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3 years ago

What is the quantity of motion ??

Physics

1 answer:

horsena [70]3 years ago

3 0

Answer:

The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly. In other words, rather than defining the quantity of motion of a given object as simply the kinematic velocity v of the object, he defined it as the product mv, where m is the mass of the object.

Explanation:

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Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom

Irina-Kira [14]

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i - .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

7 0

3 years ago

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0

3 years ago

Read 2 more answers

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

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3 years ago

What region of the spectrum best corresponds to light with a wavelength equal to:_____ a. The diameter of a hydrogen atomb. The

zavuch27 [327]

We will make the comparison between each of the sizes against the known wavelengths.

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In the case of height, it fluctuates in a person around $10 ^ 0$ to $10 ^ 1$ m, which falls to the wavelength of a radio wave.