Answer:
Magnification, m = 3
Explanation:
It is given that,
Focal length of the lens, f = 15 cm
Object distance, u = -10 cm
Lens formula :
v is image distance
Magnification,
So, the magnification of the lens is 3.
Answer:
d) 14Ω
Explanation:
series resistors add so therefore 2+4+8=14
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N
Answer:
resultant is a single force that can replace the of a number of forces , equililbrant is a force that is exactly opposite to resultant
Explanation:
The answer is 5.88 · 10⁻⁷<span> m.</span>
To calculate this we will use the light equation:
v = λ · f,
where:
v - the speed of light (units: m/s)
<span>λ - the wavelength of the ray (units: m)
</span>f - the frequency of the ray (units: Hz = 1/s <span>since Hz means cycles per second (f=1/T))
</span>
It is given:
f = 5.10 · 10¹⁴ Hz = 5.10 · 10¹⁴<span> 1/s
v = 2.998 </span>· 10⁸<span> m/s
</span><span>λ = ?
</span>
If v = λ · f, then λ = v ÷ f:
λ = 2.998 · 10⁸ m/s ÷ 5.10 · 10¹⁴ 1/s
= 0.588 · 10⁸⁻¹⁴ · m
= 0.588 · 10⁻⁶ m
= 5.88 · 10⁻⁷ m