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3 years ago

What factors affect attractive force

2 answers:

8 0

Two Factors That Affect How Much Gravity Is on an Object. Gravity is the force that gives weight to objects and causes them to fall to the ground when dropped. Two major factors, mass and distance, affect the strength of gravitational force on an object.

Dmitriy789 [7]3 years ago

5 0

attractive force only means force pulling two things together but factors affecting the force depend on the kind of force. gravity is an attractive force n the other helper gave the factors.

another attractive force is magnetic force and how strong the magnet will determine how strong the attractive force.

electrostatic force can also be attractive and the amount of electric charge will determine the force.

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The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

The work is equal to

W = 5886 [J]

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

3 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

Which statement about the carbon cycle is most true?

scZoUnD [109]

B answr is right there

3 0

3 years ago

water flows through a horizontal pipe with a cross-sectional area of 4m^2 at a speed of 5m/s with a pressure of 300,000pa at poi

LenKa [72]

The velocity at point B is 10 m/s with a pressure of 262500 Pa

<h3>Bernoulli equation</h3>

According to the continuity equation:

A₁V₁ = A₂V₂

Where A is the area and V is velocity, ρ = density of water = 1000 kg/m³

Hence:

4(5) = 2(V₂)

V₂ = 10 m/s

Using Bernoulli equation:

$P_1+\rho gh_1+\frac{1}{2}\rho V_1^2= P_2+\rho gh_2+\frac{1}{2}\rho V_2^2\\\\Hence:\\\\300000+\rho gh+(0.5)*1000*5^2=P_2+\rho gh+(0.5)*1000*10^2\\\\P_2=262500\ Pa\\$

The velocity at point B is 10 m/s with a pressure of 262500 Pa

Find out more on Bernoulli equation at: brainly.com/question/14082066

6 0

3 years ago

A certain sound is recorded by a microphone. The same microphone then detects a second sound, which is identical to the first on

katrin [286]

Answer:

Loudness of the second sound is more than the first one.

Explanation:

There are two sounds, the second sound is identical to first but the loudness of second is more than the first one.

As the frequency is same so the itch is same for both the sounds.

As the loudness depends on the amplitude of the sound so the loudness of the second sound is more than the first sound.

8 0

3 years ago