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TEA [102]
3 years ago
9

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

ce is 10 N, so that work in = work out?
Physics
1 answer:
steposvetlana [31]3 years ago
5 0

Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

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The gravity from the person's hand is weaker than the gravity from the pull of the earth

Explanation:

The gravity from the person's hand is weaker than the gravity from the pull of the earth

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Of all of the types of forest biomes, tropical rain forests contain the most biodiversity, even though they do not have the most
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Explain how the "natural frequency" of objects must be considered/analyzed in places like concert halls and airplanes.
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If you are talking about sound frequency you need to consider what area you are in because in a concert hall it is big and helps the sound spread but in an airplane it is the opposite. 
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2 years ago
calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.
Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference  = 110v

Current  = 20A

Unknown:

Power  = ?

Solution:

To solve this problem, we use the expression below:

        Power  = IV

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  This is therefore 2.2kW

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8 0
3 years ago
An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice
Rom4ik [11]

Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

8 0
2 years ago
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