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PtichkaEL [24]
3 years ago
9

the weight of a body is more at polar regions of the earth than that at the equatorial regions? why give reason​

Physics
1 answer:
ella [17]3 years ago
3 0

Answer:

The weight of an object at the Earth's South Pole is slightly more than its weight at the Equator because the polar radius of the Earth is slightly less than the equatorial radius. Though the mass of an object remains constant, its weight varies according to its location.

Explanation:

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Charlie Brown kicks a football at 24.5 m/s at 35.0. What is the maximum height of the ball?
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Answer:

d = 10.076 m

Explanation:

We need to obtain the velocity of the ball in the y direction

Vy  = 24.5m/s * sin(35) = 14.053 m/s

To obtain the distance, we use the formula

vf^2 = v0^2 -2*g*d

but vf = 0

d = -vo^2/2g

d = (14.053)^2/2*(9.8) = 10.076 m

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How many atoms of oxygen in the chemical formula 2 Ca(CIO2)2?<br> O 2<br> O 4<br> O 6<br> O 8
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Answer:

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Explanation:

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2 years ago
A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int
Inessa05 [86]

Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of

<em>F</em> = <em>m a</em>   →   <em>a</em> = <em>F </em>/ <em>m</em>

so that the cart's final speed is

<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>

<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>

<em />

If we force is halved, so is the accleration:

<em>a</em> = <em>F</em> / <em>m</em>   →   <em>a</em>/2 = <em>F</em> / (2<em>m</em>)

So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give

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3 0
3 years ago
Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field
IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is \mu_0 = 1.32*10^{-6} \ Tm/A

The percentage error is  % error = 5.25%

Explanation:

From the question we are told that

            The slope is  s= 3.9\  G/A , and

Generally  1 \ gauss =  10^{-4 } tesla

               So  s = 3.9 *10^ {-4} T /A

From the relation given in the question

                        \mu_0 = \frac{2R}{N} [\frac{B}{I} ]

 Where R is the radius of the coil  =\frac{Diameter \ of \ coil }{2} = 0.017m

             N is the number of loops of the coil = 10

Now from the question we are told that

                  s = \frac{B}{I}

substituting into the equation above

               \mu_0 = \frac{2R }{N} s

Substituting values

              \mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}

                   = 1.326 *10^ {-6} \ Tm/A

Generally the % error is mathematically represented as

                    %error = \frac{Measured \  value - accepted \ value}{accepted \  value } *100

Given that the accepted value is \mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A

      Hence substituting values

                      %error  = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100

                                  = 5.24

6 0
3 years ago
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