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Agata [3.3K]
3 years ago
13

Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent’s face,

if the glove and face compress 7.50 cm during a blow in which the 7.00-kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the days when no gloves were used and the knuckles and face would compress only 2.00 cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove?
Physics
1 answer:
Semmy [17]3 years ago
8 0

(a) -4667 N

First of all, we can calculate the acceleration of the arm and the glove, using the following equation:

v^2 - u^2 = 2ad

where

v = 0 is the final speed

u = 10 m/s is the initial speed

a is the acceleration

d = 7.50 cm = 0.075 m is the distance through which the arm and the glove move before coming to a stop

Solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(10.0 m/s)^2}{2(0.075 m)}=-666.7 m/s^2

And since the know the mass of the arm+glove:

m = 7.00 kg

We can now calculate the force exerted:

F=ma=(7.00 kg)(-666.7 N)=-4,667 N

(b) -17500 N

We can repeat the problem, but this time the stopping distance is different:

d = 2.00 cm = 0.02 m

So the acceleration is

a=\frac{v^2-u^2}{2d}=\frac{0-(10.0 m/s)^2}{2(0.02 m)}=-2500 m/s^2

and so the force is

F=ma=(7.00 kg)(-2500 m/s^2)=-17,500 N

(c) Yes

The force exerted when the glove is used is

F = 4667 N

We see that this force corresponds approximately to the weight of an object of mass m=476 kg, in fact:

W=mg=(476 kg)(9.81 m/s^2)=4670 N

Which is quite a lot. Therefore, the force even when gloves are used seems enough to cause damage.

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u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]

u x v = i(2+0) - j(2 - 1) + k(0 - 2)

u x v = 2i - j - 2k

Now the magnitude of u x v is calculated as follows:

| u x v | = \sqrt{2^2 + (-1)^2 + (-2)^2}

| u x v | = \sqrt{4 + 1 + 4}

| u x v | = \sqrt{9}

| u x v | = 3

Therefore, the magnitude of u x v is 3

(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e

û = \frac{u X v}{|u X v|}        

u x v = 2i - j - 2k        [<em>calculated in (a) above</em>]

|u x v| = 3                   [<em>calculated in (a) above</em>]

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Listed following are several fictitious stars with their luminosities given in terms of the Sun's luminosity (LSun) and their di
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Answer:

1. Nismo: L_{a}=0.1244

2. Ferdinand: L_{a}=0.0796, Shelby: L_{a}=0.0796

3. Enzo: L_{a}=0.0398

4. Lotus: L_{a}=0.0199

Explanation:

The luminosity of a star is the amount of light it emits from its surface. The luminosity is an intrinsic property of the star. Apparent brightness is how bright the star appears to a detector on Earth. Apparent brightness is not an intrinsic property of the star; it depends on the location of the observer.

As light travels towards an observer, it spreads out and covers a larger area, reducing the intensity. Thus the apparent brightness is inversely proportionate to the square of the distance between the star and observer.

Apparent brightness can be calculated by the formula below:

L_{a}=\frac{L}{4*pi*D^{2}}

Where, L_{a} is apparent brightness, L is luminosity, D is the distance between the star and observer

Enzo: 200 LSun , 20 ly

L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{200}{4*pi*20^{2}}

L_{a}=0.0398

Ferdinand: 400 LSun , 20 ly

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L_{a}=0.0796

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L_{a}=0.1244

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L_{a}=\frac{L}{4*pi*D^{2}}

L_{a}=\frac{400}{4*pi*40^{2}}

L_{a}=0.0199

Shelby: 100 LSun , 10 ly

L_{a}=\frac{L}{4*pi*D^{2}}

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L_{a}=0.0796

Apparent brightness of stars is summarized as below. Absolute values are considered as units are unspecified

Enzo: L_{a}=0.0398

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Lotus: L_{a}=0.0199

Shelby: L_{a}=0.0796

Organized in order of apparent brightness

1. Nismo: L_{a}=0.1244

2. Ferdinand: L_{a}=0.0796, Shelby: L_{a}=0.0796

3. Enzo: L_{a}=0.0398

4. Lotus: L_{a}=0.0199

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