-- If the object is moving with speed of 10.954 meters per second, then
it has 300J of kinetic energy no matter where it may be located.
-- If the object is 6.118 meters above somewhere, then it has 300J of
gravitational potential energy relative to that place.
Given :
Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.
To Find :
The magnitude and direction of
a) A + B .
b) A - B.
Solution :
Let , direction in north is given by +j and east is given by +i .
So ,
and 
Now , A + B is given by :


Direction of A+B is 45° north of west .
Also , for A-B :


Direction of A-B is 45° south of west .
( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)
Hence , this is the required solution .
The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps
So we want to know how can we detect infrared rays without an instrument. Infrared rays or heat, are a part of electromagnetic spectrum. We have specialized nerve cells in our skin called thermoreceptors that can detect differences in temperature that are produced by infrared part of EM spectrum.
Answer:
(a) The energy of the photon is 1.632 x
J.
(b) The wavelength of the photon is 1.2 x
m.
(c) The frequency of the photon is 2.47 x
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x
)
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x
Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x
Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x
) = (6.6 x
* 3 x
)/ λ
λ = 
= 1.213 x 
Wavelength of the photon is 1.2 x
m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x
= 6.6 x
x f
f = 
= 2.47 x
Hz
Frequency of the emitted photon is 2.47 x
Hz.