On an electromagnetic spectrum, one of its noticeable trends is that the wavelength increases with decreasing energy and the wavelength decrease with increasing energy. Furthermore, gamma rays have high energy and short wavelengths while microwaves have low energy and long wavelengths.
Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
Answer:
(a): The magnitude of the electric force on the small sphere = 
(b): Shown below.
Explanation:
<u>Given:</u>
- m = mass of the small sphere.
- q = charge on the small sphere.
- L = length of the silk fiber.
= surface charge density of the large vertical insulating sheet.
<h2>
(a):</h2>
When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

<em>where,</em>
is the electrical permittivity of the free space.
The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

Thus, the magnitude of the electric force on the small sphere is given by

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.
<h2>
(b):</h2>
When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.
According to the fig.,

<em>where,</em>
= electric force on the sphere, acting along left.
= weight of the sphere, acting vertically downwards.
<em />

g is the acceleration due to gravity.
After impact velocity = 14.968 ft/s
Weight and mass of Bullet and wooden block:
Bullet: w = 1oz = 1/16 lb m = 0.001941 lb
wooden block : W = 5lb M = 0.15528 lb
velocity of block and bullet immediately after impact:
Σmv1 + ΣImp = mv2
Resolving vertical component
( m× v₀cos30⁰) + 0 = ( m+M) v'
v' = ( m× v₀cos30⁰)/ (m+M)
v' = 14.968 ft/s
Horizontal and vertical component of the impulse exerted by block on the bullet:
Here we will apply the principle of impulse and momentum.
Horizontal component:
-mv₀ cos30⁰ + RxΔt =0
RxΔt = mv₀sin30⁰
= 0.001941 × 1400sin30⁰
RxΔt = 1.3587 lb.s
Vertical component:
-mv₀cos30⁰ + RyΔt = -mv'
RyΔt = m( v₀cos30⁰-v')
RyΔt = 0.001941(1400cos30⁰ - 14.968)
= 2.32 lb.s
Learn more about impact here:
brainly.com/question/15008937
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