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2 years ago

4. Anaerobic exercise helps type 2 diabetes. a. True b. False

Physics

1 answer:

Ksenya-84 [330]2 years ago

8 0

Answer:

true

Explanation:

b/c anaerobic exercis helps type 2 diabetes

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How are the electric field lines around a positive charge affected when a second positive charge is near it?

Lesechka [4]

Answer: The field lines bend away from the second positive charge

Explanation: opposite attracts, same repulse

4 0

3 years ago

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Amy uses 20N of force to push a lawn mower 10 meters. How much work does she do? *

babunello [35]

She does 200J .

We know she uses 20N of force and 10m is the distance. We multiply both numbers and we are given our answer of 200J. Hope this was helpful. :)

4 0

3 years ago

You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds

german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

$f_{beat}=\frac{1}{T}$

Then the frequency for the previous period given (2sec) is

$f_{beat}=\frac{1}{2}$

$f_{beat} = 0.5Hz$

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

$f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz$

Hence option A is incorrect.

We can do this process for 254Hz as $f_1$ and 258 Hz for $f_2$ , then

$f_{beat} =|254Hz-258Hz|$

$f_{beat} = 4Hz$

Hence option B is incorrect.

We can also do this process for 255Hz as $f_1$ and 257 Hz for $f_2$ , then

$f_{beat} =|255Hz-257Hz|$

$f_{beat} = 2Hz$

Hence option C is incorrect.

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

$f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz$

Hence option D is incorrect.

We can also do this process for 255.75Hz as $f_1$ and 256.25 Hz for $f_2$ , then

$f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz$

Hence option E is incorrect.

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0

3 years ago

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

$T=3.74\times 10^8\ s$

(b) We know that,

$1\ year=3.154\times 10^7\ s$

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

T = 11.859 earth years

Hence, this is the required solution.

7 0

3 years ago

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A car engine supplies 2.0 x 103 joules of energy during the 10. seconds it takes to accelerate the car along a horizontal surfac

andrew11 [14]

Answer: 2. 2.0*10^2 W

Explanation:

Power = Work/Time

Power = (2.0*10^3) Joules/10 seconds

Power = 2.0*10^2 Watts

7 0

2 years ago