If an object is acted on by a constant unbalanced force, the object will move with a constant ________

Physics

2 answers:

MissTica3 years ago

7 0

There is always a net force acting on the object, so it will have constant acceleration

AveGali [126]3 years ago

4 0

Answer:

C. Acceleration

Explanation:

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

Luden [163]

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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4 years ago

A book is pushed across the table by a student. The student stops pushing on the book and the book slows down and stops. The for

VashaNatasha [74]

A. inertia

Because it has to do with the motion of something, especially if it changes its pace. In this example, the book's motion, when sliding on the table, decreased because of less force being given off from the student.

3 0

4 years ago

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An airplane travels 3,260 kilometers in 4 hours. What is the airplane’s average speed?

OLga [1]

3260÷4=815 which is you average seed

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3 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$