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1 year ago

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In the following equations, the distance x is in meters, the time tin seconds. and the velocity v is in meters per second. whata

re the SI units of the constants c1 andc2?
(A) x = c1 +c2t
(B) x = .5c1t^2
(C) v^2 = 2c1x
(D) x = c1cosc2t
(E) v^2 =2c1v-(c2x)^2

2 answers:

Sav [38]1 year ago

4 0

Explanation:

Given that,

The distance x is in meters.

The time t is in seconds.

The velocity v is in meter/ second.

We need to calculate the SI units the constants c₁ and c₂

(A). $x =c_{1}+c_{2}t$

Put the unit in to the equation

$m= m+m/s\times s$

Here, $c_{1}=m$

$c_{2}=m/s$

So,

$m = m$

(B). $x=0.5c_{1}t^2$

Put the unit in to the equation

$m=0.5\times m/s^2\times s^2$

Here, $c_{1}=m/s^2$

So, $m=0.5 m$

(C). $v^2=2c_{1}x$

Put the unit in to the equation

$m^2/s^2=2\times m/s^2\times m$

Here, $c_{1}=m/s^2$

So, $m^2/s^2=2m^2/s^2$

(D). $x=c_{1}\cos c_{2}t$

Put the unit in to the equation

$m=m\times\cos(\dfrac{1}{s}\times s)$

Here, $c_{1} =m$

$c_{2}=\dfrac{1}{s}$

$\cos(c_{2}t)$ =dimension less

So, $m=m$

(E). $v^2=2c_{1}v-(c_{2}x)^2$

Put the unit in to the equation

$m^2/s^2=2\times m/s\times m/s-(\dfrac{1}{s^2}\times m^2)$

Here, $c_{1}=m/s$

$c_{2}=\dfrac{1}{s}$

So, $m^2/s^2=m^2/s^2$

Hence, This is the required solution.

lyudmila [28]1 year ago

3 0

Answer:

A) c₁ = m, c₂ = m/s

B) c₁ = m/s²

C) c₁ = m/s²

D) c₁ = m/s c₂ = °

E) c₁ = m/s , c₂ = /s

Explanation:

A) x = c₁ + c₂t

⇒m = m + (m/s)s (Only same units can be added)

⇒m = m

So, c₁ = m, c₂ = m/s

B) x = 0.5c₁t²

⇒m = 0.5 (m/s²)s²

⇒m = m

So, c₁ = m/s²

C) v² = 2c₁x

⇒m²/s² = 2 (m/s²)m

⇒m²/s² = m²/s²

So, c₁ = m/s²

D) x = c₁ cos(c₂)t

⇒m = (m/s) cos(°)s

⇒m = m

So, c₁ = m/s c₂ = °

E) v² = 2c₁v-(c₂x)²

⇒m²/s² = 2(m/s)(m/s)-(1/s²)(m²)

⇒m²/s² =m²/s²

So, c₁ = m/s , c₂ = /s

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