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2 years ago

Which of the following is an action-at-a-distance force? friction tension gravity air resistance

Physics

2 answers:

tatuchka [14]2 years ago

8 0

Answer:

Action-at-a-Distance Forces. Frictional Force. Gravitational Force. Tension Force ... The force of gravity on earth is always equal to the weight of the object as ... The friction force is the force exerted by a surface as an object moves across it or ... The force of air resistance is often observed to oppose the motion of an object

Explanation:

PilotLPTM [1.2K]2 years ago

4 0

Answer:

I believe the answer is gravity.

Explanation:

Because with friction you need to be rubbing multiple object together, in tension two objects must be pulling against each other, and in air resistance the air must be touching the object either pushing or pulling it. While in gravity the mass of an object is pulling another object toward it, the objects don't have to be touching each other making it an at-a-distance force.

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Can someone help on this I'm really stuck

castortr0y [4]

Here, "Wavelength is same for both waves" it is the distance between two crests or two consecutive troughs, so, it is constant for both of them, you can easily figure it out.

In short, Your Answer would be "Wavelength"

Hope this helps!

5 0

2 years ago

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

2 years ago

A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th

vivado [14]

Answer:

20 cm to the right of the center or 20+50 = 70 cm from the left side.

Explanation:

The length of meter stick is 1 m = 100 cm

Balance point on 50 cm

From the center the 20 N weight is 50-20 = 30 cm

Torque is obtained when force is multiplied with the distance

As the force is conserved we have

$\dfrac{20}{30}\times 30=20\ cm$

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

7 0

2 years ago

Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.

n200080 [17]

Answer:

The answer to your question is the letter A) F = 9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

F = k q₁q₂ /r²

-Substitution

F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

F = 3.512 x 10⁻⁷ / 0.381

-Result

F = 9.227 x 10⁻⁷ N ≈ 9.23 x 10⁻⁷ N

8 0

3 years ago

A cow’s mass is 410 kg and a car’s mass is 565 kg. What is the difference between their weights?

solmaris [256]

B. 1520 is the difference between their weights.

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3 years ago