Answer:
The velocity of the star is 0.532 c.
Explanation:
Given that,
Wavelength of observer = 525 nm
Wave length of source = 950 nm
We need to calculate the velocity
If the direction is from observer to star.
From Doppler effect
Put the value into the formula
Negative sign shows the star is moving toward the observer.
Hence, The velocity of the star is 0.532 c.
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
The average speed for the whole journey is 49.5 miles per hour.
Explanation:
Step 1 :
Here, both the ways, he covers the same distance. Then, the formula to find average speed is
= 2xy / (x+y)
Step 2 :
x ----> Rate at which he travels from New York to Washington
x = 45
y ----> Rate at which he travels from New York to Washington
y = 55
Step 3 :
So, the average speed is
= (2 ⋅ 45 ⋅ 55) / (45 + 55)
= 4950 / 100
= 49.5
Answer:
Displacement after 5 seconds is 155/2 meters
Explanation:
Let X (t) represent the equation of the position, then you have to d2x / dt2 = 5.
Applying the fundamental theorem of the calculation dx/dt = 5t + vo. The speed equation is V (t) = 5t + vo. Since the initial velocity is 30m/s, V (0) = 5 (0) + vo = 30. Therefore, V (t) = dx/dt = 5t + 30. Applying again the fundamental theorem of the calculation X (t) = 5t^2 / 2 + 30t + xo.
Displacement in 5 seconds is given by X (5) - X (0).
X (5) - X (0) = 5 (5)^2/2 +3 (5) + Xo - 5 (0)^2/2 -3 (0) -Xo = 155/2
Displacement after 5 seconds is 155/2 meters
The mass of the cold water, given the data from the question is 500 g
<h3>Data obtained from the question</h3>
- Mass of warm water (Mᵥᵥ) = 200 g
- Temperature warm water (Tᵥᵥ) = 75 °C
- Temperature of cold water (T꜀) = 5 °C
- Equilibrium temperature (Tₑ) = 25 °C
- Specific heat capacity of the water = 4.184 J/gºC
- Mass of cold water (M꜀) =?
<h3>How to determine the mass of the cold water </h3>
Heat loss = Heat gain
MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)
200 × 4.184 (75 – 25) = M꜀ × 4.184(25 – 5)
41840 = M꜀ × 83.68
Divide both side 83.68
M꜀ = 41840 / 83.68
M꜀ = 500 g
Learn more about heat transfer:
brainly.com/question/6363778
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