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Alika [10]
2 years ago
14

A proud new Jaguar owner drives her car at a speed of 35 m/s into a corner. The coefficients of friction between the road and th

e tires are 0.70 (static) and 0.40 (kinetic). What is the minimum radius of curvature for the corner in order for the car not to skid
Physics
1 answer:
trapecia [35]2 years ago
7 0

Answer:

178.6 m

Explanation:

Since the car moves in a circular path, it experiences a centripetal force, F = mv²/r where m = mass of car, v = speed of car = 35 m/s and r = radius of curvature of path.

Now, for the car not to skid, this centripetal force must be equal to the frictional force, F' acting in the opposite direction.

So, F' = μN where μ = coefficient of static friction(since the car does not move in this direction) and N = normal force = mg where m = mass of car and g = acceleration due to gravity = 9.8m/s²

F' = μmg

Since F = F'

mv²/r = μmg

dividing both sides by m, we have

v²/r = μg

multiplying both sides by r, we have

v² = μgr

dividing both sides by μg, we have

r = v²/μg

Here we use μ = coefficient of static friction(since the car does not move in this direction) = 0.70. Substituting the other variables into the equation, we have

r = v²/μg

r = (35 m/s)²/(0.70 × 9.8m/s²)  

r = 1225 m²/s²/6.86m/s²)  

r = 178.6 m

So, the minimum radius of curvature of the corner is 178.6 m

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Can someone please help a struggling physics student?
zhenek [66]
<h3><u>Part A:</u></h3>

<u><em>What is the maximum height the ball will reach in the air?</em></u>

Kinematics equation used:

  • v_f^2=v_i^2+2ad, where v_f is final velocity, v_i is initial velocity, a is acceleration, and d is distance travelled. From SI units, velocity should be in m/s, acceleration should be in m/s^2, and distance should be in m

We're given that the initial velocity is 12.0 m/s in the y-direction. At the maximum height, the vertical velocity of the ball will be 0 m/s, otherwise it would not be at maximum height. This is our final velocity.

The only acceleration in the system is acceleration due to gravity, which is approximately 9.8\:\mathrm{ m/s^2}. However, the acceleration is acting down, whereas the ball is moving up. To express its direction, acceleration should be plugged in as -9.8\:\mathrm{m/s^2}. We have three variables, and we are solving for the fourth, which is distance travelled. This will be the maximum height of the ball.

Substitute v_i=12, v_f=0, a=-9.8 to solve for d:

0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}

<u><em>What is the velocity of the ball when it hits the ground?</em></u>

This question tests a physics concept rather than a physics formula. The vertical velocity of the ball when it hits the ground is equal in magnitude but opposite in direction to the ball's initial vertical velocity. This is because the ball spends equal time travelling to its max height as it does travelling from max height to the ground (ball is accelerating from initial velocity to 0 and then from 0 to some velocity over the same distance and time). Since the ball has an initial vertical velocity of +12.0 m/s, its velocity when it hits the ground will be \boxed{-12.0\text{ m/s}}. (The negative sign represents the direction. Because velocity is a vector, it is required.)

<h3><u>Part B:</u></h3>

<u>**Since my initial answer exceeds the character limit, I've attached the first question to Part B as an image. Please refer to the attached image for the answer and explanation to the first question of Part B. Apologies for the inconvenience.**</u>

<u><em>What is the direction of the velocity of the ball when it hits the ground? Express your answer in terms of the angle (in degrees ) of the ball's velocity with respect to the horizontal direction (see figure).</em></u>

This question uses a similar concept as the second question of Part A. The vertical velocity of the ball at launch is equal in magnitude but opposite in direction to the ball's final velocity. The horizontal component is equal in both magnitude and direction throughout the entire launch, since there are no horizontal forces acting on the system. Therefore, the angle below the horizontal of the ball's velocity when it hits the ground is equal to the angle of the ball to the horizontal at launch.

To find this, we need to use basic trigonometry for a right triangle. In any right triangle, the tangent/tan of an angle is equal to its opposite side divided by its adjacent side.

Let the angle to the horizontal at launch be \theta. The angle's opposite side is represented by the vertical velocity at launch (12.0 m/s) and the angle's adjacent side is represented by the horizontal velocity at launch (2.3 m/s). Therefore, we have the following equation:

\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}

Take the inverse tangent of both sides:

\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})

Simplify using \arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ}):

\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}

We can express our answer by saying that the direction of the velocity of the ball when it hits the ground is \boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}} or \boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}.

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3 years ago
When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms w
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is n =7

Explanation:

    From the question we are told that

          The value of m = 2

            For every value of m, n = m+ 1, m+2,m+3,....

           The modified version of  Balmer's formula is \frac{1}{\lambda}  = R [\frac{1}{m^2} - \frac{1}{n^2}  ]

             The Rydberg constant has a value of R = 1.097 *10^{7} m^{-1}

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

   

For m = 2 and n =3

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2}  ]

                          \lambda = \frac{1}{1523611.1112}

                             \lambda = 656nm

For m = 2 and n = 4

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2}  ]

                          \lambda = \frac{1}{2056875}

                             \lambda = 486nm

For m = 2 and n = 5

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2}  ]

                          \lambda = \frac{1}{2303700}

                             \lambda = 434nm

For m = 2 and n = 6

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2}  ]

                          \lambda = \frac{1}{2422222}

                             \lambda = 410nm

For m = 2 and n = 7

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2}  ]

                          \lambda = \frac{1}{2518622}

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So the value of n is  7

7 0
3 years ago
If a typical smelt weighs 225 g, what is the total mass of pcbs in a smelt in the great lakes? If a typical smelt weighs 225 g,
photoshop1234 [79]

Answer;

= 2.34 x 10^-4 g pcbs

Explanation;

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Therefore;

The amount of PCB in a smelt in great leak can be calculated by multiplying the mass of smelt by the pcbs present in a smelt

= 225 g * ( 1.04/100000)

= 2.34 x10^-4 g pcbs

7 0
3 years ago
In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of m
lara31 [8.8K]

Answer:

F=3.61\times 10^{-47}\ N

Explanation:

Mass of a proton, m_p=1.67\times 10^{-27}\ kg

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The distance between the electron and the proton is, r=5.3\times 10^{-11}\ m

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

F=G\dfrac{m_em_p}{r^2}

Where G is the universal Gravitational constant

F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N

So, the force between the electron and proton is 3.61\times 10^{-47}\ N.

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2 years ago
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Answer:

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q = 1.741 x 10 ⁻²¹ C

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Ef = 101650 V/m

So, we can write  

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q = m*g / E f

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m = ρ * v

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v = 2.298 x 10 ⁻ ¹⁶ m³

q =  ρ * v * g / Ef

q =  81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

The charge on the drop  is

q = 1.741 x 10 ⁻²¹ C

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