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2 years ago

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Felipe walks from the house to his truck on the way to work. He walks 20m to the truck and another 60m in his truck for a total

of 20s. What is Felipes average velocity over the 20s period? What is Felipes average speed over the 20s period?

2 answers:

Wewaii [24]2 years ago

4 0

Answer: the answer is 4.0 for both

Explanation: on khan academy !!!

ANEK [815]2 years ago

3 0

Answer:

Felipe's average speed over the 20s period is 14.4 km/h.

Explanation:

To determine the speed at which Felipe has walked, the distance traveled and the time it took him must be considered.

Thus, Felipe walked 80 meters (20 towards the truck, and 60 inside it) in a total of 20 seconds. That is, he traveled 4 meters per second (80/20 = 4).

Now, to transfer said speed to kilometers per hour, the following calculation must be performed:

1 hour = 60 minutes

1 = 60 seconds

Number of seconds in an hour: 60 x 60 = 3600

Thus, Felipe has walked at an average speed of 14,400 meters per hour (4 x 3,600 = 14,400). So, since a kilometer consists of 1000 meters, its average speed in kilometers per hour was 14.4 km / h (14,440 / 1,000 = 14.4).

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60

Explanation:

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2 years ago

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In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t

mina [271]

Answer:

n = 2.06 moles

Explanation:

The absolute pressure at depth of 27 inches can be calculated by:

Pressure = Pressure read + Zero Gauge pressure

Zero Gauge pressure = 14.7 psi

Pressure read = 480 psi

Total pressure = 480 psi + 14.7 psi = 494.7 psi

P (psi) = 1/14.696 P(atm)

So, Pressure = 33.66 atm

Temperature = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

T = 298.15 K

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

33.66 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K

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2 years ago

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (<u>becarefull, gravity at this point is no longer</u> $9.8m/s^{2}$ <u>because we are not in the surface anymore</u>)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

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Answer:

5.59 m/s

Explanation:

We are given;

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Force = 615 N

Time(t) = 1 s

Now, the formula for force is;

Force = mass x acceleration

Thus;

615 = 110 × acceleration

\Acceleration(a) = 615/110 = 5.591 m/s²

Now, using Newton's first law of motion, we can find acceleration (a). Thus;

v = u + at

v = 13.41 + (5.591 × 1)

v ≈ 19 m/s

So,the change in velocity is;

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Answer:

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no of electrons in charge

Q = ne

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n. = 1.0125 x 10 ^19

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2 years ago