The force that acts on all objects, all the time on Earth is gravitational force.
The force that surface exert on an object perpendicularly is normal reaction.
<h3>What force acts on all objects, all the time on Earth?</h3>
- Force due to gravity is gravitational pull on objects due to its position on earth's surface.
The force due to gravity on object's is calculated by applying Newton's second law of motion as follows;
F = mg
where;
- m is the mass of the object
- g is acceleration due to gravity
The force that surface exert on an object perpendicularly is normal reaction.
Thus, the force that acts on all objects, all the time on Earth is gravitational force.
Learn more about force of gravity here: brainly.com/question/2537310
Answer:
B
Explanation:
Speed is the magnitude of the velocity vector, so it can never be negative.
Answer:
The focal length of eyepiece is 3.68 cm.
Explanation:
Given that,
Distance = 19 cm
Focal length = 5.5 mm
Magnification = 200
Object distance = -25 cm
We need to calculate the focal length
Using formula of magnification
Put the value into the formula
Hence, The focal length of eyepiece is 3.68 cm.
Answer:
Physical Properties of Sodium
Atomic number 11
Melting point 97.82°C (208.1°F)
Boiling point 881.4°C (1618°F)
Volume increase on melting 2.70%
Latent heat of fusion 27.0 cal/g
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Home Periodic table Elements Sodium
Sodium - Na
Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium
Atomic number
11
Atomic mass
22.98977 g.mol -1
Electronegativity according to Pauling
0.9
Density
0.97 g.cm -3 at 20 °C
Melting point
97.5 °C
Boiling point
883 °C
Vanderwaals radius
0.196 nm
Ionic radius
0.095 (+1) nm
Isotopes
3
Electronic shell
[Ne] 3s1
Energy of first ionisation
495.7 kJ.mol -1
Answer:
The magnification of an astronomical telescope is -30.83.
Explanation:
The expression for the magnification of an astronomical telescope is as follows;
Here, M is the magnification of an astronomical telescope, 
is the focal length of the eyepiece lens and 
is the focal length of the objective lens.
It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.
Put 
and 
in the above expression.
M=-30.83
Therefore, the magnification of an astronomical telescope is -30.83.
