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IgorC [24]
4 years ago
8

Jane is pulling a chain of 2 boxes with a massless rope at an angle of 20 degrees above the horizontal with a force of 30N. Jane

and the boxes are all on horizontal ground. The boxes are connected horizontally by a length of massless rope. The box closest to Jane is 45 kg and the box farthest from Jane is 30kg.
a. Draw a picture of the situation.
b. Draw a free-body diagram for each box.
c. What is the normal force acting on the 45kg box?
d. What is the acceleration of the boxes?
e. What is the tension in the rope connecting the boxes?
Physics
1 answer:
Rudik [331]4 years ago
8 0
The answer might be D
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Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138
ira [324]

Answer: 276 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:  

A=A_{o}.2^{\frac{-t}{h}} (1)  

Where:  

A=\frac{1}{4}A_{o} is the final amount of the material

A_{o} is the initial amount of the material  

t is the time elapsed  

h=138 days is the half life of polonium-210

Knowing this, let's substitute the values and find t from (1):

\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}} (2)  

\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}} (3)  

\frac{1}{4}=2^{\frac{-t}{138 days}} (4)  

Applying natural logarithm in both sides:

ln(\frac{1}{4})=ln(2^{\frac{-t}{138 days}}) (5)  

-1.386=-\frac{t}{138days}ln(2) (6)  

Clearing t:

t=276days (7)  

7 0
3 years ago
Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface
mina [271]
We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = \frac{1}{2} m*v^2
GPE (Gravitational Potential Energy) = m*g*h

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
TE =  \frac{1}{2} m*v^2 + m*g*h

Solving:
TE = \frac{1}{2} m*v^2 + m*g*h
TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50
TE =  \frac{2.0*100}{2} + 1000
TE =  \frac{200}{2} + 1000
TE = 100 + 1000
\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark


5 0
3 years ago
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A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of
wlad13 [49]
The car's average acceleration would be 1.25m/s^2 or 1.25meters/second/second. That looks to be the fourth one you've listed.
7 0
3 years ago
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PLS HELP WITH THIS FOR BRAINLIST IF ITS RIGHT
Zolol [24]

Answer:

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5 0
4 years ago
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi
Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

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b) momentum of ball after collision

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c) change in momentum

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    Δ P = -1.696 -1.535

    Δ P = -3.231 kg.m/s

d) using conservation of momentum

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  M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

  0.331 u' = 3.232

     u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0
4 years ago
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