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Maslowich
1 day ago
6

To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?

Physics
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A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while
Keith_Richards [3271]
To address this problem, Boyle's Law must be applied, which states that the initial and final pressures and volumes are related as follows: Where, P₀ and V₀ represent the initial pressure and volume, while P and V refer to the final pressure and volume. The endpoint pressure in this scenario is atmospheric pressure. Thus, using the given equation, we can find the volume the lungs would occupy at the surface.
4 0
2 months ago
Sally is pushing a shopping cart with a force of 20 N. Because the wheels are stuck, the friction caused by the ground is exerti
ValentinkaMS [3465]

The resultant force acting on the shopping cart is 12 N to the right.

This is a scenario related to Newton's Laws. A Free Body Diagram can be utilized to illustrate all the forces acting on the object. There are four forces at play:

  1. Gravity (g) acting downward.
  2. Normal force (N) acting upward.
  3. The force exerted by Sally (Fp) acting to the right.
  4. Friction force (Ff) acting to the left.

The first two forces (1 and 2) counteract one another as they are equal, while the overall force can be determined by combining forces (3) and (4)

F=Fp+Ff=20N+ (-8N)=12N

The positive value indicates that the shopping cart is moving towards the right.

Have a good day!

0 0
2 months ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
Softa [3030]
J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
7 0
2 months ago
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A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote
Keith_Richards [3271]
1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
U=q V
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is q=4.0 \mu C = 4.0 \cdot 10^{-6}C, and the potential energy is U=5.2 \cdot 10^{-3} J, thus we may rearrange the earlier formula to find the electric potential at the tip:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V

2) Using this same formula, if the charge changes to q=2.0 \mu C = 2.0 \cdot 10^{-6} C, the resulting electric potential will be:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V
8 0
2 months ago
Read 2 more answers
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