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schepotkina [342]

3 years ago

13

PLEASE HELP A cup of water and a swimming pool full of water are at 24°C. The _____ of the swimming pool is higher. 1.heat conte

nt 2.potential energy 3.temperature 4.kinetic energy

1 answer:

Dimas [21]3 years ago

4 0

Answer:

4. kinetic energy

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A 3.0 X 10^3 grams rock swings in a circle with a diameter of 1000 cm. Given the constant speed around the circle is 17.90 mph (

marissa [1.9K]

Answer:

a = 12.8 m/s^2

Explanation:

To find the centripetal acceleration you use the following formula:

$a_c=\frac{v^2}{r}$ (1)

v: tangential speed of the rock = 17.90mph

r: radius of the orbit = 1000cm/2 = 500cm = 0.5m

You first change the units of the tangential velocity in order to replace the values of v and r in the equation (1):

$17.90mph*\frac{1609.34m}{1\ m}*\frac{1\ h}{3600\ s}=8m/s$

Then, you use the equation (1):

$a_c=\frac{(8m/s)^2}{5m}=12.8\frac{m}{s^2}$

hence, the centripetal acceleration is 12.8 m/s^2

3 0

3 years ago

A client with hypertension who weighs 72.4 kg is receiving an infusion of nitroprusside (Nipride) 50 mg in D5W 250 ml at 75 ml/h

Mkey [24]

To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.

$\frac{x}{75ml/hour} \rightarrow \frac{50mg}{250ml}$

$x \rightarrow \frac{50mg*75ml/hour}{250ml}$

$x \rightarrow \frac{3750mg}{250hour}$

$x \rightarrow 15\frac{mg}{hour}$

Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore

$x \rightarrow 15\frac{mg}{hour}$

$x \rightarrow 15\frac{mg}{hour}(\frac{1000mcg}{1mg})(\frac{1hour}{60min})$

$x \rightarrow 250mcg/min$

The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,

$Dose = \frac{250mcg/min}{72.4kg}$

$Dose = 3.5 \frac{mcg/min}{kg}$

Therefore the client will receive 3.5mcg/kg/min.

8 0

3 years ago

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

2 years ago

For which pair of launch angles will two identical projectiles have equal ranges?. A. 19.24°, 80.54°B. 16.42°, 74.58°C. 60.23°,

AlladinOne [14]

The kinematic equations of motion that apply here are<span>y(t)=votsin(θ)−12gt2</span>and<span>x(t)=votcos(θ)</span>Setting y(t)=0 yields <span>0=votsin(θ)−12gt2</span>. If we solve for t, we obtain, by factoring,<span>t=<span>2vsin(θ)g</span></span>Substitute this into our equation for x(t). This yields<span>x(t)=<span><span>2v2cos(θ)sin(θ)</span>g</span></span><span>This is equal to x=<span><span>v^2sin(2θ)</span>g</span></span>Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. <span> 60.23°, 29.77° </span>

4 0

3 years ago

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A worker uses a cart to move a load of bricks weighing 680 N a distance of 10 m across a parking lot.

NNADVOKAT [17]

When you are finding work, the easiest way is to use the formula.

W = F*D

Where F is the force and D is the distance. Simply take the constant force of 209N and multiply it by the distance of 10m. Which will give you 2090J

4 0

3 years ago

Read 2 more answers