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morpeh
14 days ago
13

Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288

lb football player.
Physics
2 answers:
ValentinkaMS [1.1K]14 days ago
7 0
<span>Answer
A person who weighs 220 lb has less mass than someone who weighs 288 lb, so accelerating the 220 lb player requires less force. The heavier player therefore carries greater momentum. Because 288 lb corresponds to more weight (and mass), that player has higher inertia and is harder to stop. For these reasons it is easier to tackle a 220 lb player than a 288 lb player. 
</span>
Sav [1.1K]14 days ago
6 0

Some key relationships to remember are:

Force = mass * acceleration

Weight = mass * gravity (10m/s)

Momentum = mass * velocity

Inertia is a property of mass: the tendency of an object to keep moving at the same velocity unless an outside force acts on it.

Using these formulas, if a 220 lb player and a 288 lb player run at the same speed, the 288 lb player has greater momentum and greater inertia than the 220 lb player. Consequently, changing the motion (speed or direction) of the 288 lb player requires more force (and therefore more energy) than changing the motion of the 220 lb player, which is why tackling the lighter player is easier.

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A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
Yuliya22 [1153]
Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².

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17 days ago
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A glass tube is filled with hydrogen gas.  An electric current is passed through the tube, and the tube begins to glow a pinkish
inna [987]
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The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2
serg [1198]

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!

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13 days ago
An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the oc
Ostrovityanka [942]

Answer:

F=126339.5N

Explanation:

To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:

Fw=W+Fi+F

where

Fw=   force or weight exerted by the water column above the submarine.

To calculate Fw, we can use:

Fw=h. γ. A

h=height

γ= specific weight of seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w represents the hatch weight = 200N

Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:

Fi=PA=101325x0.7=70927.5N

Finally, the force needed to open the hatch is determined by the original equation:

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N

6 0
9 days ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
Softa [913]

Answer:

a) \Delta{t} = 5.39s

b) the distance the motorcycle covers is 155 m

Explanation:

Let t_2-t_1 = \Delta{t} denote the variables. Next, we analyze the motion equation for the accelerating motorcycle alongside the constant speed of the car:

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} represents the motorcycle's speed at time 2

v_{c} is the steady velocity of the car

v_{0} indicates the initial speeds of both vehicle types at time 1

d signifies the distance separating the car and motorcycle at the initial moment

x is the distance the car travels from time 1 to time 2

Solving the equations provides:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second query, we determine x+d by applying the car’s motion equation to compute x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

3 0
12 days ago
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