The rain gauge has 2 inches of precipitation.
Qualitative or Quantitative

Answer:

t=8

Explanation:

u have solution I give solution also

don't mark plzz follow y

Answer:

im going between 2 of them b and c but i would have choose b

Answer:C

Explanation:

I did the test

a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. Then the wave speed of the sound wave is 20.23 m/s.  Hence option D is correct.

Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.

Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.

The wave speed is given by,

c = νλ

where λ is wavelength, ν is frequency.

Given,

frequency ν = 85 Hz

wavelength λ = 4.2 m

The speed of the wave,

c = 85 × 4.2 = 20.23 m/s

Hence option D is correct.

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Answer:

i think gravity created a force strong enough to push him down.but it also depends how slow or fast the bus is going

Explanation:

gravity as we all now it is pretty strong so if the bus is going fast then the person standing up will fall but if the bus is going pretty slow then it'll just nudge him.

Answer:

C. Muddy Water

Answer:

0.55 hz

Explanation:

Given that the plane fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. And the velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m.

The maximum speed attained will be

Maximum speed = 50 + 0.5 = 5.5 m/s

The Length = 100m

Speed = FL

Where F = frequency

Substitute speed and distance length into the formula

55 = 100F

F = 55/100

F = 0.55 Hz

Therefore, the highest frequency the anemometer will encounter will be 0.55 Hz

Answer:

40079 Hz

Explanation:

1. The first step is to calculate energy dissipation ∈=u^3/l and here u is fluctuating velocity

∈=(0.5^3)/100 = 0.00125 m^2/s^3

2. Find out the length scale of the small eddies

η=(viscosity/∈)^1/4

η=(1.470e-5/0.00125)^1/4 = 0.00126 m

3. The frequency associated with these small-scale eddies will be the greatest frequency the anemometer will encounter, thus:

u_max=f_max * η

u_max = u + u' = 50+0.5=50.5 m/s

f_max = u_max/η = 50.5/0.00126 = 40079 Hz

This is the heighest frequency the anemometer will encounter.

Answer:

The critical radius of the plastic insulation is 0.72 inches.

Explanation:

Given that,

Diameter = 0.091 in

Thickness = 0.02 in

Initial temperature = 90°F

Final temperature = 50°F

Heat transfer coefficient = 2.5 Btu/h.ft²°F

Material conductivity = 0.075 Btu/h.ft °F

We need to calculate the critical radius of the plastic insulation

Using formula of critical radius

[tex]r_{cr}=\dfrac{2K}{h}[/tex]

Where, k = Material conductivity

h = Heat transfer coefficient

Put the value into the formula

[tex]r_{cr}=\dfrac{2\times0.075}{2.5}[/tex]

[tex]r_{cr}=0.06\ ft[/tex]

[tex]r_{cr}=0.72\ inches[/tex]

Hence, The critical radius of the plastic insulation is 0.72 inches.

Answer:

Explanation:

If Two cylindrical resistors are made from the same material, then their resistivity will be the same.  Formula for calculating resistivity of a material is expressed as;

[tex]\rho = \frac{RA}{L} \ where \ A = \frac{\pi d^2}{4}[/tex] where;

R is the resistance

A is the cross sectional area of the material

L is the length of the material

For the shorter cylinder:

Length = L

diameter = D

[tex]\rho = \dfrac{R_s(\frac{\pi D^2}{4})}{L} \\\\\rho = \dfrac{R_s{\pi D^2}}{4L}[/tex]

For the longer cylinder:

Length = 16L

diameter = 4D

[tex]\rho = \dfrac{R_l(\frac{\pi (4D)^2}{4})}{16L} \\\\\\\rho = \dfrac{R_l(\frac{\pi (16D^2)}{4})}{16L} \\\\\rho = \dfrac{R_l{16\pi D^2}}{16L}\\\\\rho = \dfrac{R_l{\pi D^2}}{L}[/tex]

Since their resistivity are the same then;

[tex]\dfrac{R_s{\pi D^2}}{4L} = \dfrac{R_l{\pi D^2}}{L} \\\\ \dfrac{R_s}{4} = {R_l} \\\\R_s = 4R_l\\\\R_l = \frac{R_s}{4}[/tex]

Hence the resistance of the longer resistor is a quarter of the shorter resistor.

