The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.
<h3>What is magnetic force?</h3>
A magnetic force is the force that act in a magnetic field.
To calculate the magnetic force, we use the formula below.
Formula:
- F = qvB.........Equation 1
Where:
- F = magnetic force
- q = point charge
- v = Velocity of the the charge
- B = Field strength
From the question,
Given:
- q = 5.0×10⁻⁷ C
- v = 2.6×10⁵ m/s
- B = 1.8×10⁻² T
Substitute these values into equation 2
- F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
- F = 23.4×10⁻⁴
- F = 2.34×10⁻³ N
Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.
Learn more about magnetic force here: brainly.com/question/2279150
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Answer:
Density of liquid = 4730 kg/m³
Atmospheric pressure on planet X = 8401.7 N/m²
Explanation:
Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm
So, ρgh = ρ₁gh₁
ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³
The atmospheric pressure on planet X
P = ρg₁h₃ g₁ = g/4 and h₃ = 725 mm = 0.725 m
on planet X
P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²
Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
Answer:
the tempature and the map and climate
Explanation
