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3 years ago

a man throws a football straight into the air. As it rises,it slows down. Which type of energy is the football gaining?

Physics

2 answers:

Arlecino [84]3 years ago

8 0

Answer:

Potential energy

Explanation:

As the man throws a ball, the ball gains some kinetic energy. As the ball goes up the velocity of ball goes on decreasing and the height goes on increasing. It means the it loses kinetic energy and gaining potential energy.

kotykmax [81]3 years ago

5 0

It is gaining potental energy which will then transfer to knetic energy as it falls

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A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s . What is the height of the cliff? (Use 32 ft/s 2 for

bulgar [2K]

To solve this problem we will apply the linear motion kinematic equations, which describe the change in velocity, depending on the acceleration and the distance traveled, that is,

$v_f^2 = v_i^2 +2ah$

Where,

$v_f$ = Final Velocity

$v_i$ = Initial Velocity

a = Acceleration

h = height

Our values are given as,

$v_f = 88 ft/s\\v_i = 0 ft /s\\a = 32 ft/s^2\\$

Replacing we have,

$vf^2 = vi^2 + 2*a*h$

$88^2 = 0 + 2*32*h$

$h= 121 ft$

Therefore the height of the cliff is 121ft

5 0

3 years ago

Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende

shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

$V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}$

Ahora, encontremos al densidad de la placa:

$d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}$

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.

Espero que te sea de utilidad!

6 0

3 years ago

You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

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