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2 years ago

Two students have fitted their scooters with the same engine. Student A and his

Physics

1 answer:

sammy [17]2 years ago

3 0

The force exerted by student A with his scooter is 306 N and that of student B is 204 N.

<h3>Force applied by each student</h3>

The force exerted by each student is calculated from Newton's second law of motion.

F = ma

where;

m is mass
a is acceleration

F(A) = 127.5 x 2.4

F(A) = 306 N

F(B) = 120 x 1.7

F(B) = 204 N

Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.

Learn more about force here: brainly.com/question/12970081

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Answer:

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3 years ago

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2 years ago

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3 years ago

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Alex

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3 years ago

three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25

ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0

2 years ago