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horrorfan [7]
2 years ago
11

A

Physics
1 answer:
loris [4]2 years ago
7 0

Hi

The answer to this question is B. Reaction

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In an experiment refractive index of glass was observed to be 1.45,1.56,1.54,1.44,1.54and1.53. Calculate mean value of refractiv
Sever21 [200]
The mean may be calculated by summing the values of the refractive index and dividing the sum by the number of experiments. This is:
Mean = (1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53)/6
Mean = 1.51

The mean absolute error is the sum of the absolute values of errors divided by the number of trials:
MAE = (|1.45-1.51|+|1.56-1.51|+|1.54-1.51|+|1.44-1.51|+|1.54-1.51|+|1.53-1.51|)/6
MAE = 0.043

The fractional error is the MAE divided by the actual value:
Fractional error = 0.043 / 1.51
Fractional error = 43/1510

The percentage error is the fractional error multiplied by 100:
Percentage error = 2.85%
7 0
3 years ago
skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular
laiz [17]

Answer:

L=11.3\ kg-m^2/s

Explanation:

Given that,

Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

L=I\omega

Substitute all the values,

L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

So, its angular momentum is equal to 11.3\ kg-m^2/s.

8 0
3 years ago
An object is thrown 16m/s straight up from a 7m tall cliff. How much time does it take to hit the ground below
harkovskaia [24]

Answer:

0,54 sec

Explanation:

t=s/v

t=7/13 s=0,54sec

5 0
3 years ago
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re
Leya [2.2K]

Answer:

I=2.6363\ kg.m^2

Explanation:

Given:

dimension of uniform plate, (0.673\times 0.535)\ m^2

mass of plate, m=10.7\ kg

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

I_{cm}=\frac{1}{12} \times m(L^2+B^2)

where:

L= length of the plate

B= breadth of the plate

I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)

I_{cm}=0.6591\ kg.m^2

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}

s=0.4299\ m

Now using parallel axis theorem:

I=I_{cm}+m.s^2

I=0.6591+10.7\times 0.4299^2

I=2.6363\ kg.m^2

6 0
3 years ago
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