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Kisachek [45]
1 year ago
13

Anita is comparing the accepted value for a physical property to the value she measured in the laboratory. Which characteristic

of her measured value is Anita most likely trying to evaluate? accuracy mean precision range
Physics
2 answers:
Butoxors [25]1 year ago
7 0

Answer:

nita must look for the percentage error and look for the precision of its measurement.

Explanation:

When a measurement of a physical property is made in the laboratory, it must be compared with the accepted values. For this, the mean value of the laboratory measurement is calculated and the absolute or percentage error is sought with respect to the accepted value.

In conclusion Anita must look for the percentage error and look for the precision of its measurement.

sp2606 [1]1 year ago
5 0

Answer:

accuracy

Explanation:

correct on edge2020

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Taro stated that when someone hits a golf ball with a club, the amount of energy the ball has changes, the amount of energy that
Darina [25.2K]
Taro's error is when he stated that the total energy of the ball and the club system is increasing. This is not true. The total energy of the system is not increasing. According to the first law of thermodynamics, <span>total energy of a system is always constant; energy can be transformed from one form to another however it cannot be created or destroyed. </span><span>Energy is conserved. </span>So, for this problem the total energy of the system should remain constant at all times.
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1 year ago
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A beam of monochromatic light (f =5.09 ×1014 Hz) has a wavelength of 589 nanometers in air. What is the wavelength of this light
frosja888 [35]
Lucite has a refractive index of n=1.50. This means that the speed of the light in lucite is decreased according to:
v=\frac{c}{n}
where c=3 \cdot 10^8 m/s is the speed of light in air. Putting the number in the formula, we find that the speed of light in lucite is
v=\frac{3 \cdot 10^8 m/s}{1.50}=2\cdot 10^8 m/s
The frequency of the light is f=5.09 \cdot 10^{14}Hz, so now we can calculate the wavelength in lucite by using the formula:
\lambda=\frac{v}{f}=\frac{2\cdot 10^8 m/s}{5.09 \cdot 10^{14} Hz}=3.93 \cdot 10^{-7} m=393 nm
<span>Therefore, the correct answer is (2) 393 nm.</span>
7 0
1 year ago
a student moves a box across the floor by exerting 23.3 N of force and doing 47.2 J of work on the box. How far does the student
kozerog [31]
Work = Force x Distance
47.2J = 23.3N x d
d = 47.2/23.3
d = 2.0258 m

hope this helps :P
8 0
1 year ago
A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision
TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0
2 years ago
A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

\frac{1}{f} = (nglass - ni)(\frac{1}{R1} - \frac{1}{R2}).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

\frac{1}{f} = (1.5 - 1.3)(\frac{1}{10} - \frac{1}{15})

\frac{1}{f} = 0.2(\frac{1}{30})

f = \frac{30}{0.2} = 150

For air (nair = 1):

\frac{1}{f} = (1.5 - 1)(\frac{1}{10} - \frac{1}{15})

f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

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2 years ago
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