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Tresset [83]
3 years ago
14

Help with this please in the image

Physics
1 answer:
ValentinkaMS [17]3 years ago
7 0

The two forces for a 90 degree angle.


Use the Pythagorean theorem:


sqrt(8^2 + 8^2) = sqrt(64 + 64) = sqrt(128)

= 11.31


Answer: C. 11 N

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Think about lifestyle choices that you make on a daily basis. Choose at least four lifestyle choices which you could change to i
miss Akunina [59]

Answer:

A.The four lifestyle choices which I could change are:

1. The amount of physical exercise I engage in on a daily basis.

2, The number of hours I sleep.

3. The quantity of junk foods I consume daily and 

4. Cultivation of the habit of spending time alone daily.

B. My days are always very busy, so I do not engage at all in physical exercise. Adding at least a thirty minute of physical exercise to my daily tasks will improve my health considerable. Due to the nature of my day, I usually eat on the move, thus I eat junk foods everyday. Eating healthy and balanced diet will go a long way in improving my health. I sleep for three hours only everyday, from 2 am to 5 am. This is because I have to leave my house very early in the morning and I don't usually get to sleep until 2 am in the morning. Spending more hours sleeping will improve my overall health. I want to start spending at least 30 minutes alone on a daily basis in order to meditate and to reflect over the events of the day. This will be a great boost to my mental health.

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3 0
4 years ago
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Meg walks 5 blocks south and then turns around and walks 8 blocks back to the north. The distance that she walks is 13 blocks. W
Drupady [299]

Three blocks north of where she started.

4 0
4 years ago
The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter
Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration=a=\frac{1}{4}s^{\frac{1}{2}}m/s^2

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v

vdv=ads

\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}

By using formula:\int x^ndx=\frac{x^{n+1}}{n+1}+C

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1

Substitute s=2

\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)

\frac{v^2}{2}=\frac{0.50}{3}\times 1.83

v^2=2\times 0.305=0.61

v=\sqrt{0.61}=0.78m/s

Hence, the velocity of particle at s=2m=0.78m/s

3 0
4 years ago
A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose
dmitriy555 [2]

Answer:

x=0.478\ m is the compression in the spring

Explanation:

Given:

  • mass of the bullet, m=10\ g=0.01\ kg
  • mass of block, M=2\ kg
  • stiffness constant of the spring, k=19.6\ N.m^{-1}
  • initial velocity of the spring just before it hits the block, u=300\ m.s^{-1}

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

m.u=(M+m).v

0.01\times 300=(2+0.01)\times v

v=1.4925\ m.s^{-1}

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

\rm Kinetic\ energy=Spring\ potential\ energy

\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2

\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2

x=0.478\ m is the compression in the spring

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