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inna [77]
3 years ago
10

A measurement standard is defined as ____.

Physics
2 answers:
Serggg [28]3 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]3 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 ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p
Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

v_x = v cos45

in y direction

v_y = vsin45

now displacement in x direction

x = (vcos45)t + 0

displacement in y direction

y = (vsin45)t - \frac{1}{2}gt^2

now from above two equations we have

y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2

y = xtan45 - \frac{1}{2v^2cos^245}gx^2

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

<em>C. parabolic curve.</em>

8 0
3 years ago
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
3 years ago
PLEASE HELP ME
hammer [34]

Answer:

Element Is The Answer I think

6 0
3 years ago
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Which of the following can be an effect of a volcanic winter? select the two correct answers
Marysya12 [62]

it is B. Famine and C. Climate cooling

8 0
3 years ago
Read 2 more answers
Help me with the following problem
timofeeve [1]

The rope will remain taut until the particle makes 79⁰ angle.

<h3>Change in kinetic energy of the particle</h3>

The change in kinetic energy of the particle is calculated as follows;

ΔK.E = K.Ei - K.Ef

Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%

ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E

  • let the vertical displacement of the particle = h
  • horizontal length = side of the prism = a
  • hypotenuse side  = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>

K.E = P.E

0.7K.E = mgh

0.7(¹/₂mv²) = mg(Lsinθ)

0.7(v²) = 2g(Lsinθ)

from third kinematic equation;

v² = u² + 2gh

v² = 0 + 2gh

v² = 2g(a tanθ)

0.7(2g(a tanθ)) = 2g(Lsinθ)

0.7(a tanθ) = Lsinθ

0.7a/L = sinθ/tanθ

0.7a/L = cosθ

(0.7 x 0.8)/(3) = cosθ

0.1867 = cosθ

θ = cos⁻¹(0.1867)

θ = 79⁰

Thus, the rope will remain taut until the particle makes 79⁰ angle.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
2 years ago
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