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3 years ago

Ingrid is moving a box from the ground into the back of a truck. She uses 20 N of force to move the box

Mathematics

2 answers:

3241004551 [841]3 years ago

8 0

She will decrease the amount of work

Without the ramp, Ingrid would have to, in no particular order,

• push the box along the ground, which amounts of (20 N) (5 m) = 100 J of work

• lift the box up some height h with a force at least equal in magnitude to the weight of the box mg, so that the work done in the lifting is mgh (i.e. the energy required to lift the box with constant speed to a height h)

and the total work done in this way would be (100 + mgh) J.

With the ramp, Ingrid would instead push the box a length equal to the hypotenuse of the right triangle formed by the two displacements above. Recall that the hypotenuse of any right triangle is always shorter than the sum of its legs' lengths. (This eliminates the third and fourth options.) So if the length of the ramp, or total displacement of the box along the ramp, is s, then the work done by pushing up the ramp is (10 N) s = 10s J

DerKrebs [107]3 years ago

3 0

Answer:

Step-by-step explanation:

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Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$