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3 years ago

One Pro for supplements is the cost. True Or False

Physics

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

False

Explanation:

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A group of students performed a compression experiment where they placed weights on top of a cylinder of material and measured t

kolbaska11 [484]

The material that the cylinder is made from is Butyl Rubber.

<h3>What is Young's modulus?</h3>

Young's modulus, or the modulus of elasticity in tension or compression, is a mechanical property that measures the tensile or compressive strength of a solid material when a force is applied to it.

<h3>Area of the cylinder</h3>

A = πr²

$A = \pi \times (0.02)^2 = 0.00126 \ m^2$

<h3>Young's modulus of the cylinder</h3>

$E = \frac{stress}{strain} \\\\E = \frac{F/A}{e/l} \\\\E = \frac{Fl}{Ae} \\\\$

Where;

e is extension

When 5 kg mass is applied, the extension = 10 cm - 9.61 cm = 0.39 cm = 0.0039 m.

$E = \frac{(5\times 9.8) \times 0.1}{0.00126 \times 0.0039} \\\\E = 9.97 \times 10^5 \ N/m^2\\\\E = 0.000997 \times 10^9 \ N/m^2\\\\E = 0.000997 \ GPa\\\\E \approx 0.001 \ GPa$

When the mass is 50 kg,

extension = 10 cm - 7.73 cm = 2.27 cm = 0.0227 m

$E = \frac{(50\times 9.8) \times 0.1}{0.00126 \times 0.0227} \\\\E = 1.7 \times 10^6 \ N/m^2\\\\E = 0.0017 \times 10^9 \ N/m^2\\\\E = 0.0017 \ GPa\\\\E \approx 0.002 \ GPa$

The Young's modulus is between 0.001 GPa to 0.002 GPa

Thus, the material that the cylinder is made from is Butyl Rubber.

Learn more about Young's modulus here: brainly.com/question/6864866

3 0

2 years ago

Read 2 more answers

Why was the light peppered moth able to flourish prior to the Industrial Revolution?

Oksana_A [137]

Answer:

B. by blending in with the trees

5 0

2 years ago

Read 2 more answers

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

lesantik [10]

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$ .

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$ .

The length of each side of the triangle, $l = 2.9~cm$ .

Consider a equilateral triangle $\bigtriangleup ABC$ , as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$ .

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$

8 0

2 years ago

what is the wavelength of waves at the beach if they have a frequency of 0.3 Hz and they move at 7.1 m/s?

Reil [10]

Answer:

23.67 m

Explanation:

We are given;

Frequency; f = 0.3 Hz

Speed; v = 7.1 m/s

Now, formula to get the wavelength is from the wave equation which is;

v = fλ

Where λ is wavelength

Making λ the subject, we have;

λ = v/f

λ = 7.1/0.3

λ = 23.67 m

5 0

3 years ago

This grandfather clock, made entirely of Legos, has a 0.625 m long simple pendulum. What is the period of the pendulum? (include

PSYCHO15rus [73]

1. 1.59 s

The period of a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum and g the gravitational acceleration.

In this problem,

L = 0.625 m

g = 9.81 m/s^2

Substituting into the equation, we find

$T=2\pi \sqrt{\frac{(0.625 m)}{9.81 m/s^2}}=1.59 s$

2. 54,340 oscillations

The total number of seconds in a day is given by:

$t=24 h \cdot 60 min/h \cdot 60 s/min =86,400 s$

So in order to find the number of oscillations of the pendulum in one day, we just need to divide the total number of seconds per day by the period of one oscillation:

$N=\frac{t}{T}=\frac{86,400 s}{1.59 s}=54,340$

3. 0.842 m

We want to increase the period of the pendulum by 16%, so the new period must be

$T'=T+0.16T=1.16 T = 1.16 (1.59 s)=1.84 s$

Now we can re-arrange the equation for the period of the pendulum, using T=1.84 s, to find the new length of the pendulum that is required to produce this value of the period:

$L=g(\frac{T}{2\pi})^2=(9.81 m/s^2)(\frac{1.84 s}{2\pi})^2=0.842 m$

6 0

3 years ago