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MissTica
3 years ago
7

In a football game, one team kicks off to the other. At the moment the receiver catches the ball, he is 40 yards from the neares

t tackler. The receiver runs left to right at a speed of 10 yards per second (10 y/s). The tackler runs right to left at a speed of 6 yards per second.
How long does it take before they collide? _____

How far does the receiver go? _____________________
Physics
2 answers:
Talja [164]3 years ago
8 0
The answer is hopefully if i'm correct is 5 or 6 seconds 
Nikitich [7]3 years ago
6 0

Answer:

Part a)

\delta t = 2.5 s

Part b)

d = 25 yards

Explanation:

Speed of the receiver is 10 yards per second towards right

Speed of the tackler is 6 yards per second towards left

So here we have net relative speed of the two towards each other is given as

v_{12} = v_1 - v_2

v_{12} = 10 - (-6) = 16 yards/s

now we know that the distance between receiver and tackler initially is 40 Yards

Part a)

so the time taken by two to collide is given as

\delta t = \frac{\Delta x}{v}

\delta t = \frac{40}{16}

\delta t = 2.5 s

Part b)

Distance moved by the receiver in t = 2.5 s

d = vt

d = 10 (2.5)

d = 25 yards

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What is the significance of the direction of an electric field line at some point on the line?
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Answer with Explanation:

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4 years ago
An airplane traveling from San Francisco northeast to Chicago travels 1260 km in 3.5 h. What is the airplanes average velocity?
IrinaK [193]

Answer:

Average velocity = 360km/h due Northeast.

Explanation:

Given the following data;

Distance = 1260km

Time = 3.5h.

Velocity =?

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

Velocity = \frac{distance}{time}

V = \frac{d}{t}

Substituting into the above equation;

V = \frac{1260}{3.5}

Velocity, V = 360km/h.

Since velocity is a vector quantity, it must have both magnitude and direction.

<em>Hence, the average velocity of the airplane is 360km/h due Northeast. </em>

6 0
4 years ago
If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface
Alecsey [184]

Answer: h = 3R

Explanation:

Using the law of conservation of energy,

Total energy at the beginning of the launch would be equal to total energy at any point.

kinetic energy + gravitational potential energy = constant

Initial energy of the projectile =\frac{1}{2}mv_e^2-\frac{GMm}{R}... (1)

where R is the radius of the Earth, M is the mass ofthe Earth, m is the mass of the projectile.

escape velocity, v_e=\sqrt{\frac{2GM}{R}}

Total energy at height h above the Earth where speed of the projectile is half the escape velocity:

\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} ...(2)

(1)=(2)

⇒\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} = \frac{1}{2}mv_e^2-\frac{GMm}{R}

⇒G\frac{Mm}{R}-G\frac{Mm}{R+h}= \frac{1}{2}m \frac{3}{4}v_e^2 =\frac{1}{2}m \frac{3}{4} (\sqrt{\frac{2GM}{R}})^2

⇒\frac{GM(R+h-R)}{R(R+h)} = \frac{3}{4}(\frac{GM}{R})

⇒\frac{h}{R+h} = \frac{3}{4}

⇒h = 3R

Thus, at height equal to thrice radius of Earth, the speed of the projectile would reduce to half of escape velocity.  

4 0
3 years ago
Read 2 more answers
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