Answer:
585×10⁸ m
Explanation:
Distance = rate × time
d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)
d = 585×10⁸ m
D Because If Your Going To Have A Contest Its Ganna Have To Be The Same Objectives For Both Contenders
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
Answer:
In most materials, as heat energy is absorbed, the density decreases. IF a certain object is heated, it might appear bigger than usual because it expands as the molecules inside moves faster than usual. However the mass of it stays the same while the density decreases.
Answer:
Radius of cross section, r = 0.24 m
Explanation:
It is given that,
Number of turns, N = 180
Change in magnetic field, 
Current, I = 6 A
Resistance of the solenoid, R = 17 ohms
We need to find the radius of the solenoid (r). We know that emf is given by :


Since, E = IR




or

Area of circular cross section is, 


r = 0.24 m
So, the radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.