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bulgar [2K]
3 years ago
7

Does space have plenty of oxygen ​

Physics
2 answers:
xz_007 [3.2K]3 years ago
6 0

Answer:

I don't think so because we can't breathe in space

Explanation:

saveliy_v [14]3 years ago
5 0

Answer:

No

Explanation:

There is no oxygen in space, so astronauts must bring tanks of oxygen

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An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.
motikmotik

Answer:

5 i think

Explanation:

4 0
4 years ago
A book sitting on a table is moved horizontally. Describe the
Akimi4 [234]

Frictional force and Applied force has same “magnitude” and “opposite” direction.  

Option: B  

<u>Explanation</u>:  

When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.

When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.

Formula for "frictional force" is \mu\times N

Where, \mu is coefficient of friction and N is normal force.

3 0
3 years ago
What is the measurement 111.009 mm rounded off to four significant digits?
dedylja [7]
111.0 because 111.009 rounds off to 111.01, thus rounding again off to 111.0
4 0
3 years ago
.
Dmitry [639]

Answer: 176.4 J

Explanation:

5 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
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