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Phoenix [80]
3 years ago
6

Can somebody please help me?? I know this is a personal question of what I believe in but I would want to visit it... If you cou

ld visit the International Space Station, would you want to? Tell why, and give reasons for your decision.
Physics
1 answer:
12345 [234]3 years ago
8 0
Ok. The First Reason is I would want to go is because I love Space And being able to go to the International Space Station would be cool to see how thing work and learn about space..
Hope It helps.
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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
S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr
cestrela7 [59]

The kinetic energy of this block-spring when the block has a speed (v) is given by K.E = 1/2 × (M + m/3)v².

<h3>What is kinetic energy?</h3>

Kinetic energy can be defined as a form of energy that is possessed by a person due to its motion or change in speed (acceleration).

<h3>How to calculate kinetic energy?</h3>

Mathematically, kinetic energy can be calculated by using this formula:

K.E = 1/2 × mv²

Where:

  • K.E represents the kinetic energy.
  • m represents the mass.
  • v represents the speed or velocity.

Since the mass of a segment of this spring is dm = (m/l) dx, the kinetic energy for each of its segment would be given by:

dK = 1/2 × (dm)Vx²

This ultimately implies that, the kinetic energy of this block-spring when the block has a speed (v) is given by:

K.E = 1/2 × Mv² + 1/2 × ¹∫₀((x²v²/l²)m/ldx

K.E = 1/2 × (M + m/3)v².

Read more on kinetic energy here: brainly.com/question/15848455

#SPJ4

6 0
2 years ago
The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizont
Rudiy27

Answer:

F = 22.75 lb

μ₁ = 0.15

Explanation:

The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,

F = μR

F = μW

where,

F = Smallest force needed to move dresser = ?

μ = coefficient of static friction = 0.25

W = Weight of dresser = 91 lb

Therefore,

F = (0.25)(91 lb)

<u>F = 22.75 lb</u>

<u></u>

Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:

F = μ₁W₁

where,

μ₁ = coefficient of static friction between shoes and floor

W₁ = Weight of man = 151 lb

Therefore,

22.75 lb = μ₁ (151 lb)

μ₁ = 22.75 lb/151 lb

<u>μ₁ = 0.15</u>

7 0
4 years ago
Starting velocity: 50 m/s
Taya2010 [7]

Please find attached photograph for your answer. Please do comment whether it is useful or not

6 0
3 years ago
If you want come on in bih <br> ID: 920 5432 3648<br> PW: 2c4zvQ
ElenaW [278]

Answer:

What's this to?

Explanation:

3 0
3 years ago
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