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Leviafan [203]
3 years ago
9

A wire is used as a heating element that has a resistance that is fairly independent of its temperature within its operating ran

ge. When a current I flows through the wire, the energy delivered by the heater each minute is E. For what amount of current will the energy delivered by the heater each minute be 4E?
Physics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

Double the current

Explanation:

The energy delivered by the heater is related to the current by the following relation:

E= I^{2}R t

let R * t = k ( ∴ R and t both are constant)

so E= k I^{2}

Now let:

E2= k I₂^2

E2= 4E

⇒ k I₂^2= 4* k I^{2}

Cancel same terms on both sides.

I₂^2= 4* I^{2}

taking square-root on both sides.

√I₂^2 = √4* I^2

⇒I₂= 2I

If we double the current the energy delivered each minute be 4E.

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in a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw
madam [21]

Answer:

μ=0.151

Explanation:

Given that

m= 3.5 Kg

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v= 1.36 m/s

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F- Fr = m a

22 -  μ x 3.5 x 10 = 3.5 a         ( take g =10 m/s²)

a= 6.28 - 35μ  m/s²

The final speed of the block is v

v= 1.36 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.36² = 2 x a x 0.96

a= 0.963 m/s²

a= 6.28 - 35μ  m/s²

6.28 - 35μ = 0.963

μ=0.151

3 0
3 years ago
Help ASAP
SIZIF [17.4K]
I think all of those are examples
4 0
3 years ago
An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra
WARRIOR [948]

Answer:

The inventors  claim is not real

a)  No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

     The  power input is  P_i  = 0.25 kW  =  0.25 *10^{3} \ W

      The  rate of heat transfer J  =  3050 J/s

       The temperature of the freezer content is T = 270 \ K

       The  ambient temperature is  T_a  =  293 \ K

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

      COP  =  \frac{T }{Ta - T}

substituting values

     COP  =  \frac{270 }{293 - 270}

     COP  =11.7

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

       COP  =  \frac{J}{P_i}

substituting values

       COP  =  \frac{3050}{0.25 *10^{3}}

       COP  = 12.2

Now given that the COP  of an ideal refrigerator is  less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

4 0
3 years ago
How can you define a solution to an equation?
sleet_krkn [62]
A solution is a value or a collection of values.. when substituted for the unknowns, the equation become an equality.
Example : x + 2 = 7
When we out the 5 in place of x we get: 5 + 2 = 2
8 0
3 years ago
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