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Alenkasestr [34]

2 years ago

10

A bicyclist of mass 85 kg rides in a circle at a speed of 4.5 m/s. If the radius of the circle is 8 m, what is the centripetal f

orce on the bicyclists

2 answers:

klemol [59]2 years ago

8 0

Answer:

The centripetal force on the bicyclists is 215.2 N.

Explanation:

Given that,

Mass of bicyclist = 85 kg

Speed = 4.5 m/s

Radius = 8 m

We need to calculate the centripetal force

Using formula of centripetal force

$F_{ac}=\dfrac{mv^2}{r}$

Where, m = mass

v = velocity

r = radius

Put the value into the formula

$F_{ac}=\dfrac{85\times4.5^2}{8}$

$F_{ac}=215.2\ N$

Hence, The centripetal force on the bicyclists is 215.2 N.

lord [1]2 years ago

5 0

F=m*(v^2/r)
F=85*(4.5^2/8)
F=215.2N

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Given

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t = (Xf-Xi) / ( Va)

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Answer:

a)

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b)

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Explanation:

a)

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$m_{2}$ = mass of second body = ?

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$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

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using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

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lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

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ΔU = 3.2× $10^{-6}$ J

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$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2× $10^{3}$ . 2× $10^{-9}$ . 5× $10^{-1}$ )

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

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