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TiliK225 [7]
3 years ago
9

When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe

r. (Throughout, assume that each battery has negligible internal resistance.) What power does the resistor consume if it is connected to a 3.00-V watch battery
Physics
1 answer:
Grace [21]3 years ago
8 0

Explanation:

the answer is in the above image

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The acceleration due to gravity on the moon is 1.6 m/s2. What is the gravitational potential energy of a 1200-kg lander resting
Monica [59]
Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. </span><span>The equation for </span>gravitational potential energy<span> is GPE = mgh. 

GPE = 1200(1.6)(350) = 672000 J

Hope this answers the question. Have a nice day.</span>
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Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani
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d not joining FRA and joining HOSA INSTEAD

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Which ancient idea of the nature of light did Isaac Newton build upon? Light is a wave or disturbance that travels through space
JulijaS [17]

Answer:

The answer is <u>C</u>

Explanation:

Light is a substance through which particles flow from a light source

Just took the quiz on edge.

6 0
3 years ago
The weights of soy patties sold by a diner are normally distributed. A random sample of 15 patties yields a mean weight of 3.8 o
shepuryov [24]

Complete Question

The weights of soy patties sold by a diner are normally distributed. A random sample of 15 patties yields a mean weight of 3.8 ounces with a sample standard deviation of 0.5 ounces. At the 0.05 level of significance,perform a hypothesis test to see if the true mean weight is  less than 4 ounce.

Answer:

Yes the true mean weight is less than 4 ounce

Explanation:

  From the question we are told that

     The random sample is n = 15

       The mean weight is \= x = 3.8\  ounce

        The standard deviation is  \sigma = 0.5 \ ounce

         The  level of significance is  \alpha  = 0.05

       

So

   The  null hypothesis is  H_o : \mu \ge 4

     The alternative hypothesis is H_a : \mu < 4

Generally the critical value which a bench mark to ascertain whether the null hypothesis is  true or false is mathematically represented as

            t_{0.05} = 1.79

This value is  obtained from  the critical value table

Generally the test statistics is mathematically represented as

                Test \ Statistics (ST)  = \frac{\= x - \mu }{\frac{\sigma}{\sqrt{n} }  }

           =>  ST = \frac{3.8 -4 }{\frac{0.5}{\sqrt{20} } }

                ST = - 1.79

So since ST is less than t_{0.05}  then the null hypothesis would be rejected and the alternative hypothesis would be accepted so

  Thus the true mean weight is less than 4

4 0
3 years ago
The Earth’s surface, on average, carries a net negative charge (while the clouds and lower atmosphere carry a net positive charg
LenaWriter [7]

Answer:

(a) Q = -6.765 * 10⁵ C ; σ = 1.33 * 10⁻⁹ C/m²

(b) 0.23 N/C

Explanation:

(a) Electric field at the surface of a sphere is given as:

E = kQ/r²

Where

k = Coulombs constsnt

Q = charge

r = radius of sphere

To find charge Q, we make Q subject of the formula:

Q = (E * r²)/k

Hence, charge, Q, at the surface of the earth, having radius, r = 6.371 * 10⁵ and electric field, E = 150 N/C is:

Q = [150 * (6.371 * 10⁶)²] / (9 * 10⁹)

Q = 6.765 * 10⁵ C

Since we're told that the charge at the earth's surface is negative,

Q = -6.765 * 10⁵ C

Surface charge density, σ, given as:

σ = |Q|/A

Where

|Q| = magnitude of charge

A = surface area.

Surface area, A, of the earth is given as:

A = 4πr²

A = 4π * (6.371 * 10⁶)²

A = 510064471909788 m²

σ = 6.765 * 10⁵/510064471909788

σ = 1.33 * 10⁻⁹ C/m²

(b) At a height 5km from the earth's surface, the electric field will be:

E = kQ/(r + 5km)²

r + 5km = 6376km = 6.376 * 10⁶m

=> E = (9 * 10⁹ * 6.765 * 10⁵)/(6.376 * 10⁶)²

E = 149.77 N/C

The difference between the electric field at the surface of the earth and at a height of 5km is:

159 - 149.77 N/C = 0.23 N/C

3 0
4 years ago
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