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Rom4ik [11]
3 years ago
10

PLEASE ANSWERRRRR ASAPPPPPP

Physics
2 answers:
Bad White [126]3 years ago
5 0
D Valence
.
.
^^^answer
uranmaximum [27]3 years ago
5 0
Your correct answer is Valence I just took the test!
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What information can you determine from the coefficients of this balanced chemical equation?
V125BC [204]
HCl would be the limiting factor... but idk of thats what youre looking for
6 0
3 years ago
A scuba diver sees light reflected from the water’s surface. show answer No Attempt If the index of refraction for air is 1.00 a
sasho [114]

Answer:

53.06°

Explanation:

refractive index of air = 1

refractive index of water, n = 1.33

Let the angle of polarisation is ip.

Use Brewster's law,

n = tan ip

1.33 = tan ip

ip = 53.06°

thus, the angle of polarisation is  53.06°.

3 0
3 years ago
An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri
Anettt [7]

Answer:

270 mi/h

Explanation:

Given that,

To the south,

v₁ = 300 mi/h, t₁ = 2 h

We can find distance, d₁

d_1=v_1\times t_1\\\\d_1=300\times 2\\\\d_1=600\ \text{miles}

To the north,

v₂ = 250 mi/h, d₂ = 750 miles

We can find time, t₂

t_2=\dfrac{d_2}{v_2}\\\\t_2=\dfrac{750\ \text{miles}}{250\ \text{mi/h}}\\\\t_2=3\ h

Now,

Average speed = total distance/total time

V=\dfrac{d_1+d_2}{t_1+t_2}\\\\V=\dfrac{600+750}{2+3}\\\\V=270\ \text{mi/h}

Hence, the average speed for the trip is 270 mi/h.

3 0
3 years ago
A person who weighs 69kg rides with an acceleration of +1.45 m/s2. What is the force exerted by the legs on the person's upper b
trapecia [35]

Answer:

The force exerted by the legs on the person's upper body 32.106 N.

Given:

Mass = 69 kg

Acceleration = 1.45 \frac{m}{s^{2} }

To find:

Force exerted by legs on the person's upper body = ?

Formula used:

Force = mass × acceleration

Solution:

Mass of legs = \frac{32}{100} mass of body

Mass of legs = 22.08 kg

According to Newton's second law of motion,

Force = mass × acceleration

Force = 22.08 × 1.45

Force = 32.106 N

The force exerted by the legs on the person's upper body 32.106 N.

4 0
4 years ago
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,
EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

          F₁ = 2.50 kN

Tide

         cos 30 = F₂ₓ / F₂

         sin 30 = F_{2y} / F₂

          F₂ₓ = F₂ cos 30

         F_{2y} = F₂ sin 30

         F₂ₓ = 1.20cos 30 = 1.039 kN

         F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

        Fₓ = F₁ₓ + F₂ₓ

        Fₓ = 2.50 +1.039

        Fₓ = 3,539 kN

        F_y = F_{2y}

        F_y = 0.600

to find the vector we use the Pythagorean theorem

         F = \sqrt{F_x^2 +F_y^2}

         F = \sqrt{ 3.539^2 + 0.600^2 }

         F = 3,589 kN

the address is

         tan θ = F_y / Fₓ

         θ = tan⁻¹ \frac{F_y}{F_x}

         θ = tan⁻¹  \frac{0.6}{3.539}0.6 / 3.539

         θ = 9.6º

the resultant force to two significant figures is

         F = 3.6 kN

the direction is 9.6º to the North - East

7 0
3 years ago
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