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3 years ago

The 10-kg uniform rod is pinned at end

1 answer:

3 0

Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º

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A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the

Alisiya [41]

Answer:

90J

Explanation:

The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.

5 0

2 years ago

Gas A has molecules with small mass. Gas B has molecules with larger mass. They are at the same temperature.

Julli [10]

Answer:

c)The gases have the same average kinetic energy.

Explanation:

As we know that the kinetic energy of gas is given as

$K = \frac{1}{2}mv^2$

here we know that

$v = \sqrt{\frac{3RT}{M}}$

so we have

$K = \frac{1}{2}m (\frac{3RT}{M})$

now we have

$K = \frac{3}{2}n RT$

now mean kinetic energy per molecule is given as

$K_{avg} = \frac{3}{2}KT$

so this is independent of the mass of the gas

so average kinetic energy will remain same for both the gas molecules

6 0

3 years ago

For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical

nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

7 0

3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

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

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago

M = 30.3kg<br>M = 40.17kg 9<br>R = 0.5m<br>G = 6. 67x10^11<br>F ?

Lena [83]

Answer:

m¹=30.3kg

m²=40.17kg

R=0.5m

G=6.67*10¹¹

F=Gm¹m²/R²

=160.68

4 0

3 years ago