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3 years ago

What are the coefficients in this expression?

Physics

2 answers:

skad [1K]3 years ago

7 0

6 and $\frac{1}{3}$

Explanation:

The expression:

6a - $\frac{3}{4} + 7.5 - \frac{b}{3}$

A coefficient is a numerical constant usually placed before a variable in an expression.

They are used to multiply variables.

Given equation is;

$6a - \frac{3}{4} + 7.5 - \frac{b}{3}$

There are two variables in this expression which are;

a and - b

$6 (a) - \frac{3}{4} + 7.5 - b (\frac{1}{3} )$

The coefficients in this expression are:

6 and $\frac{1}{3}$

learn more:

Quadratic equation brainly.com/question/1357167

#learnwithBrainly

bekas [8.4K]3 years ago

5 0

Answer:

6 and Negative one-third (D)

Explanation:

eDgE

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Cerrena [4.2K]

Answer:

the vertical acceleration of the case is 1.46 m/s

Explanation:

Given;

mass of the clarinet case, m = 3.07 kg

upward force applied by the man, F = 25.60 N

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the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration

F = mg + m(-a)

the acceleration is negative due to downward motion from the top of the piano.

F = mg - ma

ma = mg - F

$a = \frac{mg - F}{m} \\\\a = \frac{(3.07 \times 9.8) \ - \ 25.6}{3.07} \\\\a = 1.46 \ m/s^2$

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2 years ago

True or false? All waves need a medium in order to travel

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2 years ago

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Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving

saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

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$v_2^2=v_1^2-2gh$

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$v_2=6.26\ m/s$

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

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3 years ago

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Answer:

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Explanation:

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2 years ago

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A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

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On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

2 years ago