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tino4ka555 [31]
3 years ago
5

Does spring constant change

Physics
1 answer:
slava [35]3 years ago
8 0

Answer:

Yes it changes.

But it Spring constant k does not change when the spring is deformed

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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
4 years ago
A car with mass m-1000 kg completes a turn of radius r 540 m at a constant speed of v =25 m/s. As the car goes around the turn,
Nataly [62]

Answer:

friction coefficient on the rough road is 0.117

Explanation:

As we know that car is taking turn on flat rough road

So here we can say that required centripetal force for circular motion of car is due to frictional force

So we have

F_c = \frac{mv^2}{R}

now we have

\mu mg = \frac{mv^2}{R}

so we have

\mu = \frac{v^2}{Rg}

\mu = \frac{25^2}{540(9.81)}

\mu = 0.117

6 0
4 years ago
Please help ASAP WILL GIVE BRAINLIEST
Westkost [7]

Answer:

D

Explanation: It makes the most sense. Plz mark brainliest

5 0
3 years ago
Read 2 more answers
On a circular racetrack with a radius of 3.23 m, a certain road car set a speed record of 3.72
abruzzese [7]

Answer:

F = 1071.08 N

Explanation:

Given that,

The radius of the circular track, r = 3.23 m

Speed of the car, v = 3.72 m/s

The mass of the car, m = 250 kg

We need to find the net force acting on it. The centripetal force will act on it and it is given by the relation as follows :

F=\dfrac{mv^2}{r}\\\\F=\dfrac{250\times (3.72)^2}{3.23}\\\\F=1071.08\ N

So, the net force acting on the car is 1071.08 N.

6 0
3 years ago
A 71 kg man is walking at 2 m/s. Calculate the Kinetic Energy for the man
Westkost [7]

Answer:

<h2>142 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 71 \times  {2}^{2}  \\  =  \frac{1}{2}  \times 71 \times 4 \\  = 71 \times 2

We have the final answer as

<h3>142 J</h3>

Hope this helps you

7 0
3 years ago
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