If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.
If the angle is 45, then there is still some work involved.
The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90.
12N because you are just adding those two up on the same side
Answer:
Explanation:
Velocity of plane relative to ground V_pg = ?
Given the velocity in vector form ,
velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j
V_wg = 60 i
V_pg = V_pw +V_wg
= 120 cos30 i + 120sin30j + 60i
= 164 i + 60 j
magnitude
=251 km / h
=
Answer:
The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.
Explanation:
Given that,
The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.
A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.
We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.
Using Snell's law for red light as :
Again using Snell's law for blue light as :
The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
