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2 years ago

5

On a distance vs time graph, what does the line of an object at constantspeed look like?

1 answer:

9966 [12]2 years ago

4 0

Answer: Diagonal going up

Explanation:

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

8 0

2 years ago

A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

defon

Answer:

2940.1 joules would you burn in climbing stairs all day.

Explanation:

Work = W = F $\times$ d

going up stairs would be against force of gravity

W = mgh

where h is the height

the question is not complete because we need speed or distance

h = v $\times$ t

so assuming 1 step per second

h = 86,400 steps $\times$ 7inchs/step $\times$ 0.0254 m/inch

h = 15362 m

so from this

W = 800 N $\times$ 15362

= 12289600 J

that means YOU need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal

Kcal = $\frac{W}{(J/Kcal)}$

= $\frac{12289600}{4180}$

= 2940.1 Kcal

if you ran faster you would use more energy 2 steps per second would mean 5880 Kcal.

8 0

3 years ago

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

Which best explains why scientific theories grow stronger over time?

vampirchik [111]

If a theory is studied in let's say the 17th century, the theory has had many years to be studied and explained by many different people many different ways.

3 0

2 years ago

What is the correct equation for calculating the average atomic mass for 3 isotopes? (pls be 100%of your answer pls no guessing)

mariarad [96]

<u>Answer:</u>

<em>The correct equation for measuring the average microscopic weight for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>

<u>Explanation:</u>

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

Take the correct weight of each isotope (that will be in decimal form)

Multiply the weight of each isotope by its abundance

Add each of the results together.

<em>This gives the required average microscopic weight of the three isotopes.</em>

3 0

2 years ago