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KiRa [710]
3 years ago
14

What is the minimum number of points necessary to determine a straight line?

Physics
1 answer:
laiz [17]3 years ago
7 0
2 points are necessary
You might be interested in
How does temperature increase?
vovikov84 [41]
Increasing the temperature increases reaction rates because of the disproportionately increase in the number of high energy collisions.
It's only these collisions (possessing at least the activation energy for the reaction)
which results in a reaction!
6 0
3 years ago
Read 2 more answers
A chef places an open sack of flour on a kitchen scale. The scale reading of
Novay_Z [31]

Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below  

4 0
3 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
3 years ago
Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her h
mojhsa [17]

Answer:

Isabella will not be able to spray Ferdinand.

Explanation:

We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:

Height (h) = 1 m

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

Time (t) =.?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.9

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s

Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:

Horizontal velocity (u) = 3.5 m/s

Time (t) = 0.45 s

Horizontal distance (s) =?

s = ut

s = 3.5 × 0.45

s = 1.58 m

Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:

Ferdinand's position = 10 m

Distance travelled by the water = 1.58 m

From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.

8 0
3 years ago
An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
3 years ago
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