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3 years ago

13

To Play Ice Hockey each player needs

2 answers:

Sholpan [36]3 years ago

7 0

Answer:

If one plays ice hockey he should must have necessary equipment to support his sport. These equipment include Ice Skates,Helmet with Cage and Mouth-guard:, Hockey stick, Hockey pants,Hockey gloves, shoulder pads, elbow pads, Shin Guard:Neck guard and Jockstrap (men) or Pelvic protector (women)

Explanation:

Alexxx [7]3 years ago

5 0

Answer:

If one plays ice hockey he should must have necessary equipment to support his sport. These equipment include Ice Skates,Helmet with Cage and Mouth-guard:, Hockey stick, Hockey pants,Hockey gloves, shoulder pads, elbow pads, Shin Guard:Neck guard and Jockstrap (men) or Pelvic protector (women).

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A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

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3 years ago

Distinguish between speed and velocity?<br>atleast 4 point should write

Sergio [31]

Answer:

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3 years ago

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one

alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R= $13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ ohm

Resistance of long piece of wire= $7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire = $\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K= $\frac{\rho}{A}$ =Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'= $\frac{L}{8}$

Length of long piece of wire= $L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of wire= $\frac{\frac{L}{8}}{L}\times 100=12.5%$ %

The resistance of the short piece= $\frac{13}{8}\Omega$

The resistance of the long piece= $\frac{91}{8}\Omega$

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The solution for this problem is:

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ω2 = 0

α = - ω1/t = -2π N /t

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α = -2π 8.33 /2.6 =- 20 rad/s^2

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tatuchka [14]

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