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juin [17]
3 years ago
14

A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 110 N force applied to the

cable causes the block to reach a speed of 4.2 m/s in a distance of 2.0 m
What is the mass of the cable?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
sammy [17]3 years ago
5 0

Answer:

The mass of the cable is 4.94 kg                                                                                                                                                                            

Explanation:

It is given that,

Mass of the block, m = 20 kg

Force applied to the cable, F = 110 N

Speed of the block, v = 4.2 m/s

Distance, d = 2 m

Let a is the acceleration of the block. It can be calculated using the third equation of motion as :

v^2-u^2=2ad

(4.2)^2-(0)^2=2a\times 2

a=4.41\ m/s^2

Let m' is the mass of the cable. It can be calculated using the second law of motion as :

m'=\dfrac{F}{a}-m

m=\dfrac{110\ N}{4.41\ m/s^2}-20

m = 4.94 kg

So, the mass of the cable is 4.94 kg. Hence, this is the required solution.

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Two conducting wires of the same material are to have the same resistance. One wire is 17.0 m long and 0.44 mm in diameter. If t
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Answer:

L = 7.90  m

Explanation:

  • The general expression for the resistance of a resistor is as follows:

        R = \frac{\rho*L}{A}

  • When we are talking about a wire, of a given diameter, the above formula becomes as follows:

       R = \frac{\rho*L}{\frac{\pi*d^{2}}{4}}

  • where ρ = resistivity of the material, L= length of the wire and d= diameter of the wire.
  • If both wires are from the same material, and have the same resistance, we can say the following, simplifying common terms:

       R_{1} = R_{2}  = \frac{L_{1} }{d_{1}^{2} } =\frac{L_{2} }{d_{2}^{2} }

  • where L₁ = 17.0 m, d₁= 0.44 mm = 4.4*10⁻⁴ m, d₂= 3*10⁻⁴ m.
  • Replacing these values, we can solve for L₂, as follows:

       L_{2} = \frac{L_{1} *d_{2} ^{2}}{d_{1}^{2}}  = \frac{17m*(3e-4m)^{2} }{(4.4e-4m)^{2}} \ = 7.90 m

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3 years ago
an underground cannon launches a cannonball from ground level at a 35-degree angle. the cannonball is shot with an initial veloc
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Answer:

\displaystyle x_{max}=21.57\ m

\displaystyle y_{max}=3.78\ m

Explanation:

<u>Motion in Two-Dimensions </u>

When an object is launched with a certain angle \theta above ground level with an initial velocity \vec v_o, it describes a curve called a parabola, defined by the force of gravity which will eventually make the object return to the ground. There are two overlapping motions, the horizontal motion, at a constant speed, and the vertical motion, at changing speed because the acceleration of gravity modifies it. The velocity \vec v_o is split into two components x,y

\displaystyle v_{ox}=|vo|\ Cos\theta

\displaystyle v_{oy}=|vo|\ Sin\theta

The position of the object is also split into its components, assuming the object was launched from ground level

\displaystyle x=v_{ox}.t

\displaystyle y=v_{oy}.t-\frac{g.t^2}{2}

The maximum horizontal distance the object reaches (called range) is

\displaystyle x_{max}=\frac{2v_{ox}\ v_{oy}}{g}

The maximum height is given by

\displaystyle y_{max}=\frac{v_{oy}^2}{2g}

The question doesn't ask for anything in particular, but to guide you in the solution of your own problem, we'll compute X_{max} and Y_{max} for you. The data is

\displaystyle |vo|=15\ m/s\ ,\ \theta =35^o

Let's compute the range

\displaystyle x_{max}=\frac{(2)(15)\ cos35^o\ (15)\ sen35^o}{9.8}

\displaystyle x_{max}=\frac{211.43}{9.8}=21.57\ m

Now for the maximum height

\displaystyle y_{max}=\frac{(15\ . \ Sin35^o)^2}{2(9.8)}

\displaystyle y_{max}=\frac{74.0227}{19.6}

\displaystyle y_{max}=3.78\ m

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