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shepuryov [24]
3 years ago
8

If you dropped a ball from a height of 10 m and measured its distance to the ground every 0.1 s, what would the graph of positio

n versus time look like?
a. It would curve upward.
b. It would curve downward.
c. It would be a straight line angled downward.
d. It would be a line parallel to the horizontal axis.
Physics
1 answer:
wlad13 [49]3 years ago
5 0

If you dropped a ball from any height, and measured its distance from the ground at any regular interval while it's falling, the graph of that distance versus time would be a graph that curves downward.

-- The ball is falling down.  As time goes on, it gets closer and closer to the ground.  Its remaining distance from the ground keeps decreasing, so the line on the graph slopes down.

-- The speed of the ball keeps increasing (it accelerates) because of the gravitational force on it. As time goes on, it covers more of the remaining distance during each interval than it did in the previous interval. The downward slope of the graph keeps increasing.

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The figure above shows a rod that is fixed to a horizontal surface at pivot P The rod is initially rotating
Lisa [10]

The angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.

<h3></h3><h3>What is angular speed ?</h3>

The rate of change of angular displacement is defined as angular speed.  It is stated as follows:

ω = θ t

Where,

θ is the angle of rotation,

t is the time

ω is the angular velocity

The torque is found as;l

\rm  T = F \times r

If the force is acting on the rod  from the three point is the same, the value of the torque is depends upon the radius or the perpendicular distance.

The perpendicular distance of the right force is grater. Hence, the force acting on the right side is more, and the rod will rotate clockwise.

Both the forces are acting downwards. Thus, the resultant force is the less due to which the speed is increasing.

Hence, the angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.

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5 0
2 years ago
Acceleration of gravity on jupiter
Viefleur [7K]

The acceleration of gravity on Jupiter is listed as  <em>24.79 m/s²</em> .

That's roughly  2.53  times its value on Earth.  So if you weigh, let's say,
130 pounds on Earth, then you would weigh about 328 pounds on Jupiter.



5 0
3 years ago
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A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

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8 0
3 years ago
To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes t
QveST [7]

Answer:

(c) position

Explanation:

From the work-energy theorem, the workdone by a force on a body causes a change in kinetic energy of the body.

But, remember that the work done (W) by a force (F) on a body is the product of the force and the distance d, moved by the body caused by the force. i.e

W = F x d

This distance is a measure of the position of the body at a given instance.

Therefore, the work done is given by the force as a function of distance (or position).

3 0
3 years ago
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