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1 year ago

Which one is not a fundamental quantity ?Mass Weight LengthTime

1 answer:

7 0

A fundamental quantity is a quantity that is not composed of other quantities. In the case of the given case, the wight is a quantity that has the units of force. Forces are the product of the following units:

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3 years ago

Convert 1.5 radian into milliradians.

Alex

Answer:

1500 milliradians

Explanation:

Data provided in the question:

1.5 radians

Now,

1 radians consists of 1000 milliradians

1 milli = 1000

thus for the 1.5 radians, we have

1.5 radians = 1.5 multiplied by 1000 milliradians

1.5 radians = 1500 milliradians

Hence, after the conversion

1.5 radians equals to the value 1500 milliradians

4 0

3 years ago

A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static fricti

vfiekz [6]

Answer:

0.255

Explanation:

The following data were obtained from the question:

Force (F) = 57 N

Mass (m) = 22.8 Kg

Coefficient of static friction (µ) =...?

Next, we shall determine the normal reaction (R). This is illustrated below:

Mass (m) = 22.8 Kg

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

Normal reaction (R) =?

R = mg

R = 22.8 x 9.8

R = 223.44 N

Finally, we can obtain the coefficient of static friction (µ) as follow:

Force (F) = 57 N

Normal reaction (R) = 223.44 N

Coefficient of static friction (µ) =...?

F = µR

57 = µ x 223.44

Divide both side by 223.44

µ = 57/223.44

µ = 0.255

Therefore, the coefficient of static friction (µ) is 0.255.

7 0

3 years ago

(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model i

Anni [7]

Answer:

(a) $L=2.6742\times 10^{40}\, kg.m^2.s^{-1}$

(b) $L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}$

Explanation:

<u>For Earth we have:</u>

mass of earth, $m_E=5.97\times 10^{24}\, kg$

radius of earth, $R_E=6.38\times 10^6m$

orbital radius, $r=1.5 \times 10^{11}m$

period of rotation, $t_{rot}=24h=86400\, s$

period of revolution, $t_{rev}= 1 yr=3.156\times 10^7 s$

(a)

Angular momentum, $L=?$

∵ $L=I.\omega$ ...............................(1)

For a particle of mass m moving in a circular path at a distance r from the axis,

$I=m.r^2$

& $v=r.\omega$

Putting respecstive values in eq. (1)

$L=m_E\times r^{2}\times \omega$

$L=5.97\times 10^{24}\times (1.5 \times 10^{11})^2\times \frac{2\pi}{3.156\times 10^7}$

$L=2.6742\times 10^{40}\, kg.m^2.s^{-1}$

To model earth as a particle is reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.

(b)

For a uniform sphere of mass M and radius R and an axis through its center, $I=\frac{2}{5}M.R^2$

using eq. (1)

$L_s=(\frac{2}{5}m_E.R_E^2) \omega$

$L_s=\frac{2}{5} \times 5.97\times 10^{24} \times 6.38\times 10^6\times \frac{2\pi}{86400}$

$L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}$

6 0

3 years ago

a fan is rotating clockwise and its acceleration has a positive sign. is the angular velocity of the fan speeding up, slowing do

balu736 [363]

Answer:

The angular velocity is slowing down.

Explanation:

By convention, if a rigid body is rotating clockwise, the angular velocity is negative.
If the angular acceleration has a positive sign, since the angular acceleration and the angular velocity have opposite signs, this means that the angular velocity is slowing down.

5 0

3 years ago