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kramer
4 years ago
5

A marble accelerates from rest at a constant rate .it travels 36 m in 12.0 sec what’s is it’s final velocity and what was the ac

celeration

Physics
1 answer:
ddd [48]4 years ago
4 0

Answer:

a = 0.50 m/s²

v = 6.0 m/s

Explanation:

Step 1: Given data

  • Initial velocity (u): 0 m/s (rest)
  • Displacement of the marble (s): 36 m
  • Time elapsed (t): 12.0 s
  • Final velocity (v): ?
  • Acceleration (a): ?

Step 2: Calculate the acceleration

We will use the following suvat equation.

s = u × t + 1/2 × a × t²

s = 0 × 12.0 s + 1/2 × a × (12.0 s)²

36 m = a × 72.0 s²

a = 0.50 m/s²

Step 3: Calculate the final velocity

We will use the following suvat equation.

v = u + a × t

v = 0 + 0.50 m/s² × 12.0 s

v = 6.0 m/s

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

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A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. I
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Water level remains unchanged

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Therefore, the answer is B, the water level remains unchanged.

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8. An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object? _______________
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Answer:

Mass of object is 0.5kg

Explanation:

Given the following data;

Force = 6N

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Mass =?

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making mass (m) the subject, we have;

Mass (m) = \frac{F}{a}

Substituting into the equation;

Mass (m) = \frac{6}{12}

Mass, m = 0.5kg.

Therefore, the mass of the object is 0.5kg

5 0
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Answer: µ = ρ¹ * A¹

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µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

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By using law of indices on the right hand side of the equation, we have that

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Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0
3 years ago
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