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Vinvika [58]
3 years ago
13

A bar on a hinge starts from rest and rotates with an angular acceleration α = 15+4t, where α is in rad/s2 and t is in seconds.

Determine the angle in degrees through which the bar turns in the first 2.00 min.
Physics
1 answer:
Arisa [49]3 years ago
3 0

Answer:

230.53°

Explanation:

α = angle in rad / t^2

t =2mins = 2 × 60 = 120sec.

α = angle in rad / 2^2

α = angle in rad / 16

15+4(2) = angle in rad /4

23 = angle/4

23× 4 = angle in radian

92 = angle in radian

But π rad = 180°

1 rad = 180 /π

Therefore

92 rad = 92 ×180/π = 5270.53°

This value is far more so we subtract from 360° continuously till we get a value less than 360°

5270.53- (360×14) =5270.53- 5040

230.53°

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A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo
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Your answer is: K.E = 8.3 J

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K.E = P.E and also P.E equals to mgh

Then you substitute all the parameters into the formula  ↓

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2 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
3 years ago
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