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Contact [7]
3 years ago
8

HELPPP

Physics
1 answer:
sergey [27]3 years ago
4 0

Answer:

?

Explanation:

more explanationsplisss

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Consider a thin walled hoop of uniform material. Two axis go through the center of the hoop. One axis is perpendicular to the pl
ratelena [41]

Answer:

Explanation:

Moment of Inertia about an axis passing through its center and Perpendicular to its Plane is given by

I_{zz}=Mr^2

As all the mass is at radius therefore its moment of inertia is more than the moment of inertia about a axis parallel to the Plane

According to perpendicular axis theorem

I_{zz}=I_{xx}+I_{yy}

and I_{xx}=I_{yy} is same due to symmetry

thus I_{xx}=\frac{1}{2}\times I_{zz}

I_{xx}=\frac{Mr^2}{2}

thus Perpendicular z axis will have more moment of inertia

                 

7 0
3 years ago
Explain why adding oil to the moving parts of a machine can increase its efficiency
Vinvika [58]
Adding oil to the moving parts of machines can do that because it makes it smoother and takes the rust off of it.
6 0
3 years ago
Read 2 more answers
A force of 500 N acts horizontally on a 10,000 g body. What is its horizontal acceleration
Paladinen [302]
200n because it's 2×5=10so maybe try solving the problem like that ok does that help
8 0
4 years ago
A metal wire has a resistance of 14.00 Ω at a temperature of 25.0°C. If the same wire has a resistance of 14.55 Ω at 90.0°C, wha
aliina [53]

Answer:

13.52 Ω

Explanation:

coefficient of thermal resistance be α

R₀ , R₂₅ , R₉₀ and R₋₃₂ be resistances at 0 , 25 , 90 , and - 32 degree

R₂₅ = R₀ + α x 25

R₉₀ = R₀ + α x 90

R₉₀ - R₂₅ = 65 x α

α = (R₉₀ - R₂₅ )/ 65

= (14.55 - 14) / 65

=   .55 / 65 Ω per °C,

R₂₅ = R₀ + α x 25

14 = R₀ + (.55 / 65 )x 25

=  R₀ + .2115

R₀ = 13.7885 Ω

R₋₃₂ = R₀ - α x 32

= 13.7885 -(  .55 / 65) x 32

=  13.7885 - .27077

= 13.51773 Ω

= 13.52 Ω

7 0
4 years ago
Read 2 more answers
sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he
podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0
3 years ago
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