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victus00 [196]
3 years ago
7

What piece of equipment would a hospital most likely have as back up in case of a power failure?

Physics
2 answers:
sertanlavr [38]3 years ago
7 0

the answer to your question here is B a generator

Shalnov [3]3 years ago
4 0

B. a generator

Hope I could help!

You might be interested in
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
3 years ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
Question 2 (1 point)
eduard

Answer: Blake may have a decreased chance of disease.

Explanation:

Research has shown that on average, a person who engages in more physical activity will live longer than a person who does not with one reason for this being that engaging in physical activity helps reduce the risk of diseases such as; heart disease, diabetes, and many forms of cancer.

Obesity (risk factor) and high blood pressure are also reduced by engaging in physical activity.

6 0
3 years ago
ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is
qaws [65]

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = 2\pi\sqrt{\frac{l}{g} }

where T is time period , l is length of pendulum and g is acceleration due to gravity .

\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }  , f is frequency of oscillation

For the given case

\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }

subsequently length becomes half so

\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }

dividing

\frac{f}{f_o} = \sqrt{\frac{2}{1} }

f = \sqrt{2} f_o

frequency of oscillation becomes √2 times.

4 0
3 years ago
By method of dimension show that the following equation are homogenous.
il63 [147K]

Answer:

Proof in explanataion

Explanation:

The basic dimensions are as follows:

MASS = M

LENGTH = L

TIME = T

i)

Given equation is:

H = \frac{u^2Sin^2\phi}{2g}

where,

H = height (meters)

u = speed (m/s)

g = acceleration due to gravity (m/s²)

Sin Ф = constant (no unit)

So there dimensions will be:

H = [L]

u = [LT⁻¹]

g = [LT⁻²]

Sin Ф = no dimension

Therefore,

[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]

<u>[L] = [L]</u>

Hence, the equation is proven to be homogenous.

ii)

F = \frac{Gm_1m_2}{r^2}\\\\

where,

F = Force = Newton = kg.m/s² = [MLT⁻²]

G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²

G = [M⁻¹L³T⁻²]

m₁ = m₂ = mass = kg = [M]

r = distance = m = [L]

Therefore,

[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

<u>[MLT⁻²] = [MLT⁻²]</u>

Hence, the equation is proven to be homogenous.

8 0
3 years ago
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