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Monica [59]
4 years ago
5

If Susan exerted a force of 5 newtons and the output force of the machine is 15

Physics
1 answer:
Anon25 [30]4 years ago
4 0

Answer:

Mechanical advantage = 3

Explanation:

Mechanical advantage, MA = Output force/Input force

Output force = 15 N. Input force = 5 N

MA = Output force/Input force = 15 N/ 5 N = 3

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As you move toward a warning siren, the pitch _____.
sleet_krkn [62]

Answer:

As you move toward the siren the pitch should get louder

Explanation:

4 0
3 years ago
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The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy store
velikii [3]

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}            (you has an mistake in the formula)

         \frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \  10^{6}

         \frac{1}{C_{eq1}} = 0.1   10⁶

         C_{eq1} = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          \frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}

          \frac{1}{C_{eq2} }  = (\frac{1}{15} + \frac{1}{30} +   \frac{1}{10} ) \ 10^6

          \frac{1}{C_{eq2} } = 0.2 10⁶

          C_{eq2} = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

           

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = \frac{Q^2}{2 C_3}

          U₃ =\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = \frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}

          U₄ = 0.5 J

7 0
3 years ago
Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons
gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

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3 years ago
An element's atomic number is 34. How many protons would an atom of this element have?
Elena-2011 [213]

34. The element would have 34 protons. Also 34 electrons.

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What is the nineth plant closet to the sun
marusya05 [52]
My Very Excellent Mother Just Served Us Nachos 
(or Nine pickles if Pluto was a planet)

Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
(Pluto)

hope that helped
4 0
3 years ago
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