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3 years ago

To Play Ice Hockey each player needs

Physics

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:

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Help! Il give brainlest to who answers first

Readme [11.4K]

Answer:

1. The density of the cube is 1.03 g/mL.

2. Dish soap

Explanation:

1. Determination of the density of the cube.

From the question given above, the following data were obtained:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Thus, density can be expressed mathematically as:

Density (D) = mass (m) / volume (V)

D = m / V

With the above formula, we can obtain the density of the cube as follow:

Mass (m) of cube = 21.7 g

Volume (V) of cube = 21 mL

Density (D) of cube =?

D = m / V

D = 21.7 / 21

D = 1.03 g/mL

Thus, the density of the cube is 1.03 g/mL.

2. Determination of the layer of density the cube will settle in.

From the question given above,

Subtance >>>>>>>> Density

Vegetable oil >>>>> 0.91 g/mL

Grape juice >>>>>> 0.97 m/L

Water >>>>>>>>>>> 1 g/mL

Dish soap >>>>>>>> 1.03 g/mL

Maple syrup >>>>>> 1.37 g/mL

Comparing the density of the cube (i.e 1.03 g/mL) with those in the table able, we can conclude that the cube will settle in the DISH SOAP layer since they both have the same density.

7 0

3 years ago

Which of the following is evidence that supports the idea of uniformitarianism?

Inga [223]

D. rates of soil erosion are much lower during droughts that last several years

5 0

3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$