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emmasim [6.3K]
2 years ago
14

The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet

and x is the horizontal distance the ball travels. How far, in feet, will the ball travel across the field by the time it hits the ground? Round your answer to one decimal place.
Physics
1 answer:
Margaret [11]2 years ago
7 0

Answer:

The horizontal distance the ball travels is 0.902 meters.

Explanation:

The flight of a kicked football follows the quadratic function as :

f(x)=-0.02x^2+2.2x+2

Where

f(x) is the vertical distance in feet

x is the horizontal distance the ball travels

We need to find the distance the ball travel across the field by the time it hits the ground. At this condition,

f(x) = 0

-0.02x^2+2.2x+2=0

On solving the above quadratic equation we get, the horizontal distance the ball travels as :

x = -0.902 meters

So, the horizontal distance the ball travels is 0.902 meters. Hence, this is the required solution.

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The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
Vedmedyk [2.9K]

Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

Therefore, the  tension in the rope is 229.37 N.

3 0
2 years ago
For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou
Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0
2 years ago
Which statements identify what astronomers currently know and think will happen with our universe? Check all that apply. The big
Olegator [25]

The big bang produced dark energy, which accounts for some of the energy needed to expand the universe.

The vastness of space may contain a type of matter known as “dark matter.”

The universe is currently expanding at an accelerating rate.

Hope this helps !

7 0
2 years ago
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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>
8 0
2 years ago
The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?
Marat540 [252]
Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is. 
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2 years ago
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