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garri49 [273]
1 year ago
12

A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

hat the rope does not stretch. Use the value 9.81m/s2 for the acceleration of gravity. Calculate the tension T in the rope if the gymnast hangs motionless on the rope.
Physics
1 answer:
DiKsa [7]1 year ago
3 0

Answer:

T = 686.7N

Explanation:

For this exercise we will use Newton's second law in this case there is no acceleration,

      ∑ F = ma

      T -W = 0

The gymnast's weight is

     W = mg

We clear and calculate the tension

     T = mg

     T = 70 9.81

     T = 686.7N

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How long a time t will it take for the 133 54xe to decay so that eventually its activity decreases by a factor of 1024?
serg [7]
Base on my research the radioactive isotope X-133 has a half-life of 5 days. So base on the given it decreases by a factor of 1024, it represents 10 halvings. To get how long will it take for the 133 54xe to decay, just multiply 5 days with 10 halvings. The answer is 50 days for the 133 54xe to decay. 
8 0
2 years ago
Read 2 more answers
While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the
____ [38]

As per the question Bob drops the bag full with feathers from the top of the building.

The mass of the bag(m)= 1.0 lb

Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.

Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2


Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s

Hence time t= 1.5 s

From equation of kinematics we know that -

                S=ut + 0.5at^2     [ here S is the distance travelled]

For motion under free fall initial velocity (u)=0.

Hence   S= 0×1.5+{0.5×(-9.8)×(1.5)^2}

           ⇒ -S =0-11.025 m

            ⇒ S= 11.025 m

                   =11 m

Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .


Hence the correct option is B.

               

3 0
2 years ago
Read 2 more answers
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:
Nikitich [7]

Answer:

r ≥ R, E = Q / (4πR²ε₀)

r ≤ R, E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

Maximum at r = ⅔ R

Maximum field of E = Q / (3πε₀R²)

Explanation:

Gauss's law states:

∮E·dA = Q/ε₀

What that means is, if you have electric field vectors E passing through areas dA, the sum of those E vector components perpendicular to the dA areas is equal to the total charge Q divided by the permittivity of space, ε₀.

a) r ≥ R

Here, we're looking at the charge contained by the entire sphere.  The surface area of the sphere is 4πR², and the charge it contains is Q.  Therefore:

E(4πR²) = Q/ε₀

E = Q / (4πR²ε₀)

b) r ≤ R

This time, we're looking at the charge contained by part of the sphere.

Imagine the sphere is actually an infinite number of shells, like Russian nesting dolls.  For any shell of radius r, the charge it contains is:

dq = ρ dV

dq = ρ (4πr²) dr

The total charge contained by the shells from 0 to r is:

q = ∫ dq

q = ∫₀ʳ ρ (4πr²) dr

q = ∫₀ʳ ρ₀ (1 − r/R) (4πr²) dr

q = 4πρ₀ ∫₀ʳ (1 − r/R) (r²) dr

q = 4πρ₀ ∫₀ʳ (r² − r³/R) dr

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R) |₀ʳ

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R)

Since ρ₀ = 3Q/(πR³):

q = 4π (3Q/(πR³)) (⅓ r³ − ¼ r⁴/R)

q = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴)

Therefore:

E(4πr²) = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / ε₀

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

When E is a maximum, dE/dr is 0.

First, simplify E:

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

E = Q (4 (r³/R³) − 3 (r⁴/R⁴)) / (4πr²ε₀)

E = Q (4 (r/R³) − 3 (r²/R⁴)) / (4πε₀)

Take derivative and set to 0:

dE/dr = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = 4/R³ − 6r/R⁴

0 = 4R − 6r

r = ⅔R

Evaluating E at r = ⅔R:

E = Q (4 (⅔R / R³) − 3 (⁴/₉R² / R⁴)) / (4πε₀)

E = Q (8 / (3R²) − 4 / (3R²)) / (4πε₀)

E = Q (4 / (3R²)) / (4πε₀)

E = Q (1 / (3R²)) / (πε₀)

E = Q / (3πε₀R²)

3 0
2 years ago
A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi
Sholpan [36]

Answer:

i = 7.83 \mu A

Explanation:

Induced EMF in the coil is given by the equation

EMF = M\frac{di}{dt}

so we have

M = 31 \mu H

also we know that rate of change in current in solenoid is given as

\frac{di}{dt} = 2.5 A/s

so induced EMF of coil is given as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

now induced current in the coil will be given as

i = \frac{EMF}{R}

i = \frac{77.5 \times 10^{-6}}{9.9}

i = 7.83 \mu A

4 0
2 years ago
To eight significant figures, Avogadro's constant is 6.0221367×10^(23)mol−1. Which of the following choices demonstrates correct
Zepler [3.9K]

Answer with Explanation:

We are given Avogadro's constant =6.0221367\times 10^{23}mol^{-1}

There are eight significant figures.

We have to round off.

1.If we round off to four significant figures

The ten thousandth place of Avogadro's constant is less than five therefore, digits on left side of ten thousandth  place remains same and digits on right side of ten thousandth place and ten thousandth place  replace by zero.

 Then ,Avogadro's constant can be written as

6.022\times 10^{23}mol^{-1}

If we round off to 2 significant figures

Hundredth place of given number is less than 5 therefore, digits on left side of hundredth  place remains same and digits on right side of hundredth place and hundredth place replace by zero.

Then,Avogadro's constant can be written as

6.0\times 10^{23}mol^{-1}

If we round off six significant figures

6 is greater than 5 therefore, 1 will be added to 3 and digits on right side of 6 and 6 replace by zero and digits on left side of 6 remains same except 3.

Then, the Avogadro's constant can be written as

6.02214\times 10^{23}mol^{-1}

3 0
2 years ago
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