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3 years ago

The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3

Physics

1 answer:

Sav [38]3 years ago

6 0

Answer:

$a$ has units of distance

$b$ has units of distance over time

$c$ has units of distance over $time^2$

$d$ has units of distance over $time^3$

Explanation:

Since the expression for the distance is:

$x = a+b\,t+c\,t^2+d\,t^3$

then:

$a$ has units of distance

$b$ has units of distance over time

$c$ has units of distance over $time^2$

$d$ has units of distance over $time^3$

because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.

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Edwin Hubble calculated the expansion rate of the universe. On what evidence did he base his calculation?

algol [13]

Edwin Hubble calculated the expansion rate of the universe. The evidence that he base his calculation is the differences in redshift for galaxies. The answer is letter B. the red shift of galaxies was directly proportional to the distance of the galaxy from earth. It means that bodies farther away from Earth were moving away faster. The Hubble’s constant is the ratio of distance to redshift equal to 170 kilometers per second per light year of distance.

7 0

3 years ago

Read 2 more answers

An object is pulled up an incline plane at a constant velocity as the picture above shows. Calculate the tension on the rope.

WARRIOR [948]

Hi there!

$\large\boxed{545N}$

Since the object is being pulled at a constant velocity, the forces must be balanced.

Since there is no movement vertically, we must take into account the horizontal forces. We can also assume a positive acceleration to be in the direction of motion.

The acceleration and force due to gravity on an incline is:

a = gsinФ

F = MgsinФ

∑F = -MgsinФ + T

Since it is getting pulled at a constant velocity, ∑F = 0. So:

0 = -MgsinФ + T

MgsinФ = T

Solve for T by plugging in values. Let g = 10 m/s²

T = (120)(10)sin(27) ≈ 545 N

6 0

3 years ago

A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes la

zmey [24]

11.54 minutes

Explanation:

The decay rate equation is given by

$N = N_0e^{-\frac{t}{\lambda}}$

where $\lambda$ is the half-life. We can rewrite this as

$\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}$

Taking the natural logarithm of both sides, we get

$\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)$

Solving for $\lambda$ ,

$\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}$

$\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}$

$\:\:\:\:=11.54\:\text{minutes}$

4 0

2 years ago

Helppppo!!!

goldfiish [28.3K]

Answer:

Electric force = $3.15\times 10^{-11}\textrm{ N}$ .

Explanation:

Given:

Charge on one grain, $q_{1}=6\times 10^{-10}\textrm{ C}$

Charge on second grain, $q_{2}=2.3\times 10^{-15}\textrm{ C}$

Distance between the two grains is, $r=2\textrm{cm}=0.02\textrm{ m}$

Electric force acting between two charges $q_{1}\textrm{ and }q_{2}$ separated by a distance $r$ is:

$F_{e}=\frac{kq_{1}q_{2}}{r^2}$

Where, $k$ is Coulomb's constant equal to $9\times 10^9\textrm{ }Nm^2/C^2$ .

Now, plug in the given values and solve for $F{e}$ .

$F{e}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}$

Therefore, the electric force between them is $3.15\times 10^{-11}\textrm{ N}$ .

5 0

3 years ago

A speaker creates uniformly spherical sounds w/ 500 watts of power

sladkih [1.3K]

Answer:

A) I = 0.09947 W , β = 109 db , B) β = 116 db , β = 116 db , c) Δβ = 7 dB,

D) P = 50.27 W

Explanation:

A) The intensity of a spherical sound wave is

I = P / A

where A is the area of the sphere where the sound is distributed

A = 4π R²

we substitute

I = P / 4πR²

let's calculate

I = 500 / (4π 20²)

I = 0.09947 W

to express this quantity in decibels we use relate

β = 10 log (I / I₀)

The detectivity threshold is I₀ = 1 10⁻¹² W / m²

β = 10 lob (0.09947 / 10⁻¹²)

β = 10 (10.9976)

β = 109 db

B) intensity at r = 10m

I = 500 / (4π 10²)

I = 0.3979 W / m²

β = 10 log (0.3979 / 10⁻¹²)

β = 10 (11.5997)

β = 116 db

C) the change in intensity in decibles is

Δβ = β₁ - β₂

Δβ = 116 - 109

Δβ = 7 dB

D) let's find the intensity for 100 db

I = I₀ 10 (β / 10)

I = 10⁻¹² 10 (100/10)

I = 10⁻² W / m²

Thus

P = I A

P = I 4π R²

P = 10⁻² 4π 20²

P = 50.27 W

6 0

3 years ago