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iVinArrow [24]
2 years ago
9

The equation for the change in the position of a train (measured in units of length) is given by the following expression: x = ½

at² + bt³ where a and b are constants and t is time (measured in units of time). If (mass), (length), and (time) represent units of mass, length and time respectively then b must have units of:
(A) (length)/(time)
(B) (length)/(time³)
(C) (mass)(length)
(D) (length)/(mass)
(E) (mass2)(time2)
(F) 1/(mass)(time)
(G) 1/(length)(time)
(H) (time)(mass)
(I) (mass)(time2)
(J) 1/(time³)
(K) (time)/(mass)
(L) (length)/(time²)

(mass)(length2)
Physics
2 answers:
mel-nik [20]2 years ago
6 0

Answer:

(B) (length)/(time³)

Explanation

The equation x = ½ at² + bt³ has to be dimensionally correct. In other words the term bt³ and ½ at² must have units of change of position = length.

We solve in order to find the dimension of b:

[x]=[b]*[t]³

length=[b]*time³

[b]=length/time³

Arturiano [62]2 years ago
3 0

Answer:

Explanation:

The position of a train is given as a function of time

x(t) = ½at² + bt³

Where a and b are constant

We want to know the dimensional unit of b.

x is length and has a unit of metre

t is time and has a unit of seconds

b is a constant with unknown unit

a is a constant with unknown unit

Given that,

x = ½at² + bt³

x(metre) = ½ a t²(seconds)² + bt³(seconds)³

For x to be dimensionally correct, a and b must have a unit that cancel out the time

So, a•t², so cancel out the time square "a" must have a unit that cancel out the time square. Also a must have "a" unit of distance too. So, "a" will have m/s² unit

Then, at² will give

a (metre/seconds)² × t² (seconds)²

Also, for "b", same procedure but b wants to cancel out seconds cube, so instead of seconds square division, " b" will have secondds cube division I.e m / s³

Then, bt³ will give

b(metre/seconds³) × time (seconds)³

So, the unit of b is metre/seconds³

Since Length = metre

And Time = seconds

Which is Length/ Time³

So, the correct answer is B.

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Masja [62]

Answer:

1.10261 times g

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g

The acceleration is 1.10261 times g

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s

In mph

186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph

The speed of the dragster is 416.17506 mph

5 0
2 years ago
The formation of magma within Earth is NOT caused by which of the following processes? A. decompression (drop in pressure) B. ad
Lunna [17]

Answer:

D. loss of volatiles to the atmosphere

Explanation:

The name magma designates matter in a semi-fluid state - resulting from the fusion of silicates containing dispersed solid gases and minerals and other compounds that make up the rocks, at temperatures between 700 and 1200 ° C - that forms the region beneath the crust. land. When it is inside the earth it is specifically named magma and lava when it is ejected to the surface

There are three systems by which magma can be produced on earth:

<u> Temperature</u> rise by concentration of r<u>adioactive elements or by friction of lithospheric plates</u>.

<u> Pressure decrease,</u> since the melting point decreases.

Adding <u>water</u> A rock begins to melt earlier if it contains water because the –OH groups effectively break the Si-O bonds.

A rock is formed by a set of minerals, each of which has a characteristic melting point so a rock does not have a single melting point but a temperature range in which the rock melts into parts, leaving others solid parts. Between the point at which a solid rock begins to melt and the melting end (liquid point) the rock is partially molten.

The rise of magmas depends on their physical-chemical conditions (viscosity, density, volatile element content), on the tectonic peculiarities of the region where they are found and on the rocks to be traversed. Acid magmas are light and viscous, rise easily and cause large deposits. The basic magmas, of greater density, are less viscous and ascend with greater difficulty than the previous ones.

8 0
2 years ago
A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

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Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
2 years ago
A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled
VikaD [51]

Answer:

option B.

Explanation:

The correct answer is option B.

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When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

3 0
2 years ago
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likoan [24]

Answer:

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Explanation:

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a\sin \theta_n=n\lambda

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for the first minimum

a\sin \theta_1=(1)\lambda

solving for θ1:

\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})

\theta_1=0.63 degrees

for the second minimum:

a\sin \theta_2=(2)\lambda

\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})

\theta_2=1.26 degrees

So, the angular separation between them is the rest:

\Delta \theta =1.26-0.63

\Delta \theta=0.63

4 0
2 years ago
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