Ask question

Ask question

All categories

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).

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.

You might be interested in

Plz help on number 4

ki77a [65]

Forces on a Baseball. When a baseballis thrown or hit, the resulting motion of the ball is determined by Newton's laws of motion. ... Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion.

8 0

3 years ago

A push or pull is called a force. Unbalanced forces can cause objects to move in many ways but not to

galina1969 [7]

Answer:

d. float in the air

Explanation:

im not aure

5 0

2 years ago

A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas

siniylev [52]

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Rewrite the above equation in terms of 'm'. This gives,

$2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}$

Now, plug in the given values and solve for 'm'. This gives,

$m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg$

Therefore, the mass of the truck is 3331.5 kg.

3 0

3 years ago

The specific heat capacity of aluminum is 0.22 cal/g°C. How much energy needs to flow into 20.0 grams of aluminum to change its

Leokris [45]

Its C. 66 cal , that's your answer to your question.

3 0

3 years ago

Read 2 more answers

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago