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tangare [24]
2 years ago
10

In the equation y =cnat2 you wish to determine theinteger value (1,2 etc.) of the exponent n. the dimensions ofy, a, and t are k

nown. it is also known that chas no dimensions. can dimensional analysis be used to determinen?
Physics
1 answer:
kvasek [131]2 years ago
5 0

<u>NO.</u><u>  Since </u><u>"c"</u><u> has no dimensions, the value of its exponent </u><u>"n" </u><u>is irrelevant to dimensional analysis, and thus </u><u>dimension analysis </u><u>cannot be used to determine the value of </u><u>"n".</u>

Dimensional analysis is not applicable to proportionality constants, trigonometrical or exponential equations, equation consisting of sum or difference of some quantities and equation consisting of more than three fundamental quantities.

What is Dimensional analysis?

  • Dimensional Analysis is a problem-solving technique that makes use of the fact that any number or expression can be multiplied by one without changing its value.
  • It is also known as the Factor-Label Method or the Unit Factor Method. It is a practical method.

What is a dimensional analysis used for?

Dimensional analysis is a method used in engineering and the physical sciences to reduce physical properties like acceleration, viscosity, and energy to their three basic dimensions of length, mass, and time (T).

Learn more about Dimensional analysis

brainly.com/question/13078117

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Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
2 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
What is an ionic bond?
Airida [17]

A.) a bond that forms when electrons are transferred from one atom to another

4 0
1 year ago
Read 2 more answers
When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b
Nesterboy [21]

Answer:

Explanation:

Given

Displacement is \frac{1}{3} of Amplitude

i.e. x=\frac{A}{3} , where A is maximum amplitude

Potential Energy is given by

U=\frac{1}{2}kx^2

U=\frac{1}{2}k(\frac{A}{3})^2

U=\frac{1}{18}kA^2

Total Energy of SHM is given by

T.E.=\frac{1}{2}kA^2

Total Energy=kinetic Energy+Potential Energy

K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2

K.E.=\frac{8}{18}kA^2

Potential Energy is \frac{1}{8} th of Total Energy

Kinetic Energy is \frac{8}{9} of Total Energy

(c)Kinetic Energy is 0.5\times \frac{1}{2}kA^2

P.E.=\frac{1}{4}kA^2

\frac{1}{2}kx^2=\frac{1}{4}kA^2

x=\frac{A}{\sqrt{2}}                  

7 0
3 years ago
Read 2 more answers
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 41m in front of you. Your reaction time
nordsb [41]

Answer:

Maximum speed you could have and still not hit the deer = 24.07 m/s

Explanation:

Let the maximum speed you could have and still not hit the deer be y.

Your reaction time before stepping on the brakes is 0.50s.

Distance traveled during this time = 0.5y

A deer steps onto the road 41m in front of you

Remaining distance to deer = 41 - 0.5y

The maximum deceleration of your car is 10 m/s²

We have equation of motion, v² = u² + 2as

       Initial velocity, u = y m/s

       Final velocity, v = 0 m/s

       Acceleration, a = -10 m/s²

       Displacement, s = 41 - 0.5y

Substituting,

       v² = u² + 2as

        0² = y² + 2 x -10 x (41 - 0.5y)

          20(41 - 0.5y) = y²

           820 - 10 y = y²

            y² + 10 y -820 = 0

             y = 24.07 m/s or y = -34.06 m/s(not possible)

Maximum speed you could have and still not hit the deer = 24.07 m/s

           

7 0
3 years ago
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