Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
1. 1 s = 1 x 10⁶ μs
2. 1 g = 0.001 kg
3. 1 km = 1000 m
4. 1 mm = 1 x 10⁻³ m
5. 1 mL = 1 x 10⁻³ L
6. 1 g = 100 dg
7. 1 cm = 1 x 10⁻² m
8. 1 ms = 1 x 10⁻³ s
Explanation:
1.
1 x 10⁻⁶ s = 1 μs
(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs
<u>1 s = 1 x 10⁶ μs</u>
2.
1000 g = 1 kg
1 g = 1/1000 kg
<u>1 g = 0.001 kg</u>
3.
<u>1 km = 1000 m</u>
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4.
<u>1 mm = 1 x 10⁻³ m</u>
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5.
<u>1 mL = 1 x 10⁻³ L</u>
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6.
1 x 10⁻² g = 1 dg
(1 x 10⁻² x 10²) g = 1 x 10² dg
<u>1 g = 100 dg</u>
<u></u>
7.
<u>1 cm = 1 x 10⁻² m</u>
<u></u>
8.
<u>1 ms = 1 x 10⁻³ s</u>
Answer:
reliability
accuracy
Explanation:
If a reading of a measurement is consistently the same then the measurement is reliable.
If a reading of measurement is close the actual value of the measurement then the reading is accurate.
Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.
Hence, the radar has problems in its reliability and accuracy
Answer:
(a) Angular acceleration is 1.112 rad/s².
(b) Average angular velocity is 2.78 rad/s .
Explanation:
The equation of motion in Rotational kinematics is:
θ = θ₀ + 0.5αt²
Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.
(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:
13.9 = 0 + 0.5α(5)²
α = 1.112 rad/s²
(b) The equation of average angular velocity is:
ω = Δθ/Δt
ω = 
ω = 2.78 rad/s
Answer:The train travels 105 meters after applying the brakes
Explanation:If he decelerates 1.5 every minute, then he went from 28,5 m/s, to 27.0 m/s, to 25.5 m/s, to 24.0 m/s, after 4 seconds. Add all this together and youll get 105 meters moved in 4 seconds after he hit the brakes, I dont have a notebook on me though sorry :/