Ask question

Ask question

All categories

3 years ago

8

The specific heat capacity of oxygen is 918j/kg c. That means that it takes 918 J of energy to raise ____ kg of oxygen by 1 degr

ee Celsius. What number completes the sentence

1 answer:

Alborosie3 years ago

6 0

Answer:

1kg of oxygen(918joules of energy per 1kg of oxygen by 1 degree Celsius)

You might be interested in

16. Calculate the mass in kg of an object that is being accelerated 12 m/s with a force of 654 N.

Cloud [144]

(654m/s ) / (12m/s) = 54.5 kg.

3 0

3 years ago

Read 2 more answers

A friend of yours who has not taken astronomy sees a meteor shower (she calls it a bunch of shooting stars). The next day she co

Oxana [17]

Tell her meteors and stars aren't related a stars life time is extremely long and meteors are just rocks floating through space when they fly be the earth they can appear like falling stars but they aren't her fave constellation wont be going away any time soon

3 0

3 years ago

An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a

vaieri [72.5K]

363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

$Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}$

$v=\lambda \times f$ --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,

$L=\frac{n \lambda}{2}$

For first harmonic, n =1

$L=\frac{\lambda}{2}$

For second harmonic, n =2

$L=\frac{2 \lambda}{2}=\lambda$

For third harmonic, n =3

$L=\frac{3 \lambda}{2}$

$\lambda=\frac{2 L}{3}$ -------> eq. 2

Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get

$v=\frac{2 \times 0.58 \times 989}{3}=\frac{1089.24}{3}=363.08 \mathrm{m} / \mathrm{s}$

3 0

3 years ago

For main sequence stars what tends to happen to the absolute brightness of a star as the temperature rises? How do you know?

professor190 [17]

With the increase in the temperature of the star, the brightness of the stars will also increase.

<u>Explanation:</u>

The brightness and surface temperature of stars ordinarily increment with age. A star stays close to its underlying situation on the fundamental arrangement until a lot of hydrogen in the center has been devoured, at that point starts to advance into a progressively brilliant star.

The brightness of a star relies upon its structure and how far it is from the planet. Space experts characterize star brilliance as far as clear extent — how splendid the star shows up from Earth — and outright greatness — how brilliant the star shows up at a standard separation

6 0

2 years ago

After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While moving down the track at 10

quester [9]

Answer:

F = 2325 N

Explanation:

given,

mass of the car, m = 1550 Kg

initial speed, u = 10 m/s

final speed, v = 25 m/s

time, t = 10 s

Average net force = ?

acceleration of the car

$a = \dfrac{v-u}{t}$

$a = \dfrac{25-10}{10}$

a = 1.5 m/s²

we know

F = m a

F = 1550 x 1.5

F = 2325 N

Net average force applied on the car is equal to 2325 N.

6 0

3 years ago

Read 2 more answers