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dangina [55]
2 years ago
5

Assume this piston has a maximum pressure rating of 2.5 pounds-force per square inch [psi]. The force applied to the cylindrical

piston is 100 newtons [N]. What is the diameter of the piston, in units of inches [in]
Physics
1 answer:
Leni [432]2 years ago
7 0

Explanation:

Always write what you know first about the problem

The surface of the piston that has the force applied is a circle

The area of a circle is pi x r^2

Pressure is Force / Area

Write this out

P = F / (pi x r^2)

solve for r

sqrt( F / (P x Pi)) = r

But dont forget that d = 2 x r

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron,u_2=0

Radius,r_1=0

r_2=2.3 cm=\frac{2.3}{100}=0.023 m

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron,m=9.1\times 10^{-31} kg

Charge on an electron,q=-1.6\times 10^{-19} C

Velocity,v=\frac{Bqr}{m}

Using the formula

Speed of electron,v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0

Speed of second electron,v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}

v_2=1.5\times 10^8 m/s

Kinetic energy of incident electron=\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2

Kinetic energy of incident electron=0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J

Kinetic energy of incident electron=\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV

1KeV=1000eV

3 0
3 years ago
An electron in an atom has an uncertainty of 0.2 nm. If it is doubled to 0.4 nm by what factor does the uncertainty in momentum
fenix001 [56]

Answer:

The uncertainty in momentum changes by a factor of 1/2.

Explanation:

By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π(2Δx) = (h/2πΔx)/2 = Δp/2.

So, the uncertainty in momentum changes by a factor of 1/2.

4 0
3 years ago
Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th
kkurt [141]

<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

8 0
3 years ago
A student performs an exothermic reaction in a beaker and measures the temperature. If the thermometer initially reads 35 degree
Neporo4naja [7]

B 25 degrees Celsius

3 0
2 years ago
Read 2 more answers
Calculate net force <br> 532N 215N
lorasvet [3.4K]

Answer:

FN is the forces acting on a body. When the body is at rest, the net force formula is given by, FNet = Fa + Fg.

Im in 7th and thats all I know so I hope it's enough

8 0
3 years ago
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