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DaniilM [7]
3 years ago
8

How does distance and direction affect work?

Physics
2 answers:
loris [4]3 years ago
6 0
The greater the distance over which a force is applied, the greater the work done.
Sever21 [200]3 years ago
3 0
The greater the distance<span> over which a force is applied, the greater the </span>work<span> done.</span>
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An object's position in a given coordinate system is described by the vector r = t2 i - (3t + 3) j. Assume the object moves with
Ipatiy [6.2K]

Answer:

The object´s displacement vector is Δr =  8i - 6 j  

Explanation:

Hi there!

The position vector is given by the following function:

r = t²i - (3t + 3) j

Let´s find the position of the object at time t1 and t2:

At t1 = 1 s:

r1 = (1)² i - (3 · (1) + 3 )j

r1 = 1 i - 6 j

At t2 = 3 s:

r2 = (3)² i - (3 · (3) + 3) j

r2 = 9 i - 12 j

The displacement is calculated as follows:

displacement = Δr = final position - initial position = r2 - r1

r2 - r1 = 9 i - 12 j - (1 i - 6 j)

r2 - r1 = 9 i - 12 j - 1 i + 6 j

r2 - r1 = 8 i - 6 j

The object´s displacement vector is Δr =  8i - 6 j  

8 0
3 years ago
Angling of the focal track of the anode to create a large actual focal spot and a smaller effective focal spot describes (the):
Vsevolod [243]

Angling of the focal track of the anode to create a large actual focal spot and a smaller effective focal spot describes the line focus principle

<h3>What is the line focus principle?</h3>

The line focus principle states that as the anode angle is reduced, the actual focal spot also becomes small but the heat loading is increased.

It also explains the relationship between the anode surface and the effective focal spot size.

As a result of this, by angling the target, effective area of the target is made much smaller that the actual area of electron interaction.

Hence, angling of the focal track of the anode to create a large actual focal spot and a smaller effective focal spot describes the line focus principle.

Learn more about line focus principle here:

#SPJ1

3 0
2 years ago
A box with a mass of 18 kg is pushed across the floor. It has coefficient of friction of 0.39. Calculate the force of friction i
Taya2010 [7]

Answer:

68.8 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

3 0
3 years ago
Investigators should collect virtually everything from a crime scene. True or false?
Stels [109]
I think it would be TRUE but I’m not 100% sure.... please let me know if I’m right or wrong
4 0
3 years ago
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Make the following conversion. 0.00432 km = _____ mm
e-lub [12.9K]
0.00432 km = 4320 mm
5 0
4 years ago
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