1. Which of the following is correct about the sampling distribution of the sample mean

During takeoff, an airplane goes from 0 to 56 m/s in 9 s. How fast is it going after 4 s?

meriva

24 miles per second
56/9=6
6+6+6+6=24

8 0

3 years ago

A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m

Advocard [28]

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending at the moment when the passenger let go of the wallet.

$v^2-u^2=2gh$

Where $g=9.8 m/s^2$

Substitute the values

$0-u^2=2(-160)\times 9.8$

$u^2=3136$

$u=\sqrt{3136}=56m/s$

Option (E) is true

8 0

3 years ago

Your average speed on the first half of a car trip is 69.0 km/h. How fast do you have to drive on the second half of the trip to

vova2212 [387]

Answer:

13 km/h

Explanation:

Average speed = distance/time

Let the total distance and total time taken for the whole trip be d km and t hours respectively

Average speed for the whole trip = 82 km/h

d = 82t

The distance covered in the first half = d1/2

Time taken = t/2

Average speed = 69 km/h

69 = d1/2 ÷ t/2

d1 = 69t

The distance covered in the second half = d2/2

Time taken = t/2

Let the average sly for the see half be A

A = d2/2 ÷ t/2

d2 = At

d = d1 + d2

82t = 69t + At

At = 82t - 69t

At = 13t

A = 13t/t = 13 km/h

3 0

3 years ago

For Jane to see an image, light must enter her eyes. What specifically is entering her eyes when she sees an image?

Sergio [31]

Energy
because once the light hits her eyes energy flows through her body so the answer is A energy

3 0

2 years ago

Read 2 more answers

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 $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago