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3 years ago

6

Two identical cars, one on the moon and one on earth, are rounding banked curves at the same speed with the same radius and the

same angle. the acceleration due to gravity on the moon is 1/6 that of earth. how do the centripetal accelerations of each car compare?
a.the centripetal acceleration of the car on earth is less than that on the moon.

b.the centripetal acceleration of the car on earth is greater than that on the moon.

c.the centripetal accelerations are the same for both cars.

d.this cannot be determined without knowing the radius and the

1 answer:

Allisa [31]3 years ago

7 0

D because you are not given enough info

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Which statement describes a characteristic of an experimental design that

Anarel [89]

Answer:

c

Explanation:

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3 years ago

Jupiter’s Great Red Spot rotates completely every six days. If the spot is circular (not quite true, but a reasonable approximat

artcher [175]

Answer:

$v = 567.2 km/h$

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

$distance = 2\pi r$

$distance = \pi D$

now we will have the time to complete the rotation given as

$t = 6 days$

$t = 6 (24 h) = 144 h$

now the speed is given by

$speed = \frac{distance}{time}$

$speed = \frac{\pi D}{t}$

$speed = \frac{\pi(26000 km)}{144}$

$v = 567.2 km/h$

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2 years ago

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

a)

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

e)

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

7 0

3 years ago

In an eslastic ,the momentum is_______and the mechanical energy is

DIA [1.3K]

Answer:

In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.

Explanation:

There are two types of collisions:

- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.

- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)

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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.

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3 years ago