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Nataliya [291]
3 years ago
5

Who’s going faster in the attached image?

Physics
2 answers:
ioda3 years ago
8 0

Answer:

Art

Explanation:

Polly's line is linear, while arts line is going up with constant velocity. There for art is going faster.

Juliette [100K]3 years ago
6 0

There isn't a simple answer to that question.

Polly is going at a steady speed for the whole 5 seconds.  She must have started before time-zero, because AT time-zero, she's already moving at 1m/s, and she stays at that steady speed for at least the next 5 seconds that we know of.  (That's all the graph tells us.)

Art is standing at the starting line, waiting for the signal to GO.  At time zero, he's not moving at all, so Polly is moving faster than he is.  

At time-zero, (when the starter fires the pistol ?), Art <u>starts</u> moving AND building up speed.  He builds up speed steadily, going faster and faster.  His speed builds and builds all the way across the graph.  He starts out with zero speed, and he <em>accelerates</em> so that his speed grows by 1 m/s every 2 sec.  After 2 sec, he's moving at 1 m/s, and after 4 sec, he's built it up to 2 m/s.

Polly is moving at 1 m/s all across the graph, for the whole 5 sec. So she's going faster than Art is, <em>for the first 2 seconds</em>.  But Art is building his speed, faster and faster.  He reaches 1 m/s after 2 sec, and <u>after that</u> he's going faster that Polly.  

All this information that I'm saying is on the graph.  I'm just describing the graph in painful boring detail.  As you read my description, you should be able to look over at the graph every time I say something, and see it there.

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What is hydraulic pressure?
Ilia_Sergeevich [38]
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3 0
3 years ago
which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20
MA_775_DIABLO [31]
<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2


2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>
7 0
4 years ago
Read 2 more answers
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
A box has sides of 10 cm, 8.2 cm, and 3.5 cm. What is its volume?
Ksju [112]
The volume would be 287cm³. Multiply all the 3 numbers by each other
4 0
3 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c
seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2

So the volume flow rate along the pipe is

\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s

We can use the similar logic to find the cross-section area at the refinery

A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2

The radius of the pipe at the refinery is:

A_r = \pi r^2

r^2 =A_r/\pi = 0.446/\pi = 0.141

r = \sqrt{0.141} = 0.377m

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0
3 years ago
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