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liq [111]
4 years ago
14

Kyle has a mass of 54 kg and is jogging at a velocity of 3 m/s. What is Kyle’s kinetic energy?

Physics
2 answers:
olga nikolaevna [1]4 years ago
8 0

Answer:243joules

Explanation:

Mass(m)=54kg

Velocity(v)=3m/s

Kinetic energy =(m x v^2)/2

Kinetic energy =(54 x 3^2)/2

Kinetic energy =(54 x 9)/2

Kinetic energy =486/2

Kinetic energy =243joules

Nady [450]4 years ago
5 0

Answer:

D

Explanation:

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what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?​
Reil [10]

Answer:

160N

Explanation: When 80kg mass is one group . It's reaction force acting on a ground.

Weight of the object = 80*10

= 800 N

Here we are given cofficient of static friction its 0.2. It should be smaller than 1

Friction force = Reaction * Friction Cofficient

Reaction = 800N ( Considering Vertical Equilibrium )

F = 800* 0.2

F = 160N

3 0
3 years ago
Read 2 more answers
Lighter elements are combined to form the sun and stars.what is this process called? What are the two lighter elements? What pro
zhenek [66]

nucleosynthesis is the process in nuclear reaction

7 0
3 years ago
Por favor, necesito ayuda urgente 4. vectores perpendiculares de 25 y 40 unidades cada uno. Hallar gráfica y numericamente el ve
Darina [25.2K]

Answer:

4)  R = 47.17 units , 5)  R= 10,29 unidades, 6)   R= 2994,4 km ,    θ = -33,7

Explanation:

Este es un ejercicio de adición de vectores

4) como los vectores son perpendiculares.

Para encontrar la resultante podemos usar el teorema de pitoras

        R =  √ a² + b²

        R = √√ ( 25² + 40²)

        R =  47,17 unidades

5)  Este caso como el angulo es diferente de 90 debemos usar la relación de pitoras completa

        R² =  a² + b² + 2 a b cos θ

donde el angulo es entre los vectores a y b

        R² = 12² + 16² – 2 12 16 cos 40

        R²= 400 – 294,16

       R= 10,29 unidades

6) En este caso los dos desplazamientos son perpendiculares, por lo cual Usamos el teorema de Pitágoras

          R = √ (2400² + 1600²)

          R= 2994,4 km

para el angulo de este desplazamiento usamos trigonometría

           tan θ = y/x

           θ = tan⁻¹ y/x

           θ = tan⁻¹ 1600/(-2400)

           θ = -33,7  

TRASLATE

This is a vector addition exercise

4) as the vectors are perpendicular.

To find the result we can use the Poreor theorem

        R = √ a² + b²

        R = √ (25² + 40²)

        R = 47.17 units

5) This case, as the angle is different from 90, we must use the complete ratio of the pitoras

        R² = a² + b² + 2 a b cos θ

where the angle is between vectors a and b

        R² = 12² + 16² - 2 12 16 cos 40

         R² = 400 - 294.16

       R = 10.29 units

6) In this case the two displacements are perpendicular, which is why we use the Pythagorean theorem

          R = √ (2400² + 1600²)

          R = 2,994.4 km

for the angle of this displacement we use trigonometry

           tan θ = y / x

           θ = tan⁻¹ y / x

           θ = tan⁻¹ 1600 / (- 2400)

           θ = -33.7

3 0
3 years ago
If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch the same
Angelina_Jolie [31]

Answer:W=\frac{3}{4} ft-lb

Explanation:

Given

Work required to stretch 1 ft is 12 ft-lb

and we have to find work required to stretch 3 in.

i.e. \frac{1}{4} ft

12=\frac{1}{2}K\left ( 1\right )^2 ------(1)

W=\frac{1}{2}k\left ( \frac{1}{4}\right )^2-----(2)

divide (1)&(2)

\frac{12}{W}=\left ( \frac{4}{1}\right )^2

W=\frac{12}{4\times 4}

W=\frac{3}{4} ft-lb

6 0
3 years ago
a car is travelling along a straight road. the driver suddenly observes that the road ahead is flooded and applies the brakes. d
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Answer:

The brakes don't grip the tires as securely as the water prevents a stable grip. The car will skid straight along the road as the tires don't have traction along a wet surface

6 0
3 years ago
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