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Morgarella [4.7K]
3 years ago
14

2) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1.

Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0
<h3>pressure = force / area</h3>

<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>

<h3>pressure = 84 / 55</h3>

<h3>pressure = 1.53 pascals</h3>

hope that helps and please tell me if i am wrong :)

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What are the different isotopes of carbon and which isotopes are relevant for carbon dating?
GuDViN [60]
Ans: Radiocarbon dating uses carbon isotopes.

Radiocarbon dating relies on the carbon isotopes carbon-14 and carbon-12. Scientists are looking for the ratio of those two isotopes in a sample. Most carbon on Earth exists as the very stable isotope carbon-12, with a very small amount as carbon-13.
3 0
3 years ago
The pressure of a gas will increase as the volume of the container decreases, provided temperature does not change. This stateme
tigry1 [53]
The right answer for the question that is being asked and shown above is that: "Archimedes' principle." The pressure of a gas will increase as the volume of the container decreases, provided temperature does not change. This statement is called <span>Archimedes' principle</span>
8 0
3 years ago
I hand you two spheres. They have the same mass, the same radius, and the same exterior surface. I claim one is a solid sphere a
castortr0y [4]

Answer:

Explanation:

Between two spheres of equal masses having equal radius , the one which is hollow will have less moment of inertia and therefore less  radius of gyration ( k )

When they are allowed to roll over an inclined plane , they will have acceleration as follows

a = gsinθ / ( 1 + k²/r²)

For hollow sphere , k is less . therefore denominator is less and acceleration is more.

in other words , hollow sphere will have greater acceleration . So it will reach the bottom of an inclined surface in lesser time if allowed to roll over it .  

This is how we can identify a hollow sphere without breaking it open .

5 0
3 years ago
An object on a number line moved from x = 15 cm to x = 165 cm and then
olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

v_{avg}=\frac{x_{all}}{t}

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0
3 years ago
Read 2 more answers
A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not cons
igor_vitrenko [27]

Answer: (a) α = \frac{3I}{2.\pi.R^{3}}

(b) For r≤R: B(r) = μ_0.(\frac{I.r^{2}}{2.\pi.R^{3}})

For r≥R: B(r) = μ_0.(\frac{I}{2.\pi.r})

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

I_{c} = \int {J} \, dA

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr

I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr

I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}

\alpha = \frac{3I}{2.\pi.R^{3}}

For these circunstances, α = \frac{3I}{2.\pi.R^{3}}

(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:

\int\ {B} \, dl = μ_0.I_{c}

(i) First, first find I_{c} for r ≤ R:

I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr

I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr

I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}

I_{c} = \frac{I.r^{3}}{R^{3}}

Calculating B(r), using Ampere's Law:

\int\ {B} \, dl = μ_0.I_{c}

B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} ).μ_0

B(r) = (\frac{Ir^{3}}{R^{3}2.\pi.r}).μ_0

B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

For r ≤ R, magnetic field is B(r) = (\frac{Ir^{2}}{2.\pi.R^{3}} ).μ_0

(ii) For r ≥ R:

I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr

So, as calculated before:

I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}

I_{c} = I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = (\frac{I}{2.\pi.r} ).μ_0

For r ≥ R, magnetic field is; B(r) = (\frac{I}{2.\pi.r} ).μ_0.

3 0
3 years ago
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