A physical equation is given by v = a[4 - e-Bxt] where v is speed, x is displacement and t is time. The dimensions of B2 will be O [MLT-2] O [M-LT-21 O [M°L-2T-2] [M-'L-2T']
Answers
β² = [ M⁰ L⁻² T⁻² ]
GIVEN: A physical equation is given by v = a[4 - e-Bxt] where v is speed, x is displacement and t is time.
TO FIND: The dimensions of β²
SOLUTION:
As we are given in the question,
V = [ 4 - e⁻βxt ]
Where,
V is speed
X is displacement
T is time.
Since
The dimensions of power of e must be one always.
βxt = 1
Implying that,
β = 1/[LT]
β = [ LT ]⁻¹
β = [ M⁰ L⁻¹ T⁻¹ ]
Squaring both sides.
β² = [ M⁰ L⁻² T⁻² ]
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Answer:
β² = [M⁰ L⁻² T⁻²]
Explanation:
GIVEN: A physical equation can be written as v = a[4 - e-Bxt], where v denotes speed, x denotes displacement, and t is time.
THE SIZE OF 2 IS TO BE FOUND
SOLUTION:
the information provided in the query,
V = [4 - e⁻βxt]
Where,
V is speed.
X is the displacement.
Time is now.
Since
A constant unity in the dimensions of e's power is required.
βxt = 1
that's implied,
β = 1/[LT]
β = [LT]⁻¹
β= [M⁰ L⁻¹ T⁻¹]
both sides are balanced.
β² = [M⁰ L⁻² T⁻²]
Additional information:
The dimension of speed calculated:
Distance is expressed as [M0L1T0] in dimensions.
[M0L0T1] is the dimension formula for time.
= [M0L1T-1]
Speed can be expressed as [M0L1T-1] in dimensions.
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