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maxonik [38]
3 years ago
15

Emma is running for the bus. She runs 100 metres in 25 seconds. Calculate her speed

Physics
2 answers:
Afina-wow [57]3 years ago
8 0

Answer:

25km/h

is that the full question

beks73 [17]3 years ago
6 0
Speed = distance/time

s=d/t

s=100/25 = 4m/s

4m/s = 14.4kph (km/h)
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In a 2-dimensional coordinate system, the x- and y-axes are typically _____.
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3 years ago
A small plane flies at a speed of 102 km/h in still air. Suppose the wind blows out from the west (with the air moving east) at
Semmy [17]

Answer:

2.68 hours

Explanation:

A.) Suppose the wind blows out from the west (with the air moving east). The pilot should then head her plane to northwest direction to move directly north.

B.) Given that plane flies at a speed of 102 km/h in still air. And the wind blows out from the west (with the air moving east) at a speed of 46 km/h.

The plan resultant speed can be calculated by using pythagorean theorem.

Resultant Speed = Sqrt( 102^2 + 46^2 )

Resultant Speed = Sqrt( 12520)

Resultant speed = 111.89 km/h

From the definition of speed,

Speed = distance/time

Where distance = 300 km

Substitute the resultant speed and the distance into the formula.

111.89 = 300/time

Time = 300/111.89

Time = 2.68 hours

Therefore, it take her 2.68 hours to reach a point 300 km directly north of her srarting point

7 0
3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

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C.

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