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Archy [21]
3 years ago
6

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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4 0
3 years ago
The electrons that produce the picture in a
vredina [299]
Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled

Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²

Answer: 1.95 x 10¹⁴ m/s²

4 0
3 years ago
What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t
yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity =  ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

Required Information:

time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R =  ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R =  (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

6 0
3 years ago
Latent heat of fusion refers to which changes of state?
raketka [301]

The latent heat of fusion refers to the solid to liquid or liquid to solid states.

Answer: Option C

<u>Explanation: </u>

It is known that the inter conversion process from the states of solid to liquid is referred as fusion. So, for these conversions, the external energy in the heat form should be supplied to solid.

This external energy should be greater than the latent heat of solid in order to successfully break the bonds to form liquid. So the change in the enthalpy of the reaction while conversion from solids to liquids are termed as latent heats of fusion.

Even the inter-conversion from liquid to solid state will undergo change in enthalpy where the heat will be released and that is termed as latent heats of solidification. It is found that latent heat of solidification is equal in magnitude but opposite in direction with the latent heats of fusion.

7 0
3 years ago
A source for conducting formal research would include_
Wittaler [7]
C. written reports !
5 0
3 years ago
Read 2 more answers
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