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3 years ago

14

The uniform bar of mass m and length l is balanced in the vertical position when the horizontal force p is applied to the roller

at
a. determine the bar's initial angular acceleration and the acceleration of its top point
b.

1 answer:

WARRIOR [948]3 years ago

7 0

Which of the following is NOT a factor in efficiency?<span><span>A.metabolism</span><span>B.type of movement</span><span>C.muscle efficiency</span><span>D.digestion</span></span>

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Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago

1. Write the mass number for an atom with atomic number 26 and 32 neutrons.

Strike441 [17]

1. Mass number = Protons + Neutrons
Mass number = 26 + 32 = 58

2. Let the element be Xy
58Xy26

3. The proton number of Strontium is 38.
Mass = 38 + 50 = 88

4. 88Sr38

5. Let the atom be Ab
72Ab32

5 0

2 years ago

A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m

Ugo [173]

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

6 0

2 years ago

Juanita lifts a round box and a square box to a shelf. The gravitational potential energy (GPE) for the round box increases by 5

Vinil7 [7]

Juanita did more work on the square box

Explanation:

According to the law of conservation of energy, the work done in lifting an object is equal to the increase in gravitational potential energy of the object. This is due to the fact that the work done on the object is converted into potential energy.

The potential energy of an object (GPE) is given by

$GPE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object

In this problem, we have two objects:

- The roud box is lifted and its GPE increases by 50 J --> this means that the work done by Juanita on the box is 50 J

- Thr square box is lifted and its GPE increases by 100 J --> this means that the work done by Juanita on the box is 100 J

Therefore, Juanita did more work on the square box.

Learn more about work and potential energy:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

3 years ago

Help for physics please!

Archy [21]

Answer:

g and a

Explanation:

3 0

3 years ago