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vivado [14]
3 years ago
9

A balloon used in surgical procedures is cylindrical in shape. as it expands outward, assume that the length remains a constant

. find the rate of change of surface area with respect to radius when the radius is (answer can be left in terms of π).
Physics
1 answer:
DochEvi [55]3 years ago
6 0
A cylinder's volume is π r² h, and its surface area is 2π r h + 2π r².  Since there is no length and radius. Let is denote the length by 80 mm and 0.30 mm radius.
Surface area, A = 2 (pi) r * 80 = 160(pi)r 
dA / dr = 160 (pi)
160.0π mm^2/mm is the answer.
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Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Gnesinka [82]

Answer:

Q1 = 7.25*10^(-16) C 

Explanation:

We are given;

electric field strength = (1 x 10^5 N/C

drag force (F) = 7.25 x 10^(-11) N

The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.

Columbs law force of attraction or repulsion between two charges is given as;

F=(KQ1Q2)/r²

Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;

7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))

Q1 = 7.25*10^(-16) C 

5 0
3 years ago
Read 2 more answers
A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e
almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

8 0
4 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s&#10;&#10;0 = 2500 - 186544.54s&#10;&#10;Isolate S next.&#10;&#10;185644.54s= 2500&#10;&#10;s =  2500/(185644.54)&#10;&#10;s=0.0134&#10;

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
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4 0
3 years ago
Wat's the definition of microwave spectroscopy
ivann1987 [24]

Answer:

Microwave spectroscopy is the spectroscopy method that employs microwaves, i.e. electromagnetic radiation at GHz frequencies, for the study of matter

6 0
4 years ago
A Carnot heat engine has an efficiency of 0.200. If it operates between a deep lake with a constant temperature of 293.0 K and a
Dimas [21]

Answer:

366 K

Explanation:

T₀ = Constant Temperature of deep lake = 293.0 K

T = Temperature of hot reservoir  connected to carnot engine = ?

η = Efficiency of Carnot engine during the operation

Efficiency of Carnot engine is given as

\eta = 1-\frac{T_{o}}{T}

Inserting the values

0.200 = 1-\frac{293.0}{T}

T = 366 K

3 0
4 years ago
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