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Andrei [34K]
3 years ago
10

In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi

s linemen, he accelerates at a rate of 1.71 yards/s^2 to a velocity of 7.54 yards/s before he is tackled. How far did he run after he got the ball?
Physics
1 answer:
Inessa [10]3 years ago
8 0

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

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Two students walk in the same direction along a straight path, at a constant speedâone at 0.90 m/s and the other at 1.90 m/s.
mixer [17]

Answer:

456.143684211 seconds

564.3 m

Explanation:

s = Distance

v = Velocity

Time is given by

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{780}{0.9}\\\Rightarrow t=866.67\ s

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{780}{1.9}\\\Rightarrow t=410.526315789\ s

Difference in time = 866.67-410.526315789 = 456.143684211 seconds

According to the question

\dfrac{x}{0.9}-\dfrac{x}{1.9}=5.5\times 60\\\Rightarrow x(\dfrac{1}{0.9}-\dfrac{1}{1.9})=330\\\Rightarrow x=\dfrac{330}{\dfrac{1}{0.9}-\dfrac{1}{1.9}}\\\Rightarrow x=564.3\ m

The students would have to walk 564.3 m

3 0
3 years ago
A_ is a cut made through the wood
Cloud [144]

Answer:

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Explanation:

 

6 0
3 years ago
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The journey of an electron through an external circuit involves a long and slow zigzag path that is characterized by losses in e
Kitty [74]

Answer:

d_{b} = 2 d_{a}

Explanation:

The electrical resistance for a cylindrical wire is described by the expression

         R = ρ L / A

The area of ​​a circle is

     A = π r²

     r = d / 2

     A = π d²/4

We substitute

      R = ρ L  4 /π d²

Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B

    R_{a} = 4 R_{b}

We substitute

    ρ  4/π d_{a}² = 4 (ρ  4/π d_{b}²)

    1 / d_{a}² = 4 / d_{b}²

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4 0
3 years ago
A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.
Ivanshal [37]

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}\times 9.8 \times 10^2

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Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

speed = \frac{distance}{time}

v = \frac{650}{10}

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4 0
3 years ago
Describe how the three methods of thermal energy transfer may take place within the iguana’s enclosure.
sasho [114]

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat.  Conduction is the transfer of heat between substances that are in direct contact with each other. Thermal energy is transferred from hot places to cold places by convection. Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object as is the case with conduction and convection. Heat can be transmitted through empty space by thermal radiation often called infrared radiation.

Explanation:

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