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guapka [62]
2 years ago
12

a student measures the mass of a metal weight using three separate balances and collects the following data: 38.7g, 38.9g and 38

.6g. The actual mass of the weight is 34.2 g. Evaluate the accuracy and precision of this student's measurements.
Physics
1 answer:
Lesechka [4]2 years ago
6 0

The values are precise because the values are close together but they are not accurate because they are all far from the true value.

<h3>What is the accuracy and precision?</h3>

The term accuracy refers to how much close to the true or the actual value that a measurement is. The precision refer to how close together the values that are obtained in a series of measurements are.

Now, we have the values that were obtained as; 38.7g, 38.9g and 38.6g. We can see that values are precise because the values are close together but they are not accurate because they are all far from the true value.

Learn more about precision:brainly.com/question/28336863

#SPJ1

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A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
1. What is evaporation and how does it affect weather?
alina1380 [7]

Answer:

Explanation:don’t know

8 0
3 years ago
What is a period in physics?<br>​
Citrus2011 [14]

Answer:

The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.

3 0
3 years ago
What is the speed of the animal from 0-20s? <br><br> 50 m/s<br> 0.4 m/s<br> 20 m/s<br> 2.5 m/s
Shkiper50 [21]

Answer:

2.5 m/s

Explanation:

The speed of the animal is given by the ratio between the distance travelled by the animal and the time elapsed:

v=\frac{d}{t}

where d is the distance travelled and t the time elapsed. Note that this quantity is also equal to the slope of the curve.

In the time interval 0-20 s, we have

d = 50 m - 0 m = 50 m

t = 20 s - 0 s = 20 s

So, the speed is

v=\frac{50 m}{20 s}=2.5 m/s

3 0
3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
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