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3 years ago

What are two examples of common units for each of the above measurements

Physics

1 answer:

kap26 [50]3 years ago

3 0

Density: g/mL, kg/cubic meter

Volume: L, teaspoon

Mass: g, MeV/sq. C

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Two ice skaters approach each other at right angles. Skater A has a mass of 32.7 kg and travels in the x direction at 2.3 m/s. S

UkoKoshka [18]

Answer:

Final speed, v = 1.25 m/s

Explanation:

Given that,

Mass of skater A, $m_A=32.7\ kg$

Initial speed of skater A, $u_A=2.3i\ m/s$ (x axis)

Mass of skater B, $m_B=93.3\ kg$

Initial speed of skater B, $u_B=1.51j\ m/s$ (y axis)

It is mentioned that the two skaters collide and cling together. It is case of inelastic collision in which momentum remains conserved. Let V is the final speed of the couple. It is given by :

$m_Au_A+m_Bu_B=(m_A+m_B)V$

$32.7\times 2.3i+93.3\times 1.51j=(32.7+93.3)V$

$V=\dfrac{75.21i+140.88j}{126}$

$V=(0.59i+1.11j)\ m/s$

$|V|=1.25\ m/s$

So, the final speed of the couple is 1.25 m/s. Hence, this is the required solution.

3 0

3 years ago

As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, b

katovenus [111]

The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

<h3>What do you mean by sound radiates?</h3>

Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.

(Lw) = 10·log (W/Wo) dB

Given:

sound level, $\beta= 101 dB$

Area, A = $22\;m^{2}$

Time, $\triangle t = 20\;min=1200\;s$

Intensity, $I=1\times 10^{-12}\;W/m^{2}$

$r=1\;km=1000\;m$

(a)

We know that, Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Solving the above equation for sound intensity,

$I=I_{o} \times 10^{\frac{\beta}{10} }$

$I=1 \times 10^{-12} \times 10^{\frac{101}{10} }$

$I=0.0126\;W/m^{2}$

Therefore, The sound energy is,

$E=P\times \triangle t$

Substitute $P=I \times A$ in the above equation,

$E=I \times A \times \triangle t$

$E=0.0126 \times 22 \times 1200$

$E=332.6\;J$

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

$A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi$

Substitute the known value in the above equation ,

$A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi$

$A_{hemisphere} = 6283185\;m^{2}$

Sound Intensity is,

$I = \frac{P}{A_{hemisphere}}$

Substitute $P=I \times A$ in the above equation,

$I = \frac{I \times A}{A_{hemisphere}}$

Substitute the known value in the above equation,

$I = \frac{0.0126 \times 22}{6283185}$

$I = 4.4 \times 10^{-8}\;W/m^{2}$

Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Substitute the known value in the above equation,

$\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )$

$\beta=46.4\;dB$

To learn more about sound radiates, Visit:

brainly.com/question/20360072

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8 0

2 years ago

An object of mass 2kg is on an incline where there is an applied force of 15N. The

seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is $2.46 m/s^2$

Explanation:

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

The weight of the object, labelled as $mg$ (where m is the mass of the object and g is the acceleration of gravity), acting downward
The applied force, $F_a$ , acting up along the plane
The force of friction, $F_f$ , acting down along the plane

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

$F_a - F_f - mg sin \theta = ma$