A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
1 answer:
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Answer:
μ = 0.408
Explanation:
given,
speed of the automobile (u)= 20 m/s
distance = 50 m
final velocity (v) = 0 m/s
kinetic friction = ?
we know that,
v² = u² + 2 a s
0 = 20² + 2 × a × 50
a = 4 m/s²
We know
F = ma = μN
ma = μ mg
a = μ g
μ = 0.408
hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408
Answer:
I = 69.3 μA
Explanation:
Current through the straight wire, I = 3.45 A
Number of turns, N = 5 turns
Diameter of the coil, D = 1.25 cm
Resistance of the coil,
Distance of the wire from the center of the coil, d = 20 cm = 0.2 m
The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.
Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil
Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345
ΔB = -0.000001725
Induced current,
E = -N (Δ∅)/Δt
Δ∅ = A ΔB
Area, A = πr²
diameter, d = 0.0125 m
Radius, r = 0.00625 m
A = π* 0.00625²
A = 0.0001227 m²
Δ∅ = -0.000001725 * 0.0001227
Δ∅ = -211.6575 * 10⁻¹²
E = -N (Δ∅)/Δt
Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms
I = E/R
I = 0.0000693 A
I = 69 .3 * 10⁻⁶A
I = 69.3 μA
<u>Answer:</u>
Maximum height reached = 35.15 meter.
<u>Explanation:</u>
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g
So range of projectile,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So
Now we have
So maximum height reached = 35.15 meter.