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3 years ago

What is the difference between O and O2

1 answer:

4 0

O and O2 are very different because one has 2 oxygen's and one has only 1.
Though they are both oxygen, they may bond differently.<span />

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A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul

jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, $m_a$ = 126 kg

Speed he acquires, $v_{a}$ = 2.70 m/s

Mass of the space capsule, $m_{c}$ = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

$P_f = m_av_a + m_cv_c$

$P_I$ = 0

$m_av_a + m_cv_c = 0$

$v_c =\frac{- m_a v_a}{m_c}}\\\\$

$= \frac{126* 2.70}{1800}$

$= - 0.189$ m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

= 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

= 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = $\frac{1}{2} m v^2$

$= \frac{1}{2}$ × 126 × $(2.70) ^2$

= 459.27 J

The Kinetic Energy of the capsule;

K.E = $\frac{1}{2} m v^2$

= $\frac{1}{2}$ ×1800× $(0.189) ^2$

= 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0

2 years ago

400 N of load can be overcome by an effort of 50 N by using a lever. Calculate the mechanical advantage of the lever.

mars1129 [50]

Answer:

load (l)=400N

Effort(E)=50N

mechanical advantage (MA)= load ÷Effort

(ma)=400÷50

(ma)=8

Explanation:

I copy pasted from the answer from the same question. Remember to first check if ur question is there

4 0

3 years ago

Calculate the force it would take to accelerate a 50 kg bike at a rate of 3 m/s2.

Ede4ka [16]

ummmm it might be 300... i used a calculator

sorry if it is wrong

4 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0

3 years ago