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3 years ago

Which of the following could be a potential safety hazard of indoor recreation?

2 answers:

5 0

d. a liquid spill on the floor that was not cleaned could be a potential safety hazard of indoor recreation

A liquid spill on the floor that was not cleaned could be a potential safety hazard of indoor recreation.

NOT:

a. slippery rocks and loose gravel
b. flash floods
<span>c. traveling without a compass, map, or GPS</span>

djyliett [7]3 years ago

3 0

Answer:

THE ANSWER IS D

Explanation:

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A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g sal

muminat

Answer:

The solubility of the salt is 35.16 (g/100 g of water).

It would take 71.09 grams of water to dissolve 25 grams of salt.

The percentage of salt that dissolves is 52.7 %

Explanation:

We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.

$\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}$

Working it a little this gives us :

$x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water}$

$x = 35.16 \ g \ salt$

So, the solubility of the salt is 35.16 (g/100 g of water).

Using the rule of three, we got:

$\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}$

Working it a little this gives us :

$x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}}$

$x = 71.09 g \ water$

So, it would take 71.09 grams of water to dissolve 25 grams of salt.

Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:

$\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}$

Working it a little this gives us :

$x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water}$

$x = 5.27\ g \ salt$

This is the salt dissolved

The percentage of salt dissolved is:

$percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}$

$percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}$

$percentage \ salt \ dissolved = 52.7 \%$

3 0

3 years ago

1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.

dimulka [17.4K]

Answer:

The acceleration of the crate is $0.3356\,\frac{m}{s^2}$

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

$F=m\,a$

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

$F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}$

7 0

3 years ago

An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.

iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

a) $\vec{v}$ = 70·i

b) The package will reach ground in approximately <u>12.77 seconds</u>.

c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.

d) The path of the package based on a stationary frame of reference is <u>parabolic</u>

e) The path of the package as seen from the plane is <u>directly vertical</u> downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, $\vec{v}$ = 70·i

b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;

$\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}$

Which gives;

$\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}$

The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>

c) The vertical velocity just before the package reaches the ground, $v_y$ , is given as follows;

$v_y^2$ = 2·g·h

Therefore;

$v_y$ = √(2·g·h)

Which gives;

$v_y$ = √(2 × 9.81 × 800) ≈ 125.28

$v_y$ ≈ 125.28 m/s

Which gives; $\vec{v}$ = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating <u>directly vertical</u>

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0

2 years ago

What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h i

kondor19780726 [428]

Answer:

-352.275KJ

Explanation:

We are given that

Mass of car=1500kg

Initial speed of car =u=96 km/h= $96\times \frac{5}{18}=26.67m/s$

1km/h= $\frac{5}{18}m/s$

Final speed of car=v=56km/h= $56\times \frac{5}{18}=15.56m/s$

Distance traveled by car=s=55m

We have to find the work done by the car's braking system.

Using third equation of motion

$v^2-u^2=2as$

$(15.56)^2-(26.67)^2=2a(55)$

$-469.18=110a$

$a=\frac{-469.18}{110}=-4.27m/s^2$

Where negative sign indicates that velocity of car decreases.

Work done by a car's barking system= $w=F\times s=ma\times s$

Work done by a car's barking system= $1500\times -4.27\times 55=352275J$

Work done by a car's barking system= $-\frac{352275}{1000}=-352.275KJ$

1KJ=1000J

Where negative sign indicates that work done in opposite direction of motion.

7 0

3 years ago

Read 2 more answers

Which statement is true about the type of light used by plants to make sugar during photosynthesis? Group of answer choices

denis-greek [22]

Answer:

4. It is infrared.

Explanation:

the photosystems on the chlorophyll absorb light at 600-700nm

3 0

3 years ago