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3 years ago

5

if the forces acting upon an object are balanced, then an object must a) be moving. b) be accelerating. c) be beginning to slow

down. d) be moving at a constant velocity.

1 answer:

Dennis_Churaev [7]3 years ago

4 0

The correct answer to the question is : D) Be moving at a constant velocity.

EXPLANATION:

As per Newton's first laws of motion, every body continues to be at state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces acting on it.

Hence, it is the unbalanced force which changes the state of rest or motion of a body. Balanced force is responsible for keeping the body to be either in static equilibrium or in dynamic equilibrium.

As per the options given in the question, the last one is true for an object under balanced forces.

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A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

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3 years ago

A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and

marin [14]

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

$\lambda = \frac{v}{f}$

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

$\lambda = \frac{192}{320}$

$\lambda = 0.6m$

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

$N = \frac{L}{\lambda}$

$N = \frac{3.6}{0.6}$

$N = 6$

Therefore the number of wavelengths of the wave fit on the string is 6.

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3 years ago

Which statement about subatomic particles is not true?

Tresset [83]

Answer:

B. is the right answer

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A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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Given that,

Mass of object = 5 kg

Speed = 3 m/s

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Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

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Put the value into the fomrula

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$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

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From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

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