Answer:
Final velocity of the block = 2.40 m/s east.
Explanation:
Here momentum is conserved.
Initial momentum = Final momentum
Mass of bullet = 0.0140 kg
Consider east as positive.
Initial velocity of bullet = 205 m/s
Mass of Block = 1.8 kg
Initial velocity of block = 0 m/s
Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s
Final velocity of bullet = -103 m/s
We need to find final velocity of the block( u )
Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u
We have
2.87 = -1.442 + 1.8 u
u = 2.40 m/s
Final velocity of the block = 2.40 m/s east.
Answer:
Δy = 6.05 mm
Explanation:
The double slit phenomenon is described by the expression
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference
m = 0,±1, ±2, ...
As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea
tan θ = y / x
y = x tan θ
In the interference experiments the measured angle is very small so we can approximate the tangent
tan θ = sin θ / cos θ
cos θ = 1
tan θ = sin θ
y = x sin θ
We substitute in the destructive interference equation
d (y / x) = (m + ½) λ
y = (m + ½) λ x / d
The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it
m = 0
y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³
y₀ = 1.511 10⁻³ m
m = 2
y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³
y₂ = 7.556 10⁻³ m
The separation between these strips is Δy
Δy = y₂-y₀
Δy = (7.556 - 1.511) 10⁻³
Δy = 6.045 10⁻³ m
Δy = 6.05 mm
Your question has been heard loud and clear.
Averge velocity formula= Total distance travelled / total time taken.
Total distance=7meters
Total time taken=9 seconds.
Average velocity = 7/9= 0.77 metres/second
Average velocity= 0.77m/s
Thank you.
The magnetic field generated by a wire carrying a current I is:
where r is the distance at which the magnetic field is measured, and
is the magnetic permeability in vacuum.
The problem says that the magnetic field at a distance r=12 cm=0.12 m from the wire must be no larger than
. Substituting these values, we can find the maximum value of the current I that the wire can carry:
