
3 years ago
11

Do the data for the first part of the experiment support or

Physics
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:

your the class suru and Sahiba and Sahiba are all guddu saifi and Mam aap Aman and Sahiba are not

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What color do glacier appear if they are moving towards us?
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Subcategories: Ice field

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Glacier ice is blue because the red (long wavelengths) part of white light is absorbed by ice and the blue (short wavelengths) light is transmitted and scattered. The longer the path light travels in ice, the more blue it appears.

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Which statements describe the motion of the object represented by the graph? Check all that apply.
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An apple is held completely submerged just below the surface of water in a container. The apple is then moved to a deeper point
Tresset [83]

Answer:

Explanation:

When the apple is held submerged in water , it experiences a buoyant force due to which it floats in water . One has to apply downward force to keep it submerged. The lower the buoyant force , lower the force needed to submerge it in water.

When apple is held at much deeper point , it experience greater pressure due to column of water around it . So its size or its volume decreases . But its weight remains the same . Due to less volume , buoyant force also decreases ( buoyant force is equal to weight of displaced volume of water. )

Due to buoyant force becoming less , force needed on apple  in downward direction will also be less.

4 0
3 years ago
A sample of nitrogen gas is inside a sealed container. The volume of the container decreases while the temperature is kept const
frez [133]
I guess it’s d) isobaric mate correct me if I am wrong :D
7 0
2 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
3 years ago
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