Answer:
Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :
Put all the values,
So, the required magnetic field is equal to 
.
I think you should hold a stretch for 10-30 seconds
Answer:
23.889 Celcius
Explanation:
(75°F − 32) × 5/9 = 23.889°C
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
Answer:
F' = F
Explanation:
The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:
where,
F = Force of attraction
G = Universal gravitational costant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects
Now, if the masses and the distance between them is doubled:
<u>F' = F</u>
