All categories

3 years ago

On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?

2 answers:

5 0

<h2>Answer:</h2><h3>-47.7778</h3><h2>Explanation:</h2><h3>To translate degrees Fahrenheit to degrees Celsius, deduct 30 from the temperature in degrees Fahrenheit and divide by 2 to get the temperature in degrees Celsius. = 27 °C (In fact, 84 °F is equal to 28.89 °C—again, a close approximation!).</h3>

Evgesh-ka [11]3 years ago

3 0

Answer:

Its going to be about -47.78°C

You might be interested in

A physical quantity, G, is defined by G = (Original mass x time)/(change in mass), what is the S.I. unit of G ?

Andrei [34K]

The gravitational constant (G) in its base SI units is

3/2
m
3
k
g
/
s
2

But is often seen written as

⋅
N
⋅
2/2
m
2
/
k
g
2

Where N is the Newton unit. N=kg ⋅
⋅
m/s 2
2

4 0

2 years ago

Can someone answer pls??

SCORPION-xisa [38]

Answer:

2. the volume of the square are the same

7 0

2 years ago

Read 2 more answers

Which property of gases allows them to be stored at high concentrations in a bottle of air freshener?

Agata [3.3K]

Answer:

Explanation:

the ability for gases to compress is extremely helpful it allows tanks of oxygen to hold enough air for up to two hours and the strange thing about compression is that it allows some liquids to stay liquid at their boiling point allowing liquid nitrogen to stay liquid at room temperature

5 0

3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi

WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

$m1v1 + m2v2 = m1'v1' + m2'v2'$

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

$0 + 0 = 91*v1' + 1300*0.17$

$v1' = -1300*0.17/91 = -2.4286\ m/s$

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

$distance = 2.4286 * 7.5 = 18.2145\ meters$

8 0

3 years ago