How does the flatter design make sweeping the floor easier

This chart shows characteristics of three different types of waves.

prohojiy [21]

Answer:

1. Primary or P waves are push and pull waves

2. Secondary, S or Shear Waves are also called transverse wave

3. L or surface waves reach the earth's surface after P and S waves

7 0

3 years ago

Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs

Anestetic [448]

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

Lines per centimetre = 7000 /cm = 700000 /m

Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

λ is the wavelength of the incident light

d is the distance separating the centres of 2 adjacent slits

θ is the angle at which diffraction occurs

From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

θ = 28.9°

8 0

3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

b)

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

c)

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago

What is the acceleration of a car that travels in a straight line?

Sphinxa [80]

Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.

hope this helps!! ✨

6 0

3 years ago

Read 2 more answers

An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv

konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0

3 years ago