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liberstina [14]

2 years ago

8

A motorcycle has a constant speed of 29.6 m/s as it passes over the top of a hill whose radius of curvature is 183 m. The mass o

f the motorcycle and driver is 329 kg. Find the magnitude of (a) the centripetal force and (b) the normal force that acts on the cycle.

1 answer:

coldgirl [10]2 years ago

8 0

Answer:

a) centripetal force = m v^2 / R = 329 * (29.6 m/s)^2 / 183

F = 1580 Newtons

b) normal force = gravitational force - F = 329 * 9.8 - 1580 = 1640 N

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katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart $m_1=6 kg$

mass of second cart $m_2=3 kg$

velocity of first cart $v_1=3 m/s$

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$m_1v_1+m_2v_2=(m_1+m_2)v$

$6\times 3+3\times 0=(9)\cdot v$

$v=2 m/s$

Initial kinetic Energy $K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

$K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0$

$K.E._1=27 J$

Final Kinetic Energy

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J$

Ratio of initial Kinetic Energy to the Final Kinetic Energy

$=\frac{27}{18}=1.5$

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4 years ago

Which is not part of nuclear chemistry? A. Strong nuclear force B. Double-replacement reactions C. Radioactivity D. Nuclear deca

ziro4ka [17]

The correct choice is

B. Double-replacement reactions

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3 years ago

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brilliants [131]

sound travels through air at around 340 - 350 m/s.

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This would be the same time-lag even if it were not a major-league game,
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3 years ago

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

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