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3 years ago

How long (in ns) does it take light to travel 1.0 m in vacuum?

1 answer:

6 0

If light travels 290,000,000 meters in 1 second, simply divide 1 second by 290,000,000 to find the time it takes to travel 1 meter.

1 / 290,000,000 = 0.000000003448 seconds (<em>or 3.5 nano seconds</em>)

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Consider three scenarios in which a particular box moves downward under the pull of gravity :

luda_lava [24]

Answer:

1. True WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

WA = W h cos 0

WA = mg h

B) On a ramp without rubbing

Sin30 = h / L

L = h / sin 30

WB = F d cos θ

WB = F L cos 30

WB = mf (h / sin30) cos 30

WB = mg h ctan 30

C) Ramp with rubbing

W sin 30 - fr = ma

N- Wcos30 = 0

W sin 30 - μ W cos 30 = ma

F = W (sin30 - μ cos30)

WC = mg (sin30 - μ cos30) h / sin30

Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0

3 years ago

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

$s_1 = v_1t = 1.6t$

Similarly the distance (m) from your friend building to the restaurant:

$s_2 = v_2t = 1.2t$

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

$s_1 + s_2 = 5b$

$1.6t + 1.2t = 5b$

$2.8t = 5b$

$t = 5b/2.8 = 25b/14$

$s_2 = 1.2t = 1.2(25b/14) = 2.143b$

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

7 0

3 years ago

How does the electrostatic force compare with the strong nuclear force in the

diamong [38]

Answer:

Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force

Explanation:

There are mainly two forces acting between protons and neutrons in the nucleus:

- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by

$F_E=\frac{kq_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.

The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.

- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.

At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.

3 0

3 years ago

Read 2 more answers

Which structure contains the lowest amount of oxygen?

Alchen [17]

The answer is vena cava
Hope this helps !!

3 0

3 years ago

Read 2 more answers

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago