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3 years ago

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Using your outline and the materials you’ve gathered, write a 500- to 750-word paper using word processing software. Be sure to

proofread and revise your writing to catch any errors in grammar, spelling, logic, or organization. Add a works cited page at the end to give credit to your sources. Submit your completed paper along with this activity to your teacher for evaluation.
please do it based off of Werner Heisenberg the scientist.

1 answer:

ipn [44]3 years ago

3 0

Answer:

im half new how does this work

Explanation:

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it takes a school bus full of kids longer to stop than it does a small car with a single passenger. which law is this an example

exis [7]

Newton´s second law of motion

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4 years ago

3(NH4)2CO3 what does the 3 in front mean?

VladimirAG [237]

Answer:

3 in front means the no.of NH4 atoms

Hope it is correct ^_^

7 0

3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

3 years ago

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago

Two physics students are standing on skateboards and facing each other. Their masses, including the skateboards, are 95 kg and 5

andrey2020 [161]

Answer:

1) $\sum \vec p=0\ kg.m.s^{-1}$

2) $v_2=55.078\ m.s^{-1}$

Explanation:

Given:

mass of heavier student, $m_1=95\ kg$

mass of lighter student, $m_2=58\ kg$

velocity of the heavier student after the mutual push, $v_1=3.1\ m.s^{-1}$

1)

Since they push each other and we are neglecting any friction that means we can consider this as a perfectly elastic collision. So the momentum of the two bodies will be equal and in opposite direction.

Since momentum is a vector quantity so the total momentum of the system as a whole will be zero.

2)

<u>From the condition of elastic collision:</u>

$p_1=p_2$

$m_1.v_1=m_2.v_2$

$95\times 3.1=58\times v_2$

$v_2=55.078\ m.s^{-1}$

6 0

4 years ago