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jekas [21]
3 years ago
11

For all simple machines, when the output force is greater than the input force,

Physics
1 answer:
KengaRu [80]3 years ago
7 0
For all simple machines, when the output force is greater than the input force, <span>the input force is exerted over a larger distance than the output force.

In short, Your Answer would be Option D

Hope this helps!</span>
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An exothermic reaction is taking place in a test tube what would you expect to feel when you touch the outside of the test tube
ValentinkaMS [17]

Answer:

Heat, Let me explain,

Explanation:

An exothermic reaction is like... Lighting a candle so it produces  a small amount of heat in less than a second and then normal heat.

7 0
4 years ago
At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di
Hitman42 [59]

Answer:

The direction of the B-field is in the +y-direction.

Explanation:

The corresponding formula is

F_B = qv\times B

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

\^{x} \times \^{y} = \^{z}

3 0
4 years ago
Letting a = red spheres and b = blue spheres, write a balanced equation for the reaction. express your answer as a chemical equa
GenaCL600 [577]
Knowing that the a and b represents red and blue spheres respectively.The chemical equation we could make isL5A2 + 5B --> 4A2B + A2 + B as shown in boxes 
Simplifying this would make it:
4A2 + 4B --> 4A2B 
Thus, reducing the coefficients, the final answer is:
A2 + B --> A2B
5 0
3 years ago
Read 2 more answers
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with t
OverLord2011 [107]

Answer:

a) h = 13,205.4 m

b)  r_f = 2.12 106 m

c)        e% = 0.68%

Explanation:

a) This is an exercise we are asked to use energy conservation,

Starting point. On the surface of Mimas

        Em₀ = K = ½ m v²

Final point. Where the ball stops

       Em_{f} = U = m g h

        Em₀ = Em_{f}

        ½ m v² = m g h

         h = ½ v² / g

let's calculate

         h = ½ 41² / 0.0636

         h = 13,205.4 m

b) For this part we are asked to use the law of universal gravitation, write the energy

starting point. Satellite surface

           Em₀ = K + U = ½ m v² - GmM / r_o

final point. Where the ball stops

            Em_{f}= U = - G mM / r_f

          Em₀ = Em_{f}

          ½ m v² - G m M / r_o = - G mM / r_f

In this case all distances are measured from the center of the satellite

         1 / rf = 1 / GM (-½ v² + G M / r_o)

     

let's calculate

         1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)

         1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)

          1 / r_f = 4,714 10⁻⁷

          r_f = 1 / 4,715 10⁻⁷

          r_f = 2.12 106 m

to measure this distance from the satellite surface

          r_f ’= r_f - r_o

          r_f ’= 2.12 106 - 1.98 105

         r_f ’= 1,922 106 m

c) the percentage difference is

          e% = 13 205.4 / 1,922 106 100

          e% = 0.68%

The estimate of part a is a little low

6 0
4 years ago
A circular arc has a uniform linear charge density of 4 nC/m. 287 ◦ 2.5 m x y What is the magnitude of the electric field at the
seropon [69]

Answer:

294.0724 N/C

Explanation:

From the question, we are given the following parameters: linear charge density,α = 4 nC/m = 4 × 10^9 C/m, radius, r= 2.5 m, the angle in the counter clockwise direction from the positive x-axis= θ= 287° and the Coulomb constant,Kc= 8.98755 × 109 N · m2 /C 2.

So, we can solve this question by using the formula below;

==>E(x direction)= - (Kc ×α )/ r × {sin 287° - sin 0°} i.

==>> - (8.98755 × 10^9 × 4 × 10^-9)/2.5 × { −0.9563047559630354

- 0} I.

==> -14.4 × [−0.9563047559630354].

==> 13.8 N/C i.

E( y direction)= - (Kc ×α )/ r × ( cos 0° - cos 287°) j.

==> -14.4 × (1 - [ 0.29237170472273672]).

==>( -14.4 × 0.7076) j.

==> - 10.18944 n/C j.

==>E(x direction)^2+ E( y direction)^2.

==> (13.8^2 + (- 10.18^2) N/C.

==> 190.44 N/C. + 103.6324 N/C.

= 294.0724 N/C

7 0
3 years ago
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