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victus00 [196]
3 years ago
10

In the picture below, a doctor is having a video conference with a patient. What might be a positive effect of the technology be

ing
used?
Physics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

Sorry to say but where  is the photo???

The positive effect of technology being used might be using sethescope or checking BP rate if it is good.

Explanation:

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According to the graph of displacement vs. time, what is the object's displacement at time = 60 s?
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The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.

Explanation:

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that's fusion

Explanation:

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A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia
Bingel [31]

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

or

v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

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3 years ago
Formula of atmospheric pressure​
Alik [6]

Answer:

P_{h}=P_{0} e\frac{-mgh}{kT}

Explanation:

8 0
2 years ago
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta
creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

F_c=F_e=ma_c      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

F_e=k\frac{qq'}{r^2}             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

a_c=\frac{v^2}{r}                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}

Finally, you replace the values of all parameters in the previous expression:

r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m

The radius of the circular trajectory is 0.22m

5 0
3 years ago
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