can you fill in the blank ____ is the difference between one point of a wave to the same point in the next wave

Why are the types of shoes you wear in the kitchen important? What are the possible consequences of wearing inappropriate footwe

Nonamiya [84]

Why are the types of shoes you wear in the kitchen important?

• One of the best ways to prevent slip and fall accidents at work is to wear the right type of shoes. Footwear is paramount when it comes to safety in a kitchen setting. Grease spilled liquids, and vegetable debris can litter a chef's floor at any time, presenting major slip and fall hazards.

What are the possible consequences of wearing inappropriate footwear in the kitchen?

• Non-slip shoes greatly reduce the risk of accidents in the kitchen. Slips and falls can result in bruises, bleeding, broken bones, and concussions. If someone falls in the wrong place, he or she could receive serious burns or knife wounds.

5 0

3 years ago

A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature

Leona [35]

Answer: The specific heat of ice is 2.11 J/g°C

Explanation:

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of ice = 12.5 g

$m_2$ = mass of water = 85.0 g

$T_{final}$ = final temperature = 22.24°C

$T_1$ = initial temperature of ice = -15.00°C

$T_2$ = initial temperature of water = 25.00°C

$c_1$ = specific heat of ice = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]$

$c_1=2.11J/g^oC$

Hence, the specific heat of ice is 2.11 J/g°C

3 0

4 years ago

Match the variables to its definition

MA_775_DIABLO [31]

Answer:

See connections below

Explanation:

1 $\Rightarrow$ b

2 $\Rightarrow$ a

3 $\Rightarrow$ d

4 $\Rightarrow$ i

5 $\Rightarrow$ g

6 $\Rightarrow$ h

7 $\Rightarrow$ c

8 $\Rightarrow$ e

9 $\Rightarrow$ f

4 0

3 years ago

the speed of light in a certain medium is 0.6c. find critical angle , if the index of refraction is 1.67

baherus [9]

Answer:

$\theta_c = 36.78^o$

Explanation:

The relationship between the refractive index and the critical angle is given as follows:

$\eta = \frac{1}{Sin\ \theta_c} \\\\Sin\ \theta_c = \frac{1}{\eta}\\\\\theta_c = Sin^{-1}(\frac{1}{\eta} )$

where,

η = refractive index = 1.67

θc = critical angle =?

Therefore,

$\theta_c = Sin^{-1}(\frac{1}{1.67} )$

$\theta_c = 36.78^o$

4 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of 0,96 m from the ground.

8 0

3 years ago