What is the speed of a cyclist that rides west 88 km in 32 minutes?

Answer:

Position X.

Explanation:

To know the correct answer to the question, it is important that we know the definition of potential energy.

Potential energy can be defined as the energy possessed by an object in relation to its position. Mathematically it can be expressed as:

PE = mgh

Where

PE => is the potential energy.

m => is the mass of the object.

g => is the acceleration due to gravity.

h => is the height to which the object is located.

Considering the formula for potential energy (i.e PE = mgh), we can see clearly that the potential energy (PE) is directly proportional to the height (h). This implies that the greater the height, the greater the potential energy and the smaller the height, the smaller the potential energy.

Considering the diagram given above, we can see that the greatest height attained by the ball is at position X. Thus, the ball will have the greatest potential energy at position X.

Answer:

Explanation:

Let the balls collide after time t .

distance covered by falling ball

s₁ = v₀ t + 1/2 g t²

distance covered by rising ball

s₂ = v₀ t - 1/2 g t²

Given ,

s₁ + s₂ = D

D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²

= 2v₀ t

t = D / 2v₀

s₂ = v₀ t - 1/2 g t²

= v₀ x D / 2v₀ - (1/2) x  g x D² / 4v₀²

= D / 2 - gD² / 8 v₀²

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_{i} m_{i}[/tex]i

        x_{cm} = [tex]\frac{1}{m+m} ( 0 + l m)[/tex]

        x_{cm} = [tex]\frac{1}{2m} lm[/tex]

        x_{cm} = [tex]\frac{1}{2}[/tex]

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

Answer:

5076.5

Explanation:

Answer: 30 to 40 s

Explanation:

Answer:

the required charged is 7.06 × 10⁻¹³ C

Explanation:

Given that;

Radius = 30 microns = 30 × 10⁻⁶

Speed v = 20 m/s

length x = 1.6 cm = 0.016 m

spacing d = 1.0 mm = 0.001 m

Voltage V = 1500 V

from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.

Electric field E = V/d

E = 1500 V /  0.001 m

E = 1.5 × 10⁶ V/m

Mass of ink drop m = pv

m = 10³ kg/m³ × [tex]\frac{4}{3}[/tex]πr³

m = 1000 kg/m³ × [tex]\frac{4}{3}[/tex]π × (30 × 10⁻⁶)³

m = 1.131 × 10⁻¹⁰ Kg

Time taken to travel t =  x / sped

t = 0.016 m / 20 m/s

t = 0.0008 s

From the kinematic equation

to form the top of a letter 3 mm ( 0.003 m )high

y = [tex]\frac{1}{2}[/tex]at²

2y = at²

a = 2y/t²

we substitute

a = (2 × 0.003 m) / (0.0008 s)²

a =  9375 m/s²

Now Force F = Eq = ma

so

q = ma / E

we substitute

q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )

q = 7.06 × 10⁻¹³ C

Therefore, the required charged is 7.06 × 10⁻¹³ C

Answer:

B but I am not sure

Explanation:

Answer:

θ₂/ θ₁= 2.58

Explanation:

In this exercise you are asked to compare the angular diameters of Mars and Jupiter. The angular diameter or angle in radians is

           θ = D / R

where D is the diameter of the body, the distance from Earth to the body of interest and θ is angle in radians

The different distances are tabulated with respect to the Sun

Sun -Earth     1,496 10¹¹ m

Sun- Mars     2.28 10¹¹ m

Sun - Jupiter 7.78 10 m

The Radii of the planets are

Mars     3.37 10⁶ m

Jupiter 6.99 10⁷ m

let's calculate the angles for each body

a) Mars

       θ₁ = 2r / R'

the distance from the ground is

      R ’= D_planet - D_earth

      R ’= 2.28 10¹¹ - 1.496 10¹¹

       R ’= 0.784 10¹¹ m

let's calculate

       θ₁ = [tex]\frac{2 \ 3.37 \ 10^6 }{0.784 \ 10^{11}}[/tex]

        θ₁ = 8.6 10⁻⁵ radians

b) Jupiter

       R ’= 7.78 10¹¹ - 1.496 10¹¹

      R ’= 6.284 10¹¹ m

let's calculate

       θ₂ = [tex]\frac{2 \ 6.99 \ 10^7}{6.284 \ 10^{11}}[/tex]

        θ₂ = 2.22 10⁻⁴ radians

the ratio of the angular diameters is

       θ₂/ θ₁ = [tex]\frac{2.22 \ 10^{-4}}{8.6 \ 10^{-5}}[/tex]

        θ₂/ θ₁= 2.58

Answer:

Explanation:

Kinetic energy of the mass of 10 kg

= 1/2 m v² , m is mass and v is velocity .

= .5 x 10 x 20²

= 2000 J

The opposing force stops it . so work done by opposing force will be equal to this energy and it will be negative .

So energy transfer will be  - 2000 J .

= 2 kJ .

If distance travelled by mass is d , force 25 N will have a displacement of d . so work done by force of 25 N

= 25 x d

25 d = 2000

d = 80 m .

Hence system travels a distance of 80 m .

Answer:

software calculator is a calculator that has been implemented as a computer program, rather than as a physical hardware device. They are among the simpler interactive software tools, and, as such, they: Provide operations for the user to select one at a time.

