Answer:
The work done on the suitcase is, W = 600 J
Explanation:
Given,
The average force exerted by Jose on his suitcase, F = 60 N
Jose carried the suitcase to a distance, S = 10 m
The work done on the suitcase is given by the relation
<em>W = F x S</em>
Substituting the given values in the above equation,
W = 60 N x 10 m
= 600 J
Hence, the work done on the suitcase is, W = 600 J
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log 
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 ( 
)
β₂ - β₁ = 10 
log \frac{I_2}{I_1} = 
= 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ = 
r₂ = 0.316 m
Answer:
Yes, there is such a way.
Explanation:
If currents flow in the same direction in two or more long parallel wires, there will be an attractive force between the wires. If the current flows in different directions, there will be a repulsive force between the wires. In this case, these three parallel wires, can be be made to carry current in the same direction, creating an attractive force between all three wires.
Note that it is not possible to have at the least one of them carry current in the opposite direction and still have an attractive current between them.
Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513
Answer:
θ_p = 53.0º
Explanation:
For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º
Let's write the transmission equation
n1 sin θ₁ = ne sin θ₂
The angle to normal (vertcal) is
180 = θ2 + 90 + θ_p
θ₂ = 90 - θ_p
Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray
We replace
n1 sin θ_p = n2 sin (90 - θ_p)
Let's use the trigonometry relationship
Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p
In the law of reflection incident angle equals reflected angle,
ni sin θ_p = ns cos θ_p
n₂ / n₁ = sin θ_p / cos θ_p
n₂ / n₁ = tan θ_p
θ_p = tan⁻¹ (n₂ / n₁)
Now we can calculate it
The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water
n₂ = 1.33
θ_p = tan⁻¹ (1.33 / 1)
θ_p = 53.0º
n₂ = 1.40
θ_p = tan⁻¹ (1.40 / 1)
Tep = 54.5º
