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Sliva [168]
2 years ago
15

Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work

is needed to stretch the spring from 36 cm to 41 cm
Physics
1 answer:
dlinn [17]2 years ago
4 0

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

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A projectile fired upward from the Earth's surface will usually slow down, come momentarily to rest, and return to Earth. For a certain initial speed, however it will move upward forever, with its speed gradually decreasing to zero just as its distance from Earth approaches infinity. The initial speed for this case is called escape velocity. You can find the escape velocity v for the Earth or any other planet from which a projectile might be launched using conservation of energy. The projectile of mass m leaves the surface of the body of mass M and radius R with a kinetic energy Ki = mv²/2 and potential energy Ui = -GMm/R. When the projectile reaches infinity, it has zero potential energy and zero kinetic energy since we are seeking the minimum speed for escape. Thus Uf = 0 and Kf = 0. And from conservation of energy,
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mv²/2 -GMm/R = 0
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Are photons causing the electrons to flow off of the target? ( the target is calcium)
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Answer:

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Explanation:

5 0
3 years ago
A transformer consisting of two coils wrapped around an iron core is connected to a generator and a resistor (Resistor is connec
V125BC [204]

Answer:

The peak emf of the generator is 40.94 V.

Explanation:

Given that,

Number of turns in primary coil= 11

Number of turns in secondary coil= 18

Peak voltage = 67 V

We nee to calculate the peak emf

Using relation of number of turns and emf

\dfrac{N_{1}}{N_{2}}=\dfrac{E_{1}}{E_{2}}

E_{1}=\dfrac{N_{1}}{N_{2}}\times E_{2}

Where, N₁ = Number of turns in primary coil

N₂ = Number of turns in secondary coil

E₂ = emf across secondary coil

Put the value into the formula

E_{1}=\dfrac{11}{18}\times67

E_{1}=40.94\ V

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4 0
2 years ago
Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Strike441 [17]

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

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2 years ago
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