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2 years ago

6

Angela has been exercising regularly for two years. Right now at her house everyone is feeling sick, but she is able to help the

others to feel better. Which long-term benefit of exercise is Angela noticing?

1 answer:

Dmitry [639]2 years ago

3 0

Answer:

it make your immune system build up from the strain your puting on you body

Explanation:

excercise is good for you and builds muscle and blood cells

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Which of the following has the greatest momentum?

Jobisdone [24]

Answer:

Tortoise has the greatest momentum

Explanation:

Momentum of tourtise (p)=mass×velocity

=275×0.55

=151.25 kgm/s

4 0

3 years ago

PLS HELP FIND THE AVREGE KENETIC ENERGY!!

ANEK [815]

Answer: if you gogle it it will tell you the awenser

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3 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

3 years ago

What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength

ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

$\lambda = \frac{h}{\sqrt{2 m K}}$

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}$

λ = 1.2826 x 10^-13 m

6 0

3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

3 years ago