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nignag [31]
4 years ago
13

Which pair does NOT have an electric force between them?

Physics
2 answers:
Pepsi [2]4 years ago
7 0

Answer:

two neutral objects

Explanation:

According to the Coulomb's law in electrostatics, there is a force of attraction or repulsion between two charges. The force is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

As the two objects having charge they exerts a force on each other.

When tow opposite charges are there, the force of attraction is there and when two same charges are there the force of repulsion is there.

When the objects are neutral, then the force between them is zero.

fenix001 [56]4 years ago
5 0

Answer:

Two Neutral

Explanation:

If; One postive, and one negative they would be attracked. >-<

If; one postive and the other was postive, (or vice versa) they would repel <->

they both have an electrical charge, and force.

Neutral's are Nueatral. they have to opisition or attraction= No electrical force.

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Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh
GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=N_1-mg=N_1-690

Now, from Newton's second law:

F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=mg-N_2=690-N_2

Now, from Newton's second law:

F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

Difference=N_1-N_2=862.5-517.5=345\ N

Therefore, the difference between the up and down scale readings is 345 N.

4 0
3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
olga nikolaevna [1]

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

Angle=3^{\circ}

as the arrow travels it acquire a vertical velocity v_y

v_y=u+at

v_y=0+9.81\times t-------1

Range is given by

R=ut

where u=initial velocity

68=u\times t

t=\frac{68}{u}

substitute the value of t in eqn 1

v_y=9.81\times \frac{68}{u}

v_y\times u=9.81\times 68=667.08--------2

and tan(3)=\frac{v_y}{u}

v_y=utan(3)=0.0524u

substitute it in 2

0.0524 u^2=667.08

u^2=12,728.644

u=112.82 m/s

6 0
4 years ago
What is the meaning of physic​
irina [24]

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

5 0
3 years ago
Someone please help will mark as brainliest
forsale [732]
Getting it right because you do not want to spread false information. To be first isnt always the best, sure you might feel better being the first, but you won’t always get it right.
5 0
3 years ago
A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
3 years ago
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