You walk 10 meters to class in 20 seconds , your average speed is ___ ?

Answer:

what's the question ,so I can answer it right ?

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: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:

newton force is the state of gravity force they abstract when they came closer

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

Answer:

A. Using

E=V/d

= 600/4.49*10^-3

= 1.336 x10^5 N/C

b) F = E*q = 1.33610^5 x 1.6*10^-19

= 2.17 x 10^-14 N

c) Work = Fs distance = 2.17 x 10^-14 N (4.49-3.14)*.001= 1.35 x 10^-17 J

Answer:

total work done = -5960.8 J

Explanation:

given data

initial volume v1 = 0.22 m³

initial pressure  p1 = 86 kPa

final pressue p2 = 101.3 kPa

solution

we apply here  isothermal expansion that is express as

p1 × v1 = p2 × v2    ......................1

put here value

86 × 0.22 = 101.3 × v2

v2 = 0.1867 m³  

and

work done will be here

w1 = p1 × v1  × ln([tex]\frac{p1}{p2}[/tex])      ....................2

w1 = 86 × 10³ × 0.22 × [tex]ln(\frac{86}{101.3})[/tex]

w1 = -3.097  × 10³ J

and

it is cooled to initial volume at constant pressure  so here work done will be

w2 = p(v2 - v1)    .................3

w2 =  86 × 10³ × ( 0.1867 - 0.22 )

w2 = -2863.8 J

so

total work done is

total work done = w1 + w2

total work done = -3097 +  -2863.8

total work done = -5960.8 J

Answer:

[tex]x=119m[/tex]

Explanation:

Hello,

In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:

[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]

In such a way, we can compute the displacement via:

[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]

Best regards.

Answer:

8.6602 tons

Explanation:

We first draw the known vector forces.

2fcos30⁰

We have f to be equal to 5tons

Inserting into formula

Σfx = 2(5)cos30⁰

= 8.6602 tons

Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.

Therefore the resultant force on the ship is equal to 8.6602 tons

I hope this helps!

Please check attachment for diagram.

Explanation:

Equation for Impact

FΔt = ΔP,

F = force

Δt  = Impact of time

ΔP = Change in momentum

Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.

Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.

Road safety containers are loaded with liquid or sand as they improve the period of impact.

Answer:

10 miles per minute.

Answer:

30.56 m/s^2

Explanation:

Given that In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes.

The average acceleration that is required to accomplish this will be

Average acceleration = change in velocity / time

Average acceleration = 7700/ 4.2 × 60

Average acceleration = 7700/252

Average acceleration = 30.56 m/s^2

Answer:

D) either 436 Hz or 444 Hz

Explanation:

frequency of the tuning fork, F₁ =  440 Hz

frequency of the piano, F₂ = ?

Beat frequency, F = 4 Hz

Beat frequency is given as the difference between the frequency of the two instruments  and it is given by;

F = F₂ - F₁                    or        F =  F₁ - F₂

F₂ = F + F₁                   or        F - F₁ = - F₂

F₂ = 4 Hz + 440 Hz    or       4 - 440 = - F₂

F₂ = 444 Hz                or      - 436 = - F₂

F₂ = 444 Hz               or         F₂ = 436 Hz

Therefore, the frequency of the piano is 444 Hz or 436 Hz

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

Find: a

Δx = v₀ t + ½ at²

250 m = (0 m/s) (17 s) + ½ a (17 s)²

a = 1.73 m/s²

The acceleration of this object is 1.730 meter per seconds square.

Given the following data:

To find the acceleration of this object, we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2} at^2[/tex]

Where:

Substituting the given values into the formula, we have;

[tex]250 = 0(17) + \frac{1}{2} (a)(17^2)\\\\250 = \frac{1}{2} (289)a\\\\250 = 144.5a\\\\a = \frac{250}{144.5}[/tex]

Acceleration, a = 1.730 [tex]m/s^2[/tex]

Therefore, the acceleration of this object is 1.730 meter per seconds square.

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Answer:

Speed is 0.08 m/s.

Explanation:

Given the distance that the bird flies = 3.7 meters

The time is taken by the bird to fly the 3.7 meters = 46 seconds  

We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.  

Below is the formula to find the speed.

Speed = Distance / Time

Now insert the given value in the formula.

Speed = 3.7 / 46 = 0.08 m/s

This question is incomplete, the complete question is;

The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventh floor is 40ft of water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor

Assume 12ft of elevation per floor

Answer: 48.68 psig

Explanation:

First  we calculate the elevation of the building

hb = 27 story * 12ft per floor/story

hb =  324 ft

given that the head lost in the vertical riser hL = 40 ft

now the delivery head required in the riser on he 27th floor;

hd = 8 psig *  (2.31 ft / 1 psig)

hd = 18.46 ft

Now calculate the suction head required by balancing the energy per unit weight of water, considering pump as the control volume

hp = (hb + hL + hd) - hs

hs = hb + hL + hd - hp

where hp is the head developed by the pump (270 ft)

hb is the elevation of the 27th floor of the building ( 324 ft)

hL is the head lost in the vertical riser ( 40 ft)

hd is the head required to exist in the riser on the 27th floor (18.46 ft)

so we substitute

hs = 324 ft + 40 ft + 18.46 ft - 270 ft

hs = 112.46

so 112.46ft * (1 psig / 2.31 ft)

= 48.68 psig

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:

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 value is [tex]d = 31.45 \ m [/tex]

Explanation:

Generally the relative speed at which you are moving with respect to the person ahead of you is mathematically represented as

[tex]v_r = v_s - v_c[/tex]

substituting 1.05 m/s for [tex] v_c [/tex] and 2.75 m/s for [tex]v_s[/tex]

So

[tex]v_r = 2.75 - 1.05[/tex]

=> [tex]v_r = 1.7 \ m/s [/tex]

Generally the distance by which the person is ahead of you is mathematically represented as

[tex]d = v_r * t[/tex]

substituting 18.5 s for [tex] t [/tex]

       [tex]d =  1.7  * 18.5[/tex]

=>      [tex]d  =  31.45 \  m [/tex]

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:

20.1 m/s

Explanation:

Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.

Given that the shuttle has a constant acceleration of 4.5 m/s2. 

The total distance to cover is:

Total distance = 40.9 + 3.9 = 44.8 m

Assuming you are starting from rest. Then initial velocity U = 0

Using the 3rd equation of motion to calculate the minimum velocity.

V^2 = U^2 + 2as

V^2 = 0 + 2 × 4.5 × 44.8

V^2 = 403.2

V = sqrt (403.2)

V = 20.1 m/s

Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s

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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Whenever an object is moving at a constant rate . . .

a).  its speed is that constant rate.

b).  Its acceleration MAY BE zero, if it's also moving in a straight line.

c).  Its velocity is that constant rate if it's moving in a straight line.  Otherwise, its velocity could have many different values, depending on the path it's following.  

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:

Explanation:

Efficiency of first engine

= T₁ - T₂ / T₁  where T₁ is temperature of hot reservoir , T₂ is temperature of cold reservoir

= (889 - 657 ) / 889

= ,261 or 26.1 %

output of work = .261 x 4710 = 1229.15 J .

Efficiency of second engine

=  (657 - 406 ) / 657

= .382

Heat rejected in engine one is heat input of second engine

heat input of second engine = 4710 - 1229.15 = 3480.85

output of work of second engine

=  .382 x 3480.85 = 1329.68 J

Total work delivered by two engine

= 1229.15  +  1329.68 J

= 2558.83 J