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murzikaleks [220]
3 years ago
8

A 50 kg crate is pulled across the ice by a rope that is angled at 30 degrees above the horizontal. If the tension in the rope i

s 100 newtons, what is the magnitude of the "lifting" and "pulling" components of the force in the rope?
Physics
1 answer:
denpristay [2]3 years ago
4 0
Let:
Vx = the pulling component of force
Vy = the lifting component of force

Vy:
Sin(n°) = Vy/hypotenuse
hypotenuse * Sin(n°) = Vy
100N*sin(30°) = Vy
50N = Vy

Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
about 86.6N = Vx
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If there is no existence of capacitors in our world there would be no electrical or electronic engineering.

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2 years ago
Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern
marusya05 [52]

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)   (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}

Hence, the fraction of the maximum intensity is I = 0.636*Imax

6 0
3 years ago
A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,
iris [78.8K]

Answer:

The work done by gravity is 4.975 \: Joules

Explanation:

The data given in the question is :

Mass is 0.245 kg

Height from ground is 2.07 m

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = mass \times gravity \times (change \: in \: height)

Work done = 0.245 kg \times 9.81 m/s^{2} \times 2.07 m

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3 years ago
A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i
timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

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6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}

\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}

\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s

4)Kinetic energy before:

E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J

Kinetic energy after:

E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

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3 years ago
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Analyzing the components and decay products of school lunches.

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