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inna [77]
4 years ago
10

A measurement standard is defined as ____.

Physics
2 answers:
Serggg [28]4 years ago
5 0

<u>Answer:</u>

A measurement standard is defined as 'the exact quantity people agree to use for comparison'.

<u>Explanation:</u>

For the same type of quantity, there is a unit of measurement which is defined and adopted either conventionally or by law which is used as standard for measurement of that quantity.

This measurement standard is defined as 'the exact quantity people agree to use for comparison'.

For example, for length the standard unit of measurement is 'meter'.

Blizzard [7]4 years ago
5 0

Answer:

is an object system or experiment that bears a defined relationship to a unit of measurement of a physical quality

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A 3.00 × 10^−9-coulomb test charge is placed near
viktelen [127]

Answer:

B

Explanation:

Given:-

- The charge of the test particle q = 3.0 * 10^-9 C

- The force exerted by the metal sphere F = 6.0 * 10^-5 N

Find:-

The magnitude and direction of the electric field strength at this location?

Solution:-

- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:  

                                       F = E*q

- Using the data given we can determine E:

                                       E = F / q

                                       E = (6.0 * 10^-5) / (3.0 * 10^-9)

                                       E = 20,000 N/C

- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.

8 0
3 years ago
Read 2 more answers
A small water droplet in a mist of air is approximated as being a sphere of diameter 1.25 mil. Calculate the terminal velocity a
shtirl [24]

The terminal velocity as it falls through still air is 4.65154 in/s.

The diameter of small water droplet is 1.25 mil= 1.25×0.0254×10^-3 m

                                                                            = 3.175 × 10^-5 m

Now the viscosity of still air is η = 1.83× 10⁻⁵ Pa

 So the formula for drag force is:

                Fd = 6πηrv

  where,  v is the velocity.

Now to attain terminal velocity acceleration must be zero.

            →  W = Fd

                ρVg = 6πrηv

                ρ × 4/3 πr³×g = 6πrηv

                           v = 2/9 × ρgr³/ η

                           v =  2/9 × 10³×9.81×(3.175×10^-3) / 18.6×10^-6

                           v = 0.1181  m/s

                           v = 4.65154 in/s

Learn more about terminal velocity here:

     brainly.com/question/20409472

              #SPJ4

                   

8 0
2 years ago
A wild pig attacks a hunter with a velocity of 5 m/s. At the instant when the pig is at a distance of 100 m the hunter shoots an
castortr0y [4]

Answer:

31.55 m/s

Explanation:

Let the initial velocity of the arrow is u metre per second.

Angle of projection, θ = 40 degree

range = 100 m

Use the formula for the range.

R = \frac{u^{2}Sin2\theta }{g}

100 = u^2 Sin(2 x 40) / 9.8

100 x 9.8 = u^2 Sin 80

u^2 = 995.11

u = 31.55 m/s

4 0
3 years ago
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
,how do charged objects react???quiet urgent
son4ous [18]
Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>
8 0
3 years ago
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