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3 years ago

This force is involves the attraction between charged particles

Physics

2 answers:

Gelneren [198K]3 years ago

8 0

Answer:

The correct answer is Strong Nuclear

Explanation:

In nuclear physics and particle physics, the strong interaction is the mechanism responsible for the strong nuclear force, and is one of the four known fundamental interactions, with the others being electromagnetism, the weak interaction, and gravitation.

Nikolay [14]3 years ago

6 0

That's the 'electrostatic' force.

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Which of the following defines heredity?

lina2011 [118]

The answer to this question is A.

4 0

3 years ago

How does the change in the sa/v ratio compare with the change in distance from the center of the cell to the nearest face?

Ivanshal [37]

<span>Answer: I'm pretty sure the SA / V ratio would get smaller. Assume that the cell is more or less spherical. SA = 4(pi)r^2, while V = (3/4)(pi)r^3. The ratio = (4(pi)r^2)/((3/4)(pi)r^3), which can be simplified to 3/r. Thus, the larger r gets, the smaller the ratio becomes.</span>

3 0

3 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th

Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0

3 years ago

A woman on a bridge 82.2 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an a

solong [7]

Answer:

0.71 m/s

Explanation:

We find the time it takes the stone to hit the water.

Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.

So, substituting the values of the variables into the equation, we have

y = ut - 1/2gt²

82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²

82.2 m = 0 + (4.9 m/s²)t²

82.2 m = (4.9 m/s²)t²

t² = 82.2 m/4.9 m/s²

t² = 16.78 s²

t = √16.78 s²

t = 4.1 s

This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.

Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s

5 0

3 years ago

An embolus can cause which of the following?

QveST [7]

I think it’s C. Stroke if not then D

7 0

3 years ago