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AleksAgata [21]
3 years ago
15

The Earth rotates on its axis every __________ and revolves around the Sun every __________.

Physics
2 answers:
Ray Of Light [21]3 years ago
9 0
The earth rotates on its axis every DAY and revolves around the Sun every YEAR.
kvasek [131]3 years ago
7 0
Earth rotates about this axis once each day (approximately 24 hours)
and revolves around the sun every year ..
Hope this helped ! :)
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A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra
givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a

divide both sides by 40

a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2

5 0
3 years ago
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

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3 years ago
Um aluno pretende construir um aquecedor usando um enrolamento de fio. Experimenta e verifica que não
Vlad1618 [11]

Answer:

dddjebdhdjwopdjhgvvdr

6 0
3 years ago
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE
luda_lava [24]
Australia separated from other continents and species there evolved independently
8 0
3 years ago
The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 d
melisa1 [442]

This question is incomplete

Complete Question

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 degrees Fahrenheit. Classify the objects according to whether their molecules will speed up or slow down after being left in the kitchen for a period of time.

Drag the objects to the correct category.

Object Temperature (Fahrenheit)

ice cubes 26°F

glass of tea 60°F

cooked piece of meat 160°F

butter 55°F

pot of water 75°F

bowl of soup 140°F

Answer:

Molecules that would speed up

ice cubes 26°F

glass of tea 60°F

butter 55°F

Molecules that would slow down

cooked piece of meat 160°F

pot of water 75°F

bowl of soup 140°F

Explanation:

Molecules that are found or contained in a substances have the tendency to react in such as what that they are are sped up or slowed down. This is due to their exposure to various kinds of changes in temperature, which could be a hot temperature , cold temperature e.t.c

When a substance is exposed to a hot temperature, the molecules of that substance speed up depending on how hot the temperature is while when a substance is cooled down or exposed to cold temperature, the molecules tend to slow down.

In the question above, the temperature of the kitchen is 72 degrees Fahrenheit.

Molecules that would speed up after been left in the Kitchen for a while:

a) Ice cubes 26°F: This is because 72°F is a warm took temperature, so the ice cubes would melt causing the molecules to speed up.

b)Glass of tea 60°F: This glass of tea is at a cool temperature of 60°F, when it is kept in the Kitchen which has a temperature of 72°F for while, the temperature of the glass of tea would increase due to an increase in its rate of reaction causing the molecules of the glass of water to speed up.

c) Butter 55°F: This is because 72°F is a warm temperature, so the butter would melt, increasing its rate of reaction and causing the molecules to speed up

Molecules that would slow down

a) Cooked piece of meat 160°F : 160°F is a very hot temperature and when it is left in a the Kitchen(72°F) , the cooked meat would begin to cool down and drop in temperature from 160°F, causing the molecules to slow down.

b) Pot of water 75°F: The temperature of the kitchen and the temperature of the pot of water is the same, hence the molecules of the water would slow down.

c) Bowl of soup 140°F : This bowl of soup is very hot and when it is left in the Kitchen(72°F) , the bowl of soup would begin to cool down and drop in temperature from 140°F, slowing down the rate of reaction and causing the molecules to slow down.

3 0
3 years ago
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