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Natalka [10]
3 years ago
11

Select the correct answer

Physics
2 answers:
zheka24 [161]3 years ago
4 0

Answer:

A. a state of being aware of the environment and oneself.

zalisa [80]3 years ago
3 0

Answer: A state of being aware of the environment and oneself or the first choice.

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If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula
snow_lady [41]

Here we know that

\omega_i = - 8.4 rad/s

\alpha = - 2.8 rad/s^2

t = 1.5 s

now from kinematics we have

\omega_f = \omega_i + \alpha t

now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

\omega_f = -8.4 + (-4.2)

\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

6 0
3 years ago
The bullet's horizontal velocity and vertical velocity vectors do not affect each other and are known as
lisabon 2012 [21]
Values in physics that do not affect each other are considered Independent values
8 0
3 years ago
Read 2 more answers
Is it true or not true
maria [59]

Answer:

true

Explanation:

6 0
3 years ago
Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de G
MakcuM [25]

Answer:

The magnitude of the electric field is 1.124 X 10⁷ N/C

Explanation:

Magnitude of electric field is given as;

E = \frac{kq}{r^2} , N/C

where;

E is the magnitude of the electric field, N/C

q is the point charge, C

k is coulomb's constant, Nm²/C²

r is the distance of the point charge, m

Given;

q = 5mC = 5×10⁻³ C

r = 2m

k = 8.99 × 10⁹ Nm²/C²

Substitute these values and solve for magnitude of electric field E

E = \frac{kq}{r^2}  = \frac{(8.99 X10^9)(5X10^{-3})}{2^2}  = 1.124 X10^7 N/C

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C

7 0
3 years ago
Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t
GenaCL600 [577]
B) 14.0 N    

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is: 

 E = 0.5 M V^2 

 where 

 E = Energy 

 M = Mass 

 V = velocity   

 Substituting the known values, let's calculate the before and after energy. 

 Before: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (2.25 m/s)^2 

 E = 6.75 kg 5.0625 m^2/s^2 

 E = 34.17188 kg*m^2/s^2 = 34.17188 joules   

 After: 

 E = 0.5 M V^2 

 E = 0.5 13.5kg (1.2 m/s)^2 

 E = 6.75 kg 1.44 m^2/s^2 

 E = 9.72 kg*m^2/s^2 = 9.72 Joules   

 So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.   

 24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N   

 Rounding to 1 decimal place gives 14.0 N which matches option "B".
8 0
3 years ago
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