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Morgarella [4.7K]
2 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]2 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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Two billiard balls of equal mass are traveling straight toward each other with the same speed. They meet head-on in an elastic c
Rus_ich [418]

Answer:

0 kg m/s before and after collision

Explanation:

Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:

P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

5 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

8 0
1 year ago
What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u
Aleksandr-060686 [28]

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 ×  10^{14} Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

E=h\nu_{o}    ..................1

here E is the energy of electron per atom E=\frac{x}{N}  and h is plank constant i.e. 6.626\times10^{-34} Js  and x is  binding energy

and here N is the Avogadro constant = 6.023\times10^{23}

so E will \frac{x}{6.023\times10^{23}}  

E = 3.19\times10^{-19}  J  

so put value in equation 1 we get

\frac{x}{6.023\times10^{23}} = 2.50 ×  10^{14} × 6.626\times10^{-34} Js  

solve it we get

x = 99770.99

so  binding energy is 99771 J/mol

4 0
2 years ago
Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto
mars1129 [50]

I’ve answered this before so I know the question is missing an important given and that given is: <span>1 has an empty trailer and the other has a fully loaded one. 

So, it would be the fully loaded trailer that would take a longer distance to stop because a lot of weight is being pulled, and when the brakes are started, the fully loaded trailer is more like pushing against the truck.</span>

6 0
1 year ago
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
mote1985 [20]

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

6 0
1 year ago
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