Which change would cause the largest increase in a mountain climber’s gravitational potential energy?

Calculate the displacement of an object at 2.0 seconds when thrown straight up with an

Licemer1 [7]

Answer:

Explanation:

Velocity is the change of rate of displacement with respect to time.

velocity = displacement/time

Given

initial velocity = 15 m/s.

time taken =2 secs

Required

Displacement of the object

From the formula;

Displacement = Velocity * time

Displacement = 15 * 2

Displacement = 30m

<em>Hence the displacement of the object is 30m</em>

8 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

When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p

Shtirlitz [24]

Answer:

the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:

This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is

fr- w || = 0

The expression for friction force is that it is proportional to the coefficient of friction by normal.

fr = μ N

When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.

In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.

μ kinetic <μ static

In all this movement the normal with changed that the angle of the table remains fixed.

Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

3 0

3 years ago

What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters

Artist 52 [7]

Answer: $3.61(10)^{3} \frac{m}{s^{2}}$