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exis [7]
3 years ago
12

How are the three lines of defense the same

Physics
2 answers:
DedPeter [7]3 years ago
7 0

Answer:They all involve fighting pathogens.

Explanation:

xz_007 [3.2K]3 years ago
5 0
Here is an example of how the three lines of defense are the same:

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A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f
Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

W_f = F_fs = 15*0.5 = 7.5 J

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

E_p = mgh = 0.1*10*1.5 = 1.5 J

Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

6 0
3 years ago
An arrow leaves a bow at 60 m/s at an angle of 30 degrees to the horizon.
Orlov [11]

Answer: 11

Explanation:

4 0
3 years ago
1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant
r-ruslan [8.4K]

Answer:

a=0   v = v₀ + a t

a=0    line is horizontal

Explanation:

1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration

2, speed and relationship of a car is given by

        v = v₀ + a t

where vo is the initial velocity, a is the acceleration and tel time

in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope

6 0
3 years ago
QUESTION:
vampirchik [111]

Answer:

I don't know why you are asking me?

5 0
3 years ago
A child is sitting on the seat of a swing with ropes 5 m long. Her father pulls the swing back until the ropes make a 30o angle
morpeh [17]

Answer:

v = 3.7 m/s

Explanation:

As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:

m. g. h = 1/2 m v²

The only unknown (let alone the speed) in the equation , is the height from which the swing is released.

At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.

As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:

h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m

Replacing in the energy conservation equation, and solving for v, we get:

v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s

7 0
3 years ago
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