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Alexxx [7]
3 years ago
8

The table shows results of an experiment that was replicated. Which best describes the data? They are neither precise nor reprod

ucible. They are precise but not reproducible. They are accurate and reproducible. They are accurate but not reproducible.
Physics
2 answers:
BartSMP [9]3 years ago
8 0

Answer:

They are accurate and reproducible.

Explanation:

FrozenT [24]3 years ago
4 0

Answer:

They are precise and reproducible.

Explanation:I just took the test on Edgenuity

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A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0
3 years ago
How does exhailing remove waste from the body? Explain the systems that make this happen‚ using complete sentences​
aliina [53]

Answer:

The lungs get rid of carbon dioxide and water vapor. The liver gets rid of bile, which, in addition to breaking down fats, is partially made up of the breakdown of red blood cells. The kidneys get rid of toxins from the blood. The large intestine gets rid of undigested food

3 0
3 years ago
Which bond(s) shown are double bonds?
Nikitich [7]

Answer:

I'm pretty sure its the 2nd one.

3 0
3 years ago
Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t
Stels [109]

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

a =\frac{ v-u}{\Delta t}

Here, v is final velocity and u is the initial velocity and \Delta t is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2

Thus, the average acceleration of the car is 14 .3 \ m/s^2 in the direction to the left.

6 0
3 years ago
Hey besties, im having a mental breakdown on this question. pls help :D
forsale [732]

With "toward the wall" designated as the positive direction, the average acceleration felt the ball in the 0.0158 s of contact with the wall is

a_{\rm ave} = \dfrac{-26.412\frac{\rm m}{\rm s} - 28.4\frac{\rm m}{\rm s}}{0.0158\,\mathrm s} ≈ \boxed{-3470 \dfrac{\rm m}{\mathrm s^2}}

8 0
3 years ago
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