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azamat
3 years ago
9

Boxers attempt to move with an opponent's punch when it is thrown. In other words, a boxer moves in the same direction as their

opponent's punch. This movement may prevent a knockout blow being delivered by their opponent. Explain how.
Physics
2 answers:
Bingel [31]3 years ago
8 0

Answer: h

Explanation:

Impulse is defined as the change of momentum of an object if the object is acted by a force for an interval of time. By delaying the time, the impact of the force can be reduced.

Boxers attempt to move with an opponent's punch when it is thrown. The force applied by the opponent's punch is extended over a longer time. Then the effect of the force of the blow can be reduced. In this case, the chances of a knockout blow being delivered by their opponent can be minimized.

maxonik [38]3 years ago
5 0
When the force of the opponent's punch is extended, with time, the effect of the blow or the force of the blow is reduced thereby reducing chances for a knockout punch from the opponent to the boxer.
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a 4kg box is resting on top of a shelf that is 2m high, what is the boxs gravitational potential relative to the floor​
Aloiza [94]

Answer:

80 J

Explanation:

PE = mgh

PE = (4 kg)(9.8 m/s^2)(2 m)

PE = 78.4 J and with sig figs, it would be 80 J

6 0
2 years ago
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15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers
timofeeve [1]

The percent difference between two numbers x and y is given by

\dfrac{|y-x|}x \times 100\%

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from x to y or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

\dfrac{|6-3|}3 \times 100\% = 100\%

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

\dfrac{3-6}3\times100\%=-50\%

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take x from the left column and y from the right column.

\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%

\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%

\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%

3 0
2 years ago
It's five miles from Tim's house to Rita's house. Roughly how long is this distance in feet?
sergey [27]

Answer:

26400 ft

Explanation:

that would be my answer

hope this helps!!! :)

3 0
3 years ago
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Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor
hram777 [196]

Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs m = 300 \times 10^{-3} Kg

Velocity of snowball v = 10 \frac{m}{s}

For finding the average force,

Force is equal to the change in momentum,

   F = \frac{dP}{dt}

Here, final velocity is zero so we write,

 F = \frac{mv}{1}

Where dt = 1 sec

 F = 300 \times 10^{-30} \times 10

F = 3 N

Above value of force is due to one ball, but here given in question there are two ball,

F = 3 \times 2

F = 6 N

Therefore, the average force exerted on the window due to two snowballs is 6 N

5 0
4 years ago
A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive
Ivahew [28]

Answer:

a

The  x- and y-components of the total force exerted is

           F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

b

 The magnitude of the force is  

            |F_{31 +32}| = 10.25 *10^{-5} N

   The direction of the force is  

         \theta =327.43 ^o   Clockwise from x-axis

Explanation:

From the question we are told that

    The magnitude of the first charge is q_1 = +5.00nC = 5.00*10^{-9}C

      The magnitude of the second charge is q_2 = -2.00nC = -2.00*10^{-9}C

        The position of the second charge  from the first one is  d_{12} = 4.00i \  cm = \frac{4.00i}{100} = 4.00i *10^{-2} m

        The  magnitude of the third charge is q_3 = +6.00nC = 6.00*10^{-9}C

       The position of the third charge from the first one is  \= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} =  (4i + 3j) *10^{-2}m

                |d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m

                |d_{31}| =5 *10^{-2} m

        The position of the third charge from the second  one is

                \= d_{32} = 3j cm = 3j *10^{-2}m

               |d_{32}| =(\sqrt{ 3^2}) *10^{-2} m

               |d_{32}| =3 *10^{-2} m

The force acting on the third charge due to the first and second charge is mathematically represented as

           F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}

 Substituting values

          F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}  \\ \ +  \ \ \ \ \ \ \ \ \   \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}

            F_{31 +32} = 2.16 *10^{-5} (4i + 3j)  - 12*10^{-5} j

            F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

The magnitude of     F_{31 +32}  is mathematically evaluated as

            |F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}

             |F_{31 +32}| = 10.25 *10^{-5} N

The direction is obtained as

            tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}

              \theta = tan ^{-1} [-0.63889]

             \theta = - 32.57 ^o

             \theta = 360 - 32.57

            \theta =327.43 ^o

               

                         

5 0
3 years ago
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