Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F

F = 0.679 N
Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:

We know that the electric field is:

Where:
- k is the Coulomb constant
- q is the charge
- r is the distance from the charge to the point
So, we have:

Let's solve it for r(3).
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!
Answer:
Part a)

Part b)

Explanation:
Part a)
Level of sound = 75 dB
now we know that

here we know that

now we have


Part b)
Intensity of sound wave is given as

here we know that

so we have


now we know



now we have


Answer:
0.125 m
Explanation:
Pressure in fluids is given as the product of density, height and acceleration due to gravity and expressed as
P=hdg
Where h is the height, d is density, g is acceleration due to gravity and P is pressure.
Making h the subject of formula then
h=P/dg
Given specific gravity of a substance, its density is equal to specific gravity multiplied by density of water. Taking density of pure water as 1000 kg/m³ then the density of reference fluid will be 1.05*1000=1050 kg/m³
Substituting pressure with 1.29*10³ pa as given then taking g as 9.81 m/s² then
H=1.29*10³÷(9.81*1050)=0.1252366389981068880151448958788408329692m
Rounded off, the height is approximately 0.125 m
The horizontal component of the magnetic field is 12.6 μT.
The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.
The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.
A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.
Let B be the magnetic field and R be the radius of the circular coil.
Then the horizontal component of the Earth's magnetic field is given as:
B(h) = B(coil) = μ₀ NI / 2R
B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3
B(h) = 12.6 μT
Learn more about magnetic field here:
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