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Ksju [112]
3 years ago
15

Which muscle disorder does donna’s doctor diagnose? donna is suffering from leg pain in which the calf muscles contract involunt

arily. donna’s doctor diagnoses her wit
Physics
2 answers:
Mazyrski [523]3 years ago
8 0

Answer: spasticity ☆

Explanation:

Kay [80]3 years ago
7 0
Donna's doctor must have diagnosed her with muscle cramps. With the given symptoms above, it is likely that she is experiencing muscle cramps. Muscle cramps most likely happen when the individual experiencing it is lack of fluid intake, causing the muscles to tighten causing pain and for the muscle to contract involuntarily.
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A cube of side 0.2 m rests on the floor as shown.
andrey2020 [161]
Pressure= hqg
H=depth
q=density
g=gravity

h=0.2
q=7
g=10

0.2*7*10= 14pa

FINAL ANSWER = 14pa
4 0
2 years ago
A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/
klemol [59]

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t  where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}

-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}  where d= 10937.5m; Vo=875m/s.

Solving for t:

t1 = -11.093s   t2 = 98.59s

So, the total time of flight will be:

t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s

6 0
3 years ago
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t square
torisob [31]

Answer:

a) 2.933 m

b) 4.534 m

Explanation:

We're given the equation

v(t) = -0.4t² + 2t

If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.

See attachment for the calculations

The conclusion of the attachment will be

7.467 - 2.933 and that is 4.534 m

Thus, The distance it travels in the second 2 sec is 4.534 m

6 0
3 years ago
An object experiences an acceleration of 8.5 m/s^2 over a distance of 300 m. After that acceleration it has a velocity of 400 m/
Snezhnost [94]

Answer:

393.6m/s

Explanation:

Given parameters:

Acceleration  = 8.5m/s²

Distance  = 300m

Final velocity  = 400m/s

Unknown:

Initial velocity  = ?

Solution:

To solve this problem, we use the expression below;

             v² = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

      So;

               v²  - 2as = u²

        u²   = v²   - 2as

        u²  = 400²   - (2 x 8.5 x 300)  

         u   = 393.6m/s

7 0
3 years ago
Read 2 more answers
You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.5 m above the ground, and it is
ollegr [7]

My answer was incorrect, please disregard.

4 0
3 years ago
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