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3 years ago

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A child is pushing a shopping cart at a speed of 1.5 m/s how long will it take this child to push the cart down and I'll with th

e length of 9.3 m

2 answers:

Gelneren [198K]3 years ago

7 0

(9.3 meters) divided by (1.5 meters/sec) = 6.2 seconds .

Here's a suggestion:
When you dictate some text into a machine, take a few seconds
to check and see what the machine thinks you said.

maks197457 [2]3 years ago

4 0

The answer to this question is 6.2 seconds

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

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3 years ago

Which of the following is a component of staying positive while competing

Natalija [7]

Answer: B - complimenting others on good plays

Explanation: Reading the first words sort of give it away when staying positive you compliment, not criticize, confront angrily, or refuse.

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3 years ago

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Potential difference is measured in which units?<br> volts<br> amps<br> currents<br> watts

viktelen [127]

It is Volt
First option

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3 years ago

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The velocity of a car changes from 20 m/s east to 5 m/s east in 5 seconds. What is the acceleration of the car?

Klio2033 [76]

$acceleration = \frac{ v_{2}- v_{1} }{t} = \frac{5-20}{5} =-3m/s^{2}$

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3 years ago

Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer

sveticcg [70]

by cosine law we know that

$c^2 = a^2 + b^2 - 2 abcos\theta$

$\theta = 180 - 54.3 - 79.1 = 46.6 degree$

now using above equation

$93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6$

$4942.03 = b^2 - 85.34 b$

$b^2 - 85.34b - 4942.03 = 0$

by solving above quadratic equation we have

$b = 124.9 m$

so it is at distance 124.9 m from deer a

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3 years ago