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Alenkasestr [34]
3 years ago
11

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.0 km/h and the co

efficient of static friction between tires and track is 0.320? The bicyclist, bicycle and tires are all to the right of the center of the track. Treat the bicyclist, bicycle and tires as one object.
Physics
1 answer:
bazaltina [42]3 years ago
5 0

Answer:

Explanation:

Let the radius of track required be r.

Centripetal force will be provided by frictional force which will be equal to

m v²/ r

Frictional force = mg x μ

So

m v² /r = mg μ

r = v² / μ g =

  v = 29 km /h = 8.05 m /s

r =( 8.05 x 8.05 ) /( .32 x 9.8 ) = 20.66 m

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A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit
Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

y = \frac{1}{2} at^2+v_0t+y_0

Where,

a = acceleration

t = time

v_0 = Initial velocity

y_0 = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

x = Displacement

t = time

With the data we have we can find the time as well

v = \frac{x}{t}

t = \frac{x}{v}

t = \frac{18.6}{34}

t = 0.547s

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

y = \frac{1}{2} at^2+v_0t+y_0

y = \frac{1}{2} gt^2+0+0

y = \frac{1}{2} gt^2

y = \frac{1}{2} 9.8*0.547^2

y = 1.46m

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0
3 years ago
At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta
Evgesh-ka [11]
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3 years ago
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
How many earth years does it take mars to orbit the sun
Colt1911 [192]

This is a question that would have literally have taken two seconds to look up on google but the answer is 1.88 years.

4 0
3 years ago
Use ohms law to determine the battery voltage you would need to send 2.5 A of current through a light bulb with 3.6 Ω of resista
yan [13]

Answer:

285 or higher voltage

Explanation:

3 0
3 years ago
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