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Jobisdone [24]
2 years ago
8

Heidi (39 kg) is walking her tiny chihuahua, Chaxi (5.60 kg), on the sidewalk. To encourage Chaxi along, Heidi pulls forward wit

h a force of 9.55 N. Identify the correct reaction force in response to Heidi’s action force.
a. The friction is less than 660 N since the beam can be moved at a constant velocity.
b. There is no friction acting on the beam since it is accelerating.
c. The friction is equal to 660 N since the beam is not accelerating.
d. The friction is greater than 660 N since the beam is not in equilibrium.
Physics
1 answer:
Ivan2 years ago
4 0

Answer:

The correct reaction force in response to Heidi's action force is:

c. The friction is equal to 660 N since the beam is not accelerating.

Explanation:

Heidi's action force does not affect the beam.  Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect.  The reaction force is what makes the dog to move because it acts on it.  According to Newton's Third Law of Motion, forces always come in action-reaction pairs.  This Third Law states that for every action force, there is an equal and opposite reaction force.  This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."

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The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
2 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
A golfer takes two putts to get his ball into the hole he is on green. The first putt displaces the ball 4.5m east, and the seco
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I think its the one going it at 4.5m

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According to research, college graduates are more likely to engage in all of the following EXCEPT:
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Strange question. I’d say B)
7 0
3 years ago
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alexgriva [62]

Answer: A, B, and D,

C is incorrect because continental drift takes several centuries to move the tectonic plates witch is not an immediate impact leaving you with A,B, and D.

6 0
2 years ago
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