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elixir [45]
2 years ago
10

A hammer of mass M is moving at speed v0 when it strikes a nail of negligible mass that is stuck in a wooden block. The hammer i

s observed to drive the nail a distance L deeper into the block.
(a) Find the work W done on the hammer by the nail. Express your answer as a function of F and L, where F is the force the nail exerts on the hammer. Don't forget to consider the sign of your answer.
(b) Find the change in kinetic energy of the hammer. Express your answer in terms of M and v0.
(c) Find the magnitude F of the force that the wooden block exerts on the nail, assuming that this force is independent of the depth of penetration of the nail into the wood.
(d) Now evaluate the magnitude of the holding force of the wooden block on the nail by assuming that the force necessary to pull the nail out is the same as that needed to drive it in, which we just derived.
Assume a relatively heavy M = 0.5 kg hammer (about 18 ounces), moving with speed v0 = 10 m/s. (If such a hammer were swung this hard upward and released, it would rise 5 m). Take the penetration depth L to be 2 cm, which is appropriate for one hit on a relatively heavy construction nail. Express your answer to the nearest pound. (Note: 1 lb = 4:45 N.)
Physics
1 answer:
OleMash [197]2 years ago
8 0

Answer:

i think it would be D

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A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106
defon

Answer:

4.7\cdot 10^{-14}N

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

F=qvB

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

q=1.6\cdot 10^{-19} C

v=6.5\cdot 10^6 m/s

B=4.5\cdot 10^{-2} T

So the force experienced by the electrons is

F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N

3 0
2 years ago
Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Plank’s constant (6.63E-34
Nostrana [21]
To help you I need to assume a wavelength and then calculate the momentum.

The momentum equation for photons is:

p = h / λ , this is the division of Plank's constant by the wavelength.

Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:

p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s

p = 1.01067 * 10^ - 27 kg*m/s which  must be rounded to three significant figures.

With that, p = 1.01 * 10 ^ -27 kg*m/s

The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s

So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.
7 0
2 years ago
Read 2 more answers
The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.
Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

W=0.24\times 1\times 10^3=240\ W/m^2

Energy required by the United States is 1\times 10^{19}\ J/yr (assumed)

Power

P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W

Area

A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

1\ m^2=\frac{1}{1000\times 1000}\ km^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2

Converting to mi²

1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0
2 years ago
A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha
Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

PQ = \sqrt{4^{2}+3^{2}}=5 m

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

3 0
2 years ago
Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0
2 years ago
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