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RSB [31]
3 years ago
5

Four point masses, each of mass 1.9 $kg$ are placed at the corners of a square of side 2.2 $m$. Find the moment of inertia of th

is system about an axis that is perpendicular to the plane of the square and passes through one of the masses.

Physics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

I = 36.78 kg m^{2}

Explanation:\

Given data:

side of square = 2.2 m

mass =1.9 kg

The moment of inertia i is the total sum of the moments of inertia of the 4 point masses  and it is given as

I = I_1+I_2+I_3

= mr^{2} +mr^{2}+m(\sqrt2 r)^{2}

       = 2mr^{2}+2mr^{2}

         =4mr^{2}

       = 4*1.9*2.2^{2}

I = 36.78 kg m^{2}

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What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.
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Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

KE=\frac{1}{2}mv^2

where <em>m </em>is the mass in kilograms and <em>v</em> is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

<em>m</em>=50

<em>v</em>=60

Substitute these values into the formula.

KE=\frac{1}{2}*50*60^2

First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.

60^2=60*60=3,600

KE=\frac{1}{2}*50*3,600

Multiply 50 and 3,600

KE=\frac{1}{2}*180,000

Multiply 1/2 and 3,600, or divide 3,600 by 2.

KE=90,000

Add appropriate units. Kinetic energy uses Joules, or J.

KE=90,000 Joules

The kinetic energy of the object is 90,000 Joules

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A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

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3 years ago
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