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3 years ago

9

Concept Simulation 5.2 reviews the concepts that are involved in this problem. A car is safely negotiating an unbanked circular

turn at a speed of 16 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

1 answer:

bonufazy [111]3 years ago

5 0

Answer:

$v_W=9.23m/s$

Explanation:

The force of friction change is the pavement is dry or wet so to determine the force of friction:

$F=m*a$

$F_k=m*a_c$

$F_k=u_K*N$

$N=m*g$

$F_k=u_K*m*g=m*a_c$

$u_K*g=a_c$

$a_c=\frac{V^2}{R}$

Dry pavement

$u_{KD}*g=\frac{v_D^2}{R}$

Wet pavement

$u_{KW}*g=\frac{v_W^2}{R}$

$u_{KW}=\frac{1}{3}*u_{KD}$

Solve and reduce the factor so:

$\frac{v_W^2}{v_D^2}=\frac{\frac{1}{3}*u_{KD}}{u_{KD}}$

$v_W^2=v_D^2*\frac{1}{3}$

$v_W=v_D*\frac{1}{\sqrt{3}}=16m/s*\frac{1}{\sqrt{3}}$

$v_W=9.23m/s$

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the work required for the loading of second dart is 64 times greater as work required for loading the first dart.

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k = spring constant of the spring loaded toy dart gun

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E₁ = Work required to load the first dart

E₂ = Work required to load the second dart

Work required to load the first dart is given as

E₁ = (0.5) k x₁² = (0.5) k d²

Work required to load the second dart is given as

E₂ = (0.5) k x₂² = (0.5) k (8d)² = (64) (0.5) k d²

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3 years ago

A rocket carrying a satellite is accelerating straight up from the earth’s surface. At 1.15 s after liftoff, the rocket clears t

yarga [219]

Answer:

197.263157895 m/s

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2 years ago

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

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Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as: $\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\$

In respect to the magnetic field; $\overrightarrow{B}=(0.027 i - 0.15 j) [T]$

The magnetic force can be written as;

$\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]$

Bear in mind $q =-1.6021x10^{-19} [C]$

thus,

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.

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3 years ago

A paper airplane is thrown westward at a rate of 6 m/s. The wind is blowing at 8 m/s towards the north. Which of the following d

Andrew [12]

since airplane is thrown towards west with speed 6 m/s

while air is blowing with speed 8 m/s towards north

so here the net speed of air plane will be the resultant of airplane speed and wind speed always

SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s

so correct answer must be 1st option

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2 years ago