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Zigmanuir [339]
3 years ago
6

A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches

per second and the volume, V, is 128pi cubic inches. At what rate is the length, h, changing when the radius, r, is 1.5 inches? Note: V=pi(r^2)h
Hi, as above this is the problem I am doing. There is no answer given to check so I was wondering about a couple of things:

Is volume assumed to be a constant in this problem?

Is the height change decreasing (negative rate) or increasing (positive rate) in the final answer?

I worked out dh/dt to be 512/135 in./s
Physics
1 answer:
Luden [163]3 years ago
4 0
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec
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A race car traveling at 10. meters per second accelerates at the rate of 1.5 meters per seconds while traveling a distance of 7,
Arturiano [62]

The final speed of the car is 2) 150 m/s

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 10 m/s

a=1.5 m/s^2

s = 7,467 m

Solving for v, we find the final velocity (and speed) of the car:

v=\sqrt{u^2+2as}=\sqrt{10^2+2(1.5)(7,467)}=150 m/s

Learn more about accelerated motion:

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#LearnwithBrainly

4 0
3 years ago
Read 2 more answers
The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w
Citrus2011 [14]

Complete Question

 The complete question is shown on the first uploaded image

Answer:

The magnetic field is B_{net} = \frac{1}{4}  * mT

And the direction is  -\r k

Explanation:

      From the question we are told that

                 The magnetic field at the center is B = 1mT

Generally magnetic field is mathematically represented as

              B = \frac{\mu_o I}{2R}

We are told that it is equal to 1mT

So

                B = \frac{\mu_o I}{2R} = 1mT

From the first diagram we see that the effect of the current flowing in the circular loop is  (i.e the magnetic field generated)

                         \frac{\mu_o I}{2R} = 1mT

 This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

                   B_1 = \frac{1}{2} \frac{\mu_o I}{2R}

and  for the larger semi-circular loop  is

                 B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

    So the net magnetic field would be

                   B_{net} = B_1 - B_2

                        =  \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}

                        =\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}

                        =\frac{\mu_o I}{8R}

                        = \frac{1}{4} \frac{\mu_o I}{2R}

Recall  \frac{\mu_o I}{2R} = 1mT

    So  

             B_{net} = \frac{1}{4}  * mT

Using the Right-hand rule we see that the direction is into the page which is -k

3 0
3 years ago
Help me solve this. Someone help me pleeeaaaseee​
White raven [17]

dddddddddddddddddddddddddddfghjk

6 0
2 years ago
Help plz I’ll mark brainliest
marin [14]
I think it’s concave
3 0
3 years ago
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You are in a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball ta
notsponge [240]
B) 7.87 m/s

The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
a =  \frac{final \: velocity - initial \: velocity}{time \: taken}

Given:

• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2

We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.

3.28  =  \frac{final \: velocity - 0}{2.40}
3.28 \times 2.40 = final \: velocity
final \: velocity = 7.872 \frac{m}{ {s} }

5 0
3 years ago
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