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beks73 [17]
3 years ago
7

In a lever, the effort arm is two times as long as the load arm. The resultant force will be

Physics
2 answers:
jasenka [17]3 years ago
5 0
Answer is B. 

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

-
Charlie
ELEN [110]3 years ago
5 0
B.. twice the applied force! i just did the test
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Explanation:

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What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t
yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity =  ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

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time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R =  ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R =  (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

6 0
3 years ago
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr
xxMikexx [17]

Answer:

Radius, r=2.14\times 10^{-5}\ m

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, E=4.9\times 10^{-19}\ J

Kinetic energy is given by :

E=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E}{m}}

v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}            

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

qvB\ sin90=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }

r=2.14\times 10^{-5}\ m

So, the radius of the circular path is 2.14\times 10^{-5}\ m. Hence, this is the required solution.

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