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SSSSS [86.1K]
3 years ago
13

How has the use of computer and advance in technology changed healthcare industry

Physics
1 answer:
aev [14]3 years ago
8 0
Yes because we often used it and takes long time to spent and forgot to eat on time.
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A 1 kg pistol, initially at rest, shoots a bullet with a mass of 2.5g. After
Allisa [31]

Answer:

Use the principle of momentum

Initial momentum = final momentum

Momentum formula = Mass * Velocity

Explanation:

6 0
2 years ago
Identify the amplitude in the wave image below.*<br> F. G H J
alina1380 [7]
Answer:F
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2 years ago
For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o
jasenka [17]

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

5 0
2 years ago
Read 2 more answers
a block of wood with a mass of 2.33 kg is at rest on a frictionless pole. A .011 kg bullet is fired 722 m/s and is embedded with
MrRissso [65]
Some rewards are 2.33 miles in a hour so you have to move in 700 degrees to get the system moving faster soo 700+ 2.33 divide by 3
6 0
2 years ago
One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F
Alja [10]

Answer:

\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}

f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}

b) Wave speed

(i) Calculate the wavelength

In a  fundamental vibration, the length of the string is half the wavelength.

\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

(b) Calculate the speed s

\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}

\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}

\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

4 0
3 years ago
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