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Mars2501 [29]
3 years ago
5

A Canadian driving from Quebec to Montreal finds he has traveled 271 km. How many miles is this? There are 1.61 km in 1 mi

Physics
1 answer:
kotegsom [21]3 years ago
3 0

Answer:

168.32 mile

Explanation:

1 mile = 1.61 km

1.61 km = 1 mile

1 km = 1 / 1.61 mile

So, 271 km = 271 / 1.61 = 168.32 mile

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A scientist compares two samples of colorless gas present at the beginning and end of a lab process. She wants to determine whet
kompoz [17]

Joseph Priestley was born in Yorkshire, the eldest son of a maker of wool cloth. His mother died after bearing six children in six years. Young Joseph was sent to live with his aunt, Sarah Priestley Keighley, until the age of 19. She often entertained Presbyterian clergy at her home, and Joseph gradually came to prefer their doctrines to the grimmer Calvinism of his father. Before long, he was encouraged to study for the ministry. And study, as it turned out, was something Joseph Priestley did very well.

Aside from what he learned in the local schools, he taught himself Latin, Greek, French, Italian, German and a smattering of Middle Eastern languages, along with mathematics and philosophy. This preparation would have been ideal for study at Oxford or Cambridge, but as a Dissenter (someone who was not a member of the Church of England) Priestley was barred from England's great universities. So he enrolled at Daventry Academy, a celebrated school for Dissenters, and was exempted from a year of classes because of his achievements.

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 https://www.acs.org/content/acs/en/education/whatischemistry/landmarks/josephpriestleyoxygen.html

8 0
3 years ago
Read 2 more answers
The New Horizons probe that passed by Pluto during July 2015 is one of the fastest spacecraft ever assembled. It was moving at a
LenKa [72]

Answer:

2.91 x 10¹² sec

Explanation:

d = distance of nearest star, Proxima Centauri  = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m

v = speed of new horizon probe = 14 km/hr = 14000 m/s

t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?

Using the equation

d = v t

Inserting the values given

4.3 x 9.46 x 10¹⁵ = (14000) t

t = 2.91 x 10¹² sec

7 0
3 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
What are two examples of goods and two examples of services
julsineya [31]

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

8 0
4 years ago
20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot
Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

\Delta E = -\frac{dV}{dr}

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

\Delta E = 0

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0
3 years ago
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