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3 months ago

The total energy of a 0.050 kg object travelling at 0.70 c is:

2 answers:

8 0

The answer is 1.1025×10^15 Joules. There is no correct option. The type of energy the object holds is kinetic energy, which is related to the motion of the object. Kinetic energy is calculated using the formula KE = 1/2 MV², where M denotes the mass of the object, which is 0.05 kg in this case, and V signifies its velocity. Given that the object is moving at 0.7 times the speed of light (c), we calculate its speed as 0.7c = 0.7(3×10^8), resulting in a speed of 2.1×10^8. By plugging these values into the kinetic energy equation, we find KE = 1/2 × 0.05 × (2.1×10^8)², yielding KE = 1.1025×10^15 Joules. There is no correct option.

Maru [3.3K]3 months ago

3 0

What does the letter c represent in this context?

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A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Softa [3030]

B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.

4 0

3 months ago

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

Maru [3345]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Consequently, the temperature difference across the material will be $\Delta T = 375 K$

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

$\Delta T$ is the temperature difference we need to determine

represents the thickness of the material $\Delta x=2.5 cm =0.025 m$

To isolate $\Delta T$ from equation (1), we obtain:

$\Delta T =\frac{Q \Delta x}{Ak}$

Initially, we convert 3KW to W, resulting in:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

With all variables accounted for, we can substitute and calculate:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Thus, the temperature difference across the material will be $\Delta T = 375 K$

5 0

2 months ago

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

ValentinkaMS [3465]

Answer:

20 cm

Explanation:

The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².

Thus, solving for r gives us r = kq₁q₂/U

which leads to x - 2 = kq₁q₂/U

Then, rearranging gives x = 0.02 + kq₁q₂/U m

So, x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm

7 0

2 months ago