Answer:
No, the magnitude of the magnetic field won't change.
Explanation:
The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.
Answer:
Fog forms in a valley
Explanation: that’s the only reasonable response
Strength of the magnetic field: 20 T
Explanation:
For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by
![\epsilon=BvL](https://tex.z-dn.net/?f=%5Cepsilon%3DBvL)
where
B is the strength of the magnetic field
v is the speed of the wire
L is the length of the wire
For the wire in this problem, we have:
(induced emf)
L = 0.20 m (length of the wire)
v = 3.0 m/s (speed)
Solving for B, we find the strength of the magnetic field:
![B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cepsilon%7D%7BvL%7D%3D%5Cfrac%7B12%7D%7B%280.20%29%283.0%29%7D%3D20%20T)
Learn more about magnetic fields:
brainly.com/question/3874443
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Answer:
9. (B) ¼ Mv²
10. (A) √(3gL)
11. 20 N
12. 5 m/s²
Explanation:
9. The rotational kinetic energy is:
RE = ½ Iω²
RE = ½ (½ MR²) (v/R)²
RE = ¼ Mv²
10. Energy is conserved.
Initial potential energy = rotational energy
mgh = ½ Iω²
Mg(L/2) = ½ (⅓ ML²) ω²
g(L/2) = ½ (⅓ L²) ω²
gL = ⅓ L² ω²
g = ⅓ L ω²
ω² = 3g / L
ω = √(3g / L)
The velocity of the top end is:
v = ωL
v = √(3gL)
11. Sum of torques about the hinge:
∑τ = Iα
-(Mg) (L/2) + (T) (r) = 0
T = MgL / (2r)
T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)
T = 20 N
12. Sum of forces on the block in the -y direction:
∑F = ma
mg − T = ma
Sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a / R)
T = ½ Ma
Substitute:
mg − ½ Ma = ma
mg = (m + ½ M) a
a = mg / (m + ½ M)
Plug in values:
a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))
a = 5 m/s²
<h3>
Answer:</h3>
2.5 mg
<h3>
Explanation:</h3>
<u>We are given</u>;
- Original mass of I-131 is 40.0 mg
- Half life of I-131 is 8 days
- Number of days 32 days
We are required to determine the mass that will remain;
N = N₀ × 0.5^n
where, N is the remaining mass, N₀ is the original mass, and n is the number of half lives.
Therefore;
n = time ÷ half life
= 32 days ÷ 8 days
= 4
Therefore;
Remaining mass = 40.0 mg × 0.5^4
= 2.5 mg
Hence, the remaining mass of I-131 after 32 days is 2.5 mg