Answer:
20 m/s
Explanation:
Given:
Δy = 0.02 km = 20 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (9.8 m/s²) (20 m)
v = 19.8 m/s
Rounded to one significant figure, the final velocity is 20 m/s towards the ground.
Answer:
Explanation:
mass of object, m = 3 kg
spring constant, K = 750 n/m
compression, x = 8 cm = 0.08 m
angle of gun, θ = 30°
(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.
(b) Let v be th evelocity of ball at the tim eof launch.
by using the conservation of energy
1/2 Kx² = 1/2 mv²
750 x 0.08 x 0.08 = 3 x v²
v = 1.265 m/s
By use of the formula of maximum height


h = 0.02 m
h = 2 cm
Answer:
B = 1.353 x 10⁻³ T
Explanation:
The Magnetic field within a toroid is given by
B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.
To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,
r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.
b = 25.1 + 3 = 28.1 cm
a = 25.1 cm
r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m
B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266
B = 1.353 x 10⁻³ T
The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T
Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by

Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get

Solving the above equation we get
we N= 2.313≅ 2