It goes through your left atrium and right atrium
Answer:
They measured their train using their handspan
The component of the force in negative z-direction is -0.144 N.
The given parameters;
- <em>current in the wire, I = 2.7 A</em>
- <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
- <em>magnetic filed, B = 1.24 i</em>
The force on the segment of the wire is calculated as follows;
where;
- <em>θ is the angle wire and magnetic field</em>
<em />
The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.
The angle between the wire and the magnetic field is calculated as follows;
The magnitude of the wire length is calculated as follows;
The component of the force in negative z-direction is calculated as;
Thus, the component of the force in negative z-direction is -0.144 N.
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Does this help?
When an object is
immersed in a fluid (in this case water, but may include both liquids and
gases) the fluid exerts an upward force on the object which is called buoyancy
force or <span>up-thrust. Archimedes’ Principle states that the buoyant
force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by
that object. Thus when an object is
placed in water the rise in the water level is dictated by the mass of that
object.</span>
<span>
</span>
<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
