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3 years ago

Gina estimates that her moms truck weighs about 6,000 pounds. How can you write 6,000 using exponents?

Physics

2 answers:

Georgia [21]3 years ago

5 0

6x10^3 six times ten to the third power :)

Nookie1986 [14]3 years ago

5 0

Answer:

$6\times 10^{3} Pounds$

Explanation:

Given that the Gina moms truck weight is 6000 pounds.

And we also know that the 6000 pounds can be written as in the form.

$6000=6\times 10\times 10\times 10$

Further more it can be written as,

$6000Pounds=6\times 10^{3} Pounds$

Therefore the weight of the truck which is 6000 Pounds can be written in exponential form as $6\times 10^{3} Pounds$ .

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A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s

djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, $u=32$ m/s

Acceleration of the driver is, $a=-1.5$ m/s²

Time taken to reach final velocity is, $t=10.8$ s.

The final velocity is given using the Newton's equations of motion as:

$v=u+at$ , where, $v$ is the final velocity.

Now, plug in the given values and solve for $v$ .

$v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}$

Therefore, his final velocity is 15.8 m/s.

5 0

3 years ago

If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?

Otrada [13]

Answer: 0.2 hour

Explanation:

Speed = Distance / Time

Time = Distance / Speed

= 12/60

= 0.2 hour

8 0

3 years ago

PLEASE HELP : What happens in obese mice? (Physiology)

irina1246 [14]

Answer and

Explanation:

The gut microbiota has recently emerged as an important, and previously unappreciated, player in host physiology (1). In particular, the gut microbiota contributes to a variety of physiological and pathophysiological processes in the host including immune disorders (2–4), atherosclerosis (5), irritable bowel syndrome (6, 7), blood pressure regulation (8), and chronic kidney disease (9, 10). Bacteria residing in the human gut are an important component of human physiology: the total wet weight of gut microbes in the human has been estimated to be 175 g–1.5 kg (11, 12), and the cells of the microbiota outnumber human cells by 10:1 (1). These bacteria interact with the immune system of the host (13), and secrete a variety of metabolites, which enter host circulation and can affect a variety of physiological parameters (8, 14), reviewed in Ref. (15). In fact, metabolites produced by the gut microbiota have been found to play key roles in renal disease (16), blood pressure regulation (8), and immune disorders (2–4). Therefore, just as we consider the genetic background of an animal or an individual to be an important contributing factor to their physiology, so too must we consider the genetic background of the microbiota associated with that animal.

Gut microbiota vary greatly amongst laboratory animals, and these differences result in notable differences in experimental results. Mice of the same strain from different vendors have different microbiota profiles (17), and similarly, the same mice housed at different institutions have different microbiota profiles (18, 19). Conversely, inoculating two different inbred mouse strains with the same gut bacteria leads to differences in host gene expression between the two mouse strains (20). Clearly, there is a complex interplay between the genetics of the microbiota and that of the host organism, which has only recently begun to be appreciated.

Go to:

Gut Microbiota as an Experimental Parameter

Examples in the literature have highlighted the important and unexpected ways in which gut microbiota can affect a variety of experimental parameters. In a series of studies, Vijay-Kumar et al. (13, 21) reported that although TLR5 null animals initially had a colitis phenotype, when these mice were “rederived” and their gut microbiota altered, the colitis phenotype was greatly attenuated, and instead the null animals exhibited metabolic syndrome. In addition, Lathrop et al. put forward a model by which T-cells are educated not only by self/non-self mechanisms, but also by microbiota-derived “non-self” antigens (22). Accordingly, they found that the presence or absence of microbiota determined whether T cells would induce colitis in mice. Finally, Yang et al. reported that when the same knockout mice were housed at two different institutions, they had markedly different microbiota profiles – and the mice at one institution (MIT) were quite susceptible to colitis, whereas mice at the other institution (MHH) failed to develop any significant pathology under the same conditions (19). Unequivocally, altering gut microbiota – even by housing animals at different institutions – can have dramatic effects on the phenotype observed.

Go to:

Gut Microbiota and Obesity and Diabetes

It is important to note that not only can microbiota affect host physiology, but the gut microbiota are not necessarily stable over time. Rather, gut microbiota can change or shift as a result of experimental manipulation (in animals) or changes in lifestyle or nutrition (in humans). It is now appreciated that there are “shifts” in microbiota that occur in obesity in mice, rats, and humans (23–26). In one study, Turnbaugh et al. (25) examined human female twin pairs concordant for leanness or obesity, and found that obesity was associated with phylum-level changes in microbiota.

7 0

3 years ago

The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is

Diano4ka-milaya [45]

Answer:

f = 3.1 kHz

Explanation:

given,

length of human canal =2.8 cm = 0.028 m

speed of sound = 343 m/s

fundamental frequency = ?

The fundamental frequency of a tube with one open end and one closed end is,

$f = \dfrac{v}{4L}$

$f = \dfrac{343}{4\times 0.028}$

$f = \dfrac{343}{0.112}$

f = 3062.5 Hz

f = 3.1 kHz

hence, the fundamental frequency is equal to f = 3.1 kHz

8 0

3 years ago

Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates

IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

Sphere A : ma = 80 kg , ( 0 , 0 )

Sphere B : ma = 60 kg , ( 0.25 , 0 )

Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

Determine the angle (α) between vectors rac and rab using cosine rule:

$cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}$

Determine the angle (β) between vectors rbc and rab using cosine rule:

$cos ( \beta ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta ) = 0.6\\\\\beta = 53.13^{\circ \:}$

- Now determine the scalar gravitational forces due to sphere A and B on C:

Between sphere A and C:

Fac = G*ma*mc / rac^2

Fac = (6.674×10−11)*80*0.2 / 0.2^2

Fac = 2.67*10^-8 N

vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

Between sphere B and C:

Fbc = G*mb*mc / rbc^2

Fbc = (6.674×10−11)*60*0.2 / 0.15^2

Fbc = 3.56*10^-8 N

vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

Fc = vector Fac + vector Fbc

Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8

Fc = [ - 4.45 * 10^-8 j ] N

3 0

3 years ago