Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
Answer:
5.843 m
Explanation:
suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.
lets consider that horizontal motion
distance = speed * time
time = 40/ 37 = 1.081 s
arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.
applying motion equation
(assume g = 10 m/s²)
Arrow misses the target by 5.843m ig the arrow us split horizontally
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
Explanation:
a ) No of turns per metre
n = 450 / .35
= 1285.71
Magnetic field inside the solenoid
B = μ₀ n I
Where I is current
B = 4π x 10⁻⁷ x 1285.71 x 1.75
= 28.26 x 10⁻⁴ T
This is the uniform magnetic field inside the solenoid.
b )
Magnetic field around a very long wire at a distance d is given by the expression
B = ( μ₀ /4π ) X 2I / d
= 10⁻⁷ x 2 x ( 1.75 / .01 )
= .35 x 10⁻⁴ T
In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .
Answer:
Expression of work done is
Work done to move the sled is given as 1.94 J
Explanation:
As we know that the formula of work done is given as
here we know that
F = 6 N
d = 0.4 m
so we will have
