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Paul [167]
3 years ago
13

A football player carrying the ball runs straight ahead at the line of scrimmage and directly into a wall of defensive linemen.

The ball carrier has an initial speed of 7.36 m/s and is stopped in a time interval of 0.180 s. Find the magnitude and direction of his average acceleration.
Physics
1 answer:
Colt1911 [192]3 years ago
4 0

solution:

we have the following formula for average acceleration\\a_{ave}=\frac{v_{f}-v_{i}}{t}\\=\frac{0-(7.36)}{0.180}\\=-40.88m/s^2\\

Now the direction of this signed acceleration is towards in the direction of travel of the ball carrier.

But as this acceleration is negative we could just as validly state that the acceleration is 40.88m/s^2in the opposite direction to the ball carrier's travel.

Although both are valid, the latter statement is probably the answer you are expected to give .



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g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
2 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
Scientists believe that the earth's core is made mostly of:
leva [86]

Answer:

magma

Explanation:

4 0
3 years ago
Read 2 more answers
A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the
Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g=\frac{530}{1000}=0.530 kg

1kg=1000 g

Area=A=935 cm^2=935\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Height,h=4.4 m

C=1

\rho=1.21 kg/m^3

Width of rectangular prism,b=11.6 cm=\frac{11.6}{100}=0.116 m

1 m=100 cm

Length,l=23.2 cm=0.232 m

Area=l\times b=0.116\times 0.232=0.0269 m^2

Terminal velocity,v_t=\sqrt{\frac{2mg}{\rho CA}}

Where g=9.8 m/s^2

Using the formula

v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}

v_t=17.86 m/s

The velocity of person,v=\sqrt{2gh}

Using the formula

v=\sqrt{2\times 9.8\times 4.4}

v=9.29 m/s

4 0
2 years ago
PLS HELP IM GOING TO GIVE U BRAINLIST
Hitman42 [59]
Answer : A) nucleons

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