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3 years ago

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!

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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a ra

worty [1.4K]

Answer:

The torque on the pulley, when the system is motionless is approximately 9.81 N·m

Explanation:

The given parameters are;

The mass of the object = 2 kg

The friction between the rope and the pulley = 0

The mass of the rope, $m_r$ = 0.5 kg

The mass of the pulley, $m_p$ = 0.01 kg

The radius of the pulley, r = 0.25 m

The torque on the pulley, τ = I·α = F × D

The torque on the pulley, when the system is motionless, τ = F × D

Where;

F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N

D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m

Therefore;

τ = 19.62 N × 0.5 m = 9.81 N·m

The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.

4 0

2 years ago

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

2 years ago

A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

8 0

2 years ago

Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?

Alik [6]

Answer:

Below

Explanation:

You can use this equation to find the distance :

distance = velocity x time

distance = (26.7)(3.06)

= 81.702 m

Rounding to 3 sig figs

= 81.7 m

Hope this helps

6 0

2 years ago

What is the half-reaction that occurs at the cathode during the electrolysis of molten potassium bromide?

polet [3.4K]

The complete ionization of KBr into its constituents is:
KBr (s) ---> K+ (aq) + Br- (aq)

During electrolysis, oxidation takes place at the anode electrode. This means that an ion is stripped off its electron hence becoming more positive:
2 Br- (aq) ---> Br2 (g) + 2e-

We can see that Bromine gas Br2 is evolved at the anode.

Meanwhile at the cathode, the reduction reaction occurs. Which means that the electron from the anode electrode is used to make an ion more negative:
2K+ (aq) + 2e- ---> 2K (s) 
Hence, through reduction, solid potassium is deposited on the plate.

Half reactions:

Anode: 2 Br- (aq) ---> Br2 (g) + 2e-

Cathode: 2K+ (aq) + 2e- ---> 2K (s)

7 0

2 years ago