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Kamila [148]
2 years ago
12

~~~~~NEED HELP ASAP~~~~~

Physics
1 answer:
Romashka-Z-Leto [24]2 years ago
8 0

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

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zubka84 [21]
A hydrogen<span> atom contains 1 </span>electron<span>, 1 proton, and no neutrons based on the Periodic Table. The isotopes of the </span>hydrogen<span> atoms however, do contain neutrons.</span>
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3 years ago
HELP ASAP! <br>why is rolling friction much smaller than sliding friction? Also give an example.​
IceJOKER [234]

Answer:

Rolling friction is much smaller than sliding friction because Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

Rolling Friction example: Anything with weels (cars,skateboards) or a ball rooling.

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3 years ago
A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a
AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x 10^{24} Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = \frac{GM}{r^{2} }

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g =  \frac{(6.67 * 10^{-11}) (5.972 * 10^{24})  }{5733^{2} }

g = 12.12 m/s^{2}

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}

6 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
Identify the law, write the equation and calculate the answer to the problem below.
lyudmila [28]

Find refractive index first

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\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}

\\ \rm\Rrightarrow \mu =0.75

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\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75

\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr

\\ \rm\Rrightarrow sinr=0.94

\\ \rm\Rrightarrow r=sin^{-1}(0.94)

\\ \rm\Rrightarrow r=70^{\circ}

5 0
2 years ago
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