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3 years ago

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Superman throws a 2400 n boulder at a villain. what horizontal force must superman apply to the boulder to give a horizontal acc

eleration of 12.0 m/s2?

1 answer:

Ksenya-84 [330]3 years ago

4 0

The boulder has a weight of W=2400 N. The weight of an object is the product between its mass m and the gravitational acceleration g:
$W=mg$
Rearranging the relationship, we can calculate the mass of the boulder:
$m= \frac{W}{g}= \frac{2400 N}{9.81 m/s^2}=244.6 kg$

We are told that Superman applies a horizontal force to this object, and as a result, the acceleration of the boulder is $a=12.0 m/s^2$ . We can find the force applied by using Netwon's second law of motion:
$F=ma=(244.6 kg)(12.0 m/s^2)=2935 N$

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A wire of cross-sectional area 5.00 106 m2 has a resistance of 1.75 O. What is the resistance of a wire of the same material and

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Answer:

the resistance of the second wire is 1 ohm.

Explanation:

Given;

cross sectional area of the first wire, A₁ = 5.00 x 10⁶ m²

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cross sectional area of the second wire, A₂ = 8.75 x 10⁶ m²

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As an electron approaches a proton, the electron's force of attraction...

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A baseball is thrown at an angle of 40.0° above

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Three children are struggling and pulling on a single toy. Two of the children, Abe and Barry, are EACH (individually) pulling w

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Solution :

c). $\vec{F}_A =$ force applied by Abe

$\vec{F}_B =$ force applied by Barry

$\vec{F}_E =$ force applied by Eric

$\vec{F}_R =$ Resultant force

$\vec{F}_A$ in the vector form can be written as :

$\vec{F}_A = 0 \hat{i} + 60 \hat{j}$

$\vec{F}_B$ in the vector form can be written as :

$\vec{F}_B = 60 \hat{i} + 0 \hat{j}$

The resultant,

$\vec{F}_R= \vec{F}_A+\vec{F}_B$

$=(0 \hat i + 60 \hat j)+(60 \hat i + 0\hat j)$

$=60 \hat i + 60 \hat j$

$|\vec{F}_R| = \sqrt{60^2+60^2}$

= 84.853 N

d). As the three forces are in equilibrium, therefore,

$|\vec F_E| = |\vec F_R|$

$|\vec F_E| =84.853 \ N$

e). The direction of the force exerted by Eric is exactly opposite to the direction of the resultant force.

The direction of the resultant force is :

$\theta = \tan ^{-1}\left(\frac{F_y}{F_x}\right)$

$= \tan ^{-1}\left(\frac{60}{60}\right)$

= 45° north east

The direction of the force E is 45° west or 45° south west.

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