Answer: The calculator is a compact portable device that performs mathematical calculations. Some calculators also allow easy text editing and programming. It's also a programming software that simulates a portable calculator. Calculator applications help you make basic math calculations without leaving your screen.

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns  to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

                                               = 1/5 ∫ 3*10^-3  d(t)  + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

 The reading of the voltmeter ( V )

V = IR

   = 0.96 mA * 21 k Ω  = 20.16 v

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force [tex]N^>[/tex] (750 N)  and the tension in the Achilles tendon [tex]F^>_{Achilles}[/tex] (2225 N) must be equal to the force exerted on the foot by the tibia;

so

| [tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] | = | [tex]F^>_{Tibia}[/tex] |

so force exerted on the foot by the tibia will be;

| [tex]F^>_{Tibia}[/tex] | = |[tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] |

so we substitute IN OUR VALUES

| [tex]F^>_{Tibia}[/tex] | = 750 N + 2225 N

| [tex]F^>_{Tibia}[/tex] |  = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

Answer:

I think the answer is 63,

Answer:

Kinetic energy Temperature

Explanation:

Answer:

a = 7.749 ft/s²

Explanation:

First to all, we need to convert all units, so we can work better in the calculations.

The horizontal acceleration is asked in ft/s² so the units of speed will be the same. The Work is in BTU and we need to convert it in ft.lbf in order to get the acceleration and final speed in ft/s:

W = 10 BTU * 778.15 Lbf.ft / BTU = 7781.5 lbf.ft

Now, to get the acceleration we need to get the final speed of the object first. This can be done, by using the following expression:

W = ΔKe  (1)

And Ke = 1/2mV²

So Work would be:

W = 1/2 mV₂² - 1/2mV₁²

W = 1/2m(V₂² - V₁²)    (2)

Finally, we need to convert the mass in lbf too, because Work is in lbf, so:

m = 55 lb * 1 lbf.s²/ft / 32.174 lb = 1.7095 lbf.s²/ft

Now, we can calculate the final speed by solving V₂ from (2):

7781.5 = (1/2) * (1.7095) * (V₂² - 20²)

7781.5 = 0.85475 * (V₂² - 441)

7781.5/0.85475 = (V₂² - 400)

9103.83 + 400 = V₂²

V₂ = √9503.83

V₂ = 97.49 ft/s

Now that we have the speed we can calculate the acceleration:

a = V₂ - V₁ / t

Replacing we have:

a = 97.49 - 20 / 10

Hope this helps

Answer: [tex]26.460\times 10^3\ J[/tex]

Explanation:

Given the mass of skier m=72 kg

distance traveled d=75 m

constant speed v=3.4 m/s

If speed is constant  then there must no force acting in the direction of motion

i.e. tension force must be equal to the component of weight

[tex]T=mg\sin 30^{\circ}[/tex]

Work done is given by

[tex]\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J[/tex]

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

Answer:

13.4436

Explanation:

maybe a spectrograph ?

Answer:

 x₂ = 4.455 m

Explanation:

The average speed is defined as the displacement traveled between the time interval

         v_average =  [tex]\frac{\Delta x}{\Delta t}[/tex]

in this case we have a first stop walking west at speed v = 3.10 m / s a ​​distance of x = 5.96 km = 5.96 10³ m

Let's find the time it takes on this tour

          v = x / t

          t = x / v

          t₁ = 5.96 10³ / 3.10

          t₁ = 1.9226 10³ s

in a second to walk east at a speed of v₂ = 0.744 m / s

           v₂ = x₂ / t₂

           t₂ = x₂ / v₂

for full movement

let's assume that the eastward movement is positive

         v =[tex]\frac{- x_1 + x_2}{t_1 +t_2}[/tex]

         v = [tex]\frac{-x_1 +x_2}{t_1 + \frac{x_2}{t_2} }[/tex]

the only unknown term is the distance to the east. We replace and resolve

         1.19 = [tex]\frac{-5.96 \ 10^3 + x_2}{1.92 \ 10^3 + \frac{x_2}{0.744} }[/tex]

         1.19 (1.92 10³ + [tex]\frac{x_2}{0.744}[/tex]) = x₂ 1.92 10³ - 5.96 10³

          2.2848 10³ + 1.599 x₂ = 1.92 10³ x₂ - 5.96 10³

          x₂ (1.92 10³ - 1.599) = 2.2848 10³ + 5.96 10³

          x₂ = 8.5448 10³ / 1.918 10³

          x₂ = 4.455 m

Answer:

I cannot fully see the picture, but I'm going to have to tell you to go with towards the sun because it says summer.

Answer: the at which the bar conducts now is 5 Js⁻¹

Explanation:

Given the data in the question;

we know that; Heat transfer by conduction is given by;

Q ∝ [tex]A / l[/tex]

such that,

[tex]Q_{1}/Q_{2}[/tex] = [tex]A_{1}l_{2} / A_{2}l_{1}[/tex]

[tex]Q_{2} = Q_{1} A_{2}l_{1}/A_{1}l_{2}[/tex]

so

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r}{2}[/tex])² × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])  

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r^{2} }{4}[/tex]) × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × πr² × 1/4 × [tex]l[/tex] ) / (πr² × 1/2 × [tex]l[/tex] )

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)

[tex]Q_{2} =[/tex]  ( 10 Js⁻