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frozen [14]
2 years ago
6

The crossbar of a goal post on a professional American football field sits at a height of 3.05 meters (10 feet) above the field.

A football is placed on the ground at midfield so that it is 54.9 meters (60 yards) horizontally from the position of the crossbar. The ball is then kicked so that its initial velocity is 26.0 m/s (58.2 mph) at an angle of 35 degrees above the horizontal. Neglecting air friction, does the ball make it over the goal post and by how much?
A. Yes, the ball passes 2.83 meters above the crossbar
B. No, the ball passes 1.57 meters below the crossbar
C. No, the ball hits the ground before it reaches the crossbar
D No, the ball hits the crossbar
E. Yes, the ball passes 8.22 meters above the crossbar
Physics
1 answer:
Lemur [1.5K]2 years ago
5 0

Answer:

A. Yes, the ball passes 2.83 meters above the crossbar

Explanation:

This problem relates to projectile motion .

Horizontal component of velocity of ball = 26 cos 35 = 21.3 m /s

Vertical component of velocity = 26 sin35 = 14.9 m /s

Time to cover horizontal distance of goal post that is 54.9 m

= distance / horizontal speed

= 54.9 / 21.3

= 2.577 s .

vertical distance covered in this time

h = ut - 1/2 gt² , u is vertical component of velocity , t = 2.577 s

h = 14.9 x 2.577 - .5 x 9.8 x 2.577²

= 38.39 - 32.54

= 5.85 m

So the ball crosses the crossbar above it by 5.85 - 3.05 = 2.8 m

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7 0
2 years ago
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

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Work done by the gravity =- 3 \times 9.8 \times 4

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work done by gravity is equal to -117.6 J            

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7 0
2 years ago
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie
Anni [7]

Answer:

The inverse frequency is \dfrac{3}{80}\ s

Explanation:

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We need to calculate the inverse frequency

Using formula of frequency

\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}

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\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s

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5 0
2 years ago
The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.
grin007 [14]
<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

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So

RI = 1/0.6 = 5/3 or 1.66

3 0
2 years ago
Read 2 more answers
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