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Vanyuwa [196]
2 years ago
13

To Play Ice Hockey each player needs

Physics
2 answers:
Sholpan [36]2 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]2 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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An electric circuit can have no current when a switch is
Finger [1]

Answer:

Open

Explanation:

A switch is a part of a circuit where a connection can be made or broken. By convention, when the switch is "open", the connection is broken and current cannot pass. When the switch is "closed", the connection is complete and current can pass.

6 0
2 years ago
A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the
iren [92.7K]

Answer:

work = 1728

Power = 134

Explaination:

by using the formula,

Work(W)= Force(F)×Distance(D)

<h2> and</h2>

Power(P)= Work(W)/Time taken(T)

4 0
3 years ago
Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.
Artemon [7]
The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant =  8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
8 0
3 years ago
The time period T of a simple pendulum is given by the relation
Vanyuwa [196]

Answer:

T^2 \propto L

Explanation:

The period of a simple pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

From this equation we can write

T\propto \sqrt{L}\\T\propto \frac{1}{\sqrt{g}}

Taking the square of this equation, we get:

T^2 = (2\pi)^2 \frac{L}{g}

So we see that T^2 is proportional to L and inversely proportional to g. So, we can write:

T^2 \propto L\\T^2 \propto \frac{1}{g}

So the only correct option is

T^2 \propto L

5 0
3 years ago
Read 2 more answers
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

   The  focal length of  B  is  f_{objective } =  43.0 \ cm

    The focal length of  A  is   f_{eye} =  10.4 \  cm

The  theoretical angular  magnification is mathematically represented as

           m = \frac{f_{objective }}{f_{eye}}  =  \frac{43.0}{10.4}

            m = \frac{f_{objective }}{f_{eye}}  =  4.175

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular  magnification lies within the angular magnification range

3 0
3 years ago
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