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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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A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a
lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

  • Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
  • At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

       E_{o} = U_{o} = m*g*h_{init}  (1)

  • When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
  • We can express this final energy as follows:

       E_{f} = U_{f} + K_{f}  = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2}  (2)

  • In order to find hinit, we need to make (1) equal to (2), and solve for it.
  • In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
  • Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
  • We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
  • This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
  • If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
  • So, the centripetal force must be equal to the weight.
  • The centripetal force can be expressed as follows:

       F_{c} = m*\frac{v_{f} ^{2}}{R}  (3)

  • We have just said that (3) must be equal to the weight:

       F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)

  • Simplifying, and rearranging, we can solve for vf², as follows:

       v_{f}^{2} = R*g  (5)  

  • Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

      h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)

7 0
3 years ago
A passenger in a car is holding an open water bottle when the car stops suddenly. Why does the water in the bottle splash out wh
OlgaM077 [116]

Answer:

The answer is It has inertia

Explanation:

I just got it correct on E2020

8 0
4 years ago
Read 2 more answers
Suppose the moon were held in its orbit not by gravity but by the tension in a massless cable. You are given that the period of
Ksenya-84 [330]

Answer:

Tension in the cable = 200.655 × 10^(18) N

Explanation:

The tension in the cable will be the force on the string which is caused by when it holds against the centripetal acceleration.

Now, the formula for centripetal acceleration is given by;

a = v²/R

Now, let's find the velocity.

The circumference is 2πR

We are given that R = 3.85 x 10^(8) m

So, Circumference = 2 × π × 3.85 x 10^(8) = 24.19 × 10^(8) m

We are told the period of the moon orbit is 27.3 days

Distance travelled per day = Circumference/ period

Distance travelled per day = (24.19 × 10^(8))/27.3 = 88608058.608 m/day

Now, a day has 24 hours = 24 × 60 × 60 = 86400 seconds

Thus,

Distance travelled per seconds =

88608058.608/86400 = 1025.556 m/sec

So, from a_c = v²/r,

a_c = 1025.556²/3.85x10^8

a_c = 0.00273 m/s²

Now, Force = Mass x Acceleration = ma

We are given mass = 7.35 x 10^(22) kg

Plugging in the relevant values, we have;

F = 0.00273 x 7.35 x 10^(22)

F = 200.655 × 10^(18) N

5 0
3 years ago
A 240-volt, 2-amp motor is connected to a three-wire, 120/240-volt system. Connected between the black wire and neutral are four
pantera1 [17]

Answer:

(i)The current flow in black wire = 9.67 A (ii) The current low in the red wire is 9.68 A (iii) The current flow in neutral wire is  15.36 A (iv) when 240 volt were disconnected current  in black wire is 7.68 A (v) when 240 volt were disconnected current in red wire is 7.68 A (vi) 15.36 A (vii) 6.34 (viii) 9.68 A (ix) 12.02 A

Explanation:

Solution

The current drawn by one amp is

I =P/V

I =200/120

I= 1.67 A

(i) The current flow in the black wire  is

IBK = 4 * 1.67 A + 1A + 2A

IBK = 9.67 A

(ii) Current flow in the red wire is

IRD = 3 * 1.67 A + 1.67 A + 1A + 2A

= 8.68A + 1 A = 9.68 A

Note: Kindly find an attached copy of part of the solution to the given question above.

6 0
4 years ago
When you bring the south ends of two magnets close together, they repel each other.The strength of an electromagnet can be incre
snow_lady [41]

When south poles of the two magnets are brought close together, they will repel.

Thus, the statement is true.

The strength of an electromagnet can be increased by increasing the number of turns on the wire of the coil.

Thus, the second statement is false.

8 0
1 year ago
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