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3 years ago

Despite the use of diffrent types of measurement in our country

Physics

1 answer:

stealth61 [152]3 years ago

6 0

Answer: For standardization purposes

Explanation:

As the question mentions, there were different types of measurements in the country for the same things so measuring a thing in one type of measurement and then comparing it to another was quite cumbersome.

It also made comparing units with the rest of the world difficult. To combat this the SI system was introduced. It gave an international standard for measuring different things based on the metric system. This allowed for worldwide standardization and comparison as things could now be measured in the same units.

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After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

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3 years ago

Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

Vera_Pavlovna [14]

Answer:

A, B, C, E

Explanation:

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3 years ago

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a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

$x_1 = 50 cm$ represents an actual distance of

$d_1 = 1.0\cdot 10^5 km$

The actual distance between Mars and the Sun is 228 million km, therefore

$d_2=228\cdot 10^6 km$

On the scale model, this would corresponds to a distance of $x_2$ .

Therefore, we can write the following proportion:

$\frac{x_1}{d_1}=\frac{x_2}{d_2}$

And solving for $x_2$ , we find:

$x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m$

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3 years ago

An object is most likely to sink in water if

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Answer:

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2 years ago

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Despite the use of diffrent types of measurement in our country​

Despite the use of diffrent types of measurement in our country