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AleksandrR [38]
3 years ago
10

A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door

. Where is the pivot point and what is the maximum torque?

Physics
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

Explanation:

Given

Force applied F=80\ N

Door is d=3\ m wide

for gate to revolve Properly Pivot Point must be at center i.e. 1.5 from either end

Torque applied is T=force\times distance

Maximum torque

T_{max}=F\times \frac{d}{2}

T_{max}=80\times \frac{3}{2}

T_{max}=120\ N-m

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An object starts from rest and uniformly accelerates at a rate of 1.25 m/s2 for 7.0 seconds.
stich3 [128]

Explanation:

Since its accelerating, the velocity vs time graph is linear

For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

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1.25=vf / 7

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X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

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2 years ago
In an insulated vessel, a quantity of hot water at temperature T1 is mixed with a different quantity of cold water at temperatur
erastovalidia [21]

Answer:Water Only

Explanation:

Given

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4 0
3 years ago
In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x
Andrei [34K]

Answer:

(a) F_g=1.62*10^{-48}N

(b) F_e=3.68*10^{-9}N

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

F_g=-G\frac{m_1m_2}{r^2}

Where G is the Cavendish gravitational constant, m_1 and m_2 are the masses of the electron and the proton respectively and r is the distance between them:

F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

F_g=1.62*10^{-48}N

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N

Its magnitude is:

F_e=3.68*10^{-9}N

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3 years ago
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