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vlada-n [284]
3 years ago
8

Action and reaction forces will cancel because they are pushing on the same object toward each other. True or False

Physics
2 answers:
sweet-ann [11.9K]3 years ago
5 0

Answer:

False

Explanation:

Newton's third law states that for every action there is an equal and opposite reaction. The action-reaction pair act on different bodies. Thus, they do not cancel each other.

For example, in order to walk, we push the floor backwards. This is action force which the foot applies to the floor. The floor pushes the foot forward. This is the reaction force. The amount of force is equal but the direction is opposite. Also, the bodies on which action force and reaction force are acting are different.

Hence, the given statement is false.

vampirchik [111]3 years ago
4 0
Sometimes. If the forces are equal, they will cancel out and the net force will be zero. But if the forces are not equal, the object will move away from the stronger force because there is still a net force.
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When fuel and air are compressed in the compression stroke, ...... a. each molecule of fuel is heated to its flash point b. each
alexira [117]

None of the choices is an appropriate response.

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When the fuel/air mixture is compressed during the compression stroke,
the temperature is raised to the flash point of the mixture.  The work done
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3 years ago
It's not important to monitor your heart rate during moderate intensity activities. True False
Mama L [17]
I think it is   False 

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3 years ago
Read 2 more answers
A person's prescription for her new bifocal glasses calls for a refractive power of -0.450 diopters in the distance-vision part,
Angelina_Jolie [31]

Answer:

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Explanation:

\frac{1}{f}=-0.045

Object distance = u

Image distance = v

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=-0.045-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.045}\\\Rightarrow v=-22.22\ m

Far point

|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

\frac{1}{f}=1.75

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=1.75-\frac{1}{0.23}\\\Rightarrow v=-0.38\ m

Near point

|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m

Far point of the eye is 0.4 m

7 0
3 years ago
How are electrons represented in orbitals?
Alexxx [7]
Electrons move in atomic orbitals (or subshells). there are four different orbital shapes (s p d f). in each shell, the s subshell is at a lower energy than the p. an orbital diagram is used to determine an atom's electron configurations
8 0
3 years ago
Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,
shutvik [7]

Answer:

p_t_o_t_a_l=250kg\frac{m}{s}

Explanation:

<u>The total momentum of a system is defined by:</u>

(mv)_t_o_t=m_1v_1+m_2v_2+...

Where,

(mv)_t_o_t is the total momentum or it could be expressed also as p_t_o_t_a_l.

m_1 and m_2 represents the masses of the objects interacting in the system.

v_1 and v_2 are the velocities of the objects of the system.

<em>Remember: </em><em>The momentum is a fundamental physical magnitude of vector type.</em>

We have:

m_1=725 kg

v_1=15\frac{m}{s}\\m_2=625 kg

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be <u>negative</u>. It goes in a different direction from car A.

v_2=-17\frac{m}{s}

Then the total momentum of the system is:

p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}

5 0
3 years ago
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