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dimulka [17.4K]
2 years ago
13

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ce it travels each succeeding second would be:_________
a) constant.
b) less and less each second.
c) greater than the second before
Physics
1 answer:
ioda2 years ago
5 0

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

h=ut+\frac{1}{2}gt^2

here u=0 and t=time taken

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

for t=1 s

h_1=\frac{1}{2}g

for t=2 s

h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g

distance traveled in 2 nd sec=2g-\frac{1}{2}g=\frac{3}{2}g

for t=3 s

h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g

distance traveled in 3 rd sec=\frac{9}{2}g-2g=\frac{5}{2}g

so we can see that distance traveled in each successive second is increasing

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Recall, frictional force = normal reaction x coefficient of friction

Given that coefficient of kinetic friction = 0.35,

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The force pulling the object upwards along the inclined plane is

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A physical change occurs when a material changes shape or size but the composition of the material does not change. True or Fals
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The current theory of the structure of the
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Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

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