Answer:

a)  I = 0.0198 kg m² ,  b)    I = 21.85 kg m²

Explanation:

For this exercise we will use the definition of moment of inertia

        I = ∫ r² dm

For body with high symmetry they are tabulated

sphere  I = 2/5 m r²

bar with respect to  center of mass I = 1/12 m L²

let's calculate the mass of each body

bar

        ρ = m / V

        m = ρ V

        m = ρ l w h

where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm

        m = 32820 1 0.008 0.04

        m = 10.5 kg

Sphere

       M = ρ V

       V = 4/3 pi r³

       M = rgo 4/3 π r³

give us the density 37800 kg / m³ and the radius of 5 cm

       M = 37800 4/3 π 0.05³

       M = 19.8 kg

a) asks us for the moment of inertia of the sphere with respect to its center of mass

        I = 2/5 M r²

        I = 2/5 19.8 0.05²

        I = 0.0198 kg m²

b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes

        I = I_cm + M d2

where d is the distance from the body to the point of interest

        I_cm = 0.0198 kg m²

the distance to the pivot point is

        l = length of the bar + radius of the sphere

        l = 1 + 0.05 = 1.005 m

        I = 0.0198 + 19.8 1.05²

        I = 21.85 kg m²

Answer:

The value is [tex]v_s = 1.394 \ m/s[/tex]

Explanation:

From the question we are told that

The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]

The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]

The speed of sound is [tex]v_s = 343 \ m/s [/tex]

Generally the frequency of the note played by the first player is mathematically represented as

[tex]f_1 = f_2 + f_b[/tex]

=> [tex]f_1 = 490 + 2.0 [/tex]

=> [tex]f_1 = 492 Hz[/tex]

From the relation of Doppler Shift we have that

[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]

Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]

So

[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]

=> [tex]v_s = 1.394 \ m/s[/tex]

Answer:

Hello!

Sorry you haven't put up an image of your question! Without it we can't answer your question!

Explanation:

Maybe put up another one and it'll be answered!

:D

The Answer Is - 0.58 Kilogram's

Hope This Help's

• Answer:

0.58 kg

• Explanation:

1 kg = 1000 g

580 g = 0.58 kg

Answer:

The tension in the light rope must be equal to the weight of the bucket

Explanation:

Given that,

Constant velocity of bucket and direction of bucket in downward

We need to find the tension in the rope

Using given data,

When a bucket  moves downward with a constant velocity then the net force does not applied on the bucket.

So, The weight of the bucket will be equal to the tension in the light rope

In mathematically,

[tex]T=mg[/tex]

Where, T = tension

m = mass of bucket

g = acceleration due to gravity

Hence, The tension in the light rope must be equal to the weight of the bucket.

Answer:

The hight of the building is 38.16 m

Explanation:

These two pieces of information given, first, the watermelon is 30 m  above the ground and after 1.50 s the watermelon has been spotted. Now we are required to find the height of the building.

Use the below formula to find the height of buildings.

S = ut + ½ gt^2

30  =1.5u + (1/2) × 9.8 (1.5)^2

u = 12.65 m/sec

v^2 – u^2 = 2gs

(12.65)^2 = 2×9.8 s’

S’ = 8.16 m  

h = s + s’

h = 30 + 8.16 = 38.16 m

The hight of the building is 38.16 m.

The height of the building is 38.16 m.

Given data:

The height above the ground is, h = 30.0 m.

The time interval after observation of first spot is, t = 1.50 s.

We need to find the height of building. And since two pieces of information given, first, the watermelon is 30 m  above the ground and after 1.50 s the watermelon has been spotted. So, using the second kinematic equation of motion as,

[tex]h = ut + \dfrac{1}{2}gt^{2}[/tex]

Here, u is the initial speed. Solving as,

[tex]30 = (u \times 1.50) + \dfrac{1}{2} \times 9.8 \times (1.50)^{2}\\\\u =12.65 \;\rm m/s[/tex]

Now landing distance (s') is calculated using the third kinematic equation of motion as,

[tex]v^{2} =u^{2}+2(-g)s\\\\0^{2} =12.65^{2}+2(-9.8)s\\\\s = 8.16 \;\rm m[/tex]

Then the height of building is given as,

H = h + s

H = 30 m + 8.16 m

H = 38.16 m

Thus, we can conclude that the height of the building is 38.16 m.

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Answer:

44 years

Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44

44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.

The period of time it takes for one-half of a radioactive isotope to decay is known as the half-life. A given radioactive isotope's half-life is constant; it is unaffected by external factors and independent of the isotope's starting concentration.

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44 sec

44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.

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Explanation:

13+x=43>2x

13+x=43>2x+43>2x

13+x=86>4x

x-4x=86-13

3x=73

x=73/3

x=24.333

x=24.4

Answer: Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another. ... During the 1930s Bohr helped refugees from Nazism.

Answer:

The value is [tex]D = 205.8 \ m [/tex]

Explanation:

The time taken for the column to take a step mathematically represented as

100 steps => 1 minutes => 60 seconds

1 step => t

=> [tex]t = 0.6 \ s [/tex]

Generally the length of the column is mathematically represented as

[tex]D = v * t[/tex]

substituting 343 m/s for v we have

        [tex]D =  343 * 0.6 [/tex]

  =>    [tex]D =  205.8 \  m  [/tex]

Answer:

The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].

