Answer:
a) V = -0.227 mV
b) V = -0.5169 mV
Explanation:
a)
Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude
E = (qr) / (4πε₀R³)
As we know that
V = -
By solving above equation, we get
V = (-qr²) / (8πε₀R³)
When
R = 1.81 cm
r = 1.2 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)
V = -2.27 × 10⁻⁴ V
V = -0.227 mV
b)
When
r = R
R = 1.81 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-qR²) / (8πε₀R³)
V = (-q) / (8πε₀R)
V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))
V = -5.169 × 10⁻⁴ V
V = -0.5169 mV
Blood cell : Eukaryotic cell
and
Bacteria : Prokaryotic cell.
Answer:
So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.
Now,
We have F=ma,
and a=v/t,
so we can have another equation for it as,
Now, providing the required data to, it; ∴t =2 sec,
F=(1)×(20/2),
- So, the car would be acting the force,F of about 10 N while the car is present on the lower region of the track.
Answer:
The angle of launch is 52.49 Degree.
Explanation:
The Range R and Height H of a thrown object is calculated using the formula,
R=V₀² sin(2φ)/g
H=V₀²sin²(φ)/g
From these equations it can be written,
V₀²=R g/ sin(2φ)
V₀²=H g/ sin²(φ)
These values are equal so it can be written by equating these equations,
R g/sin(2φ)=H g/sin²(φ)
tan(φ)= 2H/R
Given H=72.3 m and R=111 m, the angle of launch is,
tan(φ)= 2*72.3/111
φ= 52.49 Degree.
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