People under going physical therapy after an injury often find it helpful to perform exercise in water? Why?

const2013 [10]

The answer should be C. Both matter and energy because physical sciences are sciences concerned with the study of inanimate natural objects, including physics, chemistry, astronomy, and related subjects.

Matter and energy fall in with chemistry

6 0

3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

A 700 kg car makes a turn going at 30 m/s with radius of

Gnoma [55]

Answer:5250 N

Explanation: ig:iihoop.vince

5 0

3 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

3 years ago

A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef

Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.

Initial horizontal velocity is: 
cos(30 degrees) * 12m/s = 10.3923m/s

15.5m / 10.3923m/s = 1.49s

So the rock was in the air for 1.49 seconds.

Now that we know that, we can use the following kinematics equation:

d = v i * t + 1/2 * a * t^2

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2.

Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s

So:

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89

d = -1.95

This means that the initial y position is 1.95 m higher than where the rock lands.

5 0

3 years ago