Explanation:

We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:

x-Direction

[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]

y-Direction

[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.

[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.

[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:

[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]

[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)

[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)

At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):

[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]

[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]

(Ec. 1) in (Ec. 2):

[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]

Now we clear the time in the resulting expression:

[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]

[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]

There are two solutions:

[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]

[tex]t_{1} \approx 1.101\,s[/tex]

[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]

[tex]t_{2}\approx 1.131\,s[/tex]

The initial velocity is cleared within (Ec. 2):

[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]

The bounds of the range of initial speed is determined hereafter:

[tex]t_{1} \approx 1.101\,s[/tex]

[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]

[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]

[tex]t_{2}\approx 1.131\,s[/tex]

[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]

[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]

The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].

Answer:

   ρ = 4.4 10⁻⁸ Ω m

Explanation:

For this exercise let's start by finding the value of the resistance using Ohm's law

          V = I R

           R = V / I

The voltage is related to the electric field

          V = E L

let's substitute

          R = E L / I

          R = 8.2 10⁻² / 20  

          R = 4.1 10⁻³ L

now we can use the resistance relation

          R = ρ L / A

the area of ​​a circular wire is

          A = π r² = π d² / 4

         ρ = R π d² / (4 L)

let's calculate

        ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)

        ρ = 4.4 10⁻⁸ Ω m

The resistivity of the wire is 4.4 10⁻⁸ Ω m.

We know that;

V = I R

R = V / I

The voltage  the electric field can be connected using the formula;

V = E L

Hence;

R = E L /I

R = 8.2 10⁻² / 20  

R = 4.1 10⁻³ L

For resistivity;

R = ρ L / A

the area can be obtained from;

A = π r² = π d² / 4

ρ = R π d² / (4 L)

ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)

ρ = 4.4 10⁻⁸ Ω m

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Answer:

FALSE

Explanation:

Answer:

F=200N

a=500m/s2​

Mass=?

Explanation:

F=ma

200=m*500

200/500=m

Mass=0.4kg

Answer:

0.0552 m/s²

Explanation:

Given:

Δx = 69.0 m

v₀ = 0 m/s

t = 50 s

Find: a

Δx = v₀ t + ½ at²

69.0 m = (0 m/s) (50 s) + ½ a (50 s)²

a = 0.0552 m/s²

Answer:

10 miles per minute.

Answer:

(a) , .  and .

(b)[tex]\vec A - \vec C=22 \hat i -10 \hat j[/tex].

(c)[tex]|\vec A - \vec B|=63.13[/tex] and the direction [tex]\theta =[/tex] 124.56°.

Explanation:

Given that,

,

and

[tex]\vec {C}=2 \hat i +43 \hat j[/tex]

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

[tex]|\vec A|=\sqrt {24^2+ 33^2}[/tex]

[tex]\Rightarrow |\vec A|=\sqrt {1665}[/tex]

.

The magnitude of the is

[tex]|\vec B|=\sqrt {55^2+ (-12)^2}[/tex]

[tex]\Rightarrow |\vec B|=\sqrt {3169}[/tex]

.

And, the magnitude of the is

[tex]|\vec C|=\sqrt {2^2+ 43^2}[/tex]

[tex]\Rightarrow |\vec C|=\sqrt {1853}[/tex]

.

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

[tex]\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)[/tex]

[tex]\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j[/tex]

[tex]\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j[/tex]

(c) The expression of is

[tex]\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)[/tex]

[tex]\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j[/tex]

[tex]\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)[/tex]

The magnitude of is

[tex]|\vec A - \vec B|=\sqrt {(-31)^2+55^2}[/tex]

[tex]\Rightarrow |\vec A - \vec B|=\sqrt {3986}[/tex]

[tex]\Rightarrow |\vec A - \vec B|=63.13[/tex]

Now, if a vector [tex]\vec V= -\alpha \hat i +\beta \hat j[/tex] in 3rd quadrant having direction [tex]\theta[/tex] with respect to [tex]\hat i[/tex] direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector [tex]\vec A - \vec C[/tex], [tex]\alpha=31[/tex] and [tex]\beta=45[/tex].

[tex]\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)[/tex]

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for [tex]\vec A +\vec B[/tex] and [tex]\vec A - \vec B[/tex] has been shown  

in the figure(b) and figure(c) recpectively.

Vector [tex]\vec A - \vec B[/tex] is in 3rd quadrant as calculated in part (c).

While Vector [tex]\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)[/tex]

[tex]\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j[/tex], which is in 1st quadrant as both the components are position has been shown in figure(b).