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yawa3891 [41]
3 years ago
12

You are going sledding with your little cousins during a well-earned break. You push your 23 kg cousin so that they have a speed

of 6 m/s at the top of the hill that is 17 m tall. What is the kinetic energy of your cousin at the top of the hill (after you push)? (in J)
Physics
2 answers:
pogonyaev3 years ago
7 0

Answer:

kinetic energy = 414 J

Explanation:

given data

mass = 23 kg  

speed = 6 m/s  

height = 17 m    

to find out

kinetic energy

solution

we get kinetic energy that is express as

kinetic energy = 0.5 × m × v²    ..........................1

here m is mass and v is speed so

put here value we get kinetic energy

kinetic energy = 0.5 × 23 × 6²

kinetic energy = 414 J

Thepotemich [5.8K]3 years ago
5 0

Answer:

414 J

Explanation:

mass, m = 23 kg

velocity, v = 6 m/s

Kinetic energy is given by

K = 0.5 mv²

K = 0.5 x 23 x 6 x 6

K = 414 J

Thus, the kinetic energy is 414 J.

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Morgarella [4.7K]

Answer:

0.166 m or 16.6 cm

Explanation:

From the question,

v = ωr......................... Equation 1

Where v = tangential speed of the string, ω = Angular speed of the string, r = Length of the rotating string.

make r the subject of the equation

r = v/ω....................... Equation 2

Given: v = 50 m/s, ω = 48 rev/s = 48(2π) rad/s = 48(2×3.14) rad/s = 301.44 rad/s.

Substitute into equation 2

r = 50/301.44

r = 0.166 m or 16.6 cm

Hence the length of the rotating string = 0.166 m or 16.6 cm

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3 years ago
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Dennis_Churaev [7]
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Read 2 more answers
Una niña está empujando un baúl. El PESO del baúl es de 230 N y el roce es de 50 N, la niña sólo logra ejercer una fuerza de 30
Gre4nikov [31]

Answer:

Su padre necesita aplicar una fuerza de 20 newtons en la misma dirección que la fuerza aplicada por su hija.

Explanation:

Asúmase que el baúl se mueve en una superficie horizontal. La fuerza de rozamiento dada por el problema es la fuerza de rozamiento estático máximo, se requiere una fuerza externa antiparalela a la fuerza de rozamiento estático máximo para que el baúl se empiece a mover. La ecuación de equilibrio de fuerzas horizontales sobre el baúl es:

\Sigma F = P - f = 0

Donde:

P - Fuerza externa aplicada sobre el baúl, medida en newtons.

f - Fuerza de rozamiento estático máximo, medida en newtons.

Se despeja la fuerza externa:

P = f

Si f = 50\,N, entonces:

P = 50\,N

Si la niña solo logra ejercer una fuerza de 30 newtons, su padre necesita aplicar una fuerza de 20 newtons paralela a aquella fuerza.

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A 200 g ball moving at 5 m/s strikes a wall perpendicular and rebounds elastically at the same speed. The impulse given the wall
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The following table lists properties of three different objects.
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