The resistance needed to be added is R
The Current is 2 ma
The voltage reading is a maximum of 50 volts.
The ma meter has an internal resistance of 40 ohms.
Formula
E = I * R
Givens
E = 50
I = 2 ms
R = R + 40
Solution
E = I * R
I = 2 ma [ 1 amp / 1000 ma] = 0.002 amp
50 = 0.002 * (R + 40) Divide by 0.002
50/0.002 = R + 40
25000 = R + 40 Subtract 40 from both sides.
R = 25000 - 40
R = 24960 Answer
Answer:
15 KJ
Explanation:
The quantity of heat (Q) required is given as:
Q = mcΔθ + mL
where m is the mass of ice, c is its specific heat capacity, L is its specific latent heat andΔθ is the change in temperature.
Given: m = 20g, temperature of ice = 0
, specific heat capacity of water = 4200 J/kg
, latent heat of fusion of ice = 3.3 x 10^(5) J/kg, temperature of water = 100
.
Q = m (cΔθ + L)
= 0.02(4200 x (100) + 330000)
= 0.02(420000 + 330000)
= 0.02 (750000)
Q = 15000
Q = 15000 Joules
Q = 15KJ
The quantity of heat needed to complete the conversion is 15 KJ.
Answer:
As the wavelength of an electromagnetic wave _decrease__ the frequency of the wave _increase_______.
Explanation:
What is the relationship between frequency and wavelength?
Wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.
That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency is:
V= fλ
where v= velocity
f= frequency
λ = wavelength
⇒ f = v/λ
also f ∝ 1/λ
For electromagnetic radiation, the speed is equal to the speed of light, c, and the equation becomes:
C= fλ
where c= Speed of light
f= frequency
λ = wavelength
⇒ f = v/λ
also f ∝ 1/λ
<span>The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it
That is X-93 so it will be
</span><span>Zirconium-93
</span>hope it helps
<span>A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away). The image of an object is found to be upright and reduced in size.</span>