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3 years ago

What are not examples of velocity

Physics

2 answers:

Nookie1986 [14]3 years ago

8 0

Answer:

Explanation:

Acceleration roughly means “change in velocity.” If this change in velocity is consistently in one direction, the body will reach a point at which its velocity is zero.

Another consideration is when you drop a basketball onto the pavement. The ball falls, hits the ground, and comes back up. Because its direction of motion changed, you know that its velocity MUST have been 0 m/s at some point during the bounce.

This doesn't mean there was no acceleration at this point.

maw [93]3 years ago

4 0

How long it takes to get somewhere

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Let f(x) =, then what is (0)?

Ad libitum [116K]

F(x) = f(0) just solve for x

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4 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

A motor raises the car up the tower from point C to point A with a constant velocity. State Newton’s First Law of Motion and exp

GarryVolchara [31]

newton 1st law: w no external force, a body will stay at rest or in constant uniform motion.

a motor raises the car up the tower. cuz it moves w/ constant vel, there is no external force: the motor force is the same n in opposity direction as the gravity force on car.

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3 years ago

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A yoyo is hanging motionless from a string. Identify and describe the forces exerted on the string.

sashaice [31]

Potential and kinetic energy along with an external force

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4 years ago

You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from

maxonik [38]

Answer:

A. There is a localization of positive charge near the door handle.

Explanation:

When on a cold morning a person wearing cotton/ polyester cloth walking on the carpet moves toward his car then due to friction between the feet and the carpet there are transfer of electrons from the carpet to our feet, and since our body is a good conductor of electricity the charges spread throughout on the surface of or body.

When the person brings his hands close to the neutral conducting door of the car it gets induced with equal intensity of opposite charge to our hands thus having a concentration of positive charges near to the hand on the car's door is developed as a result of polarization within the conductor.

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3 years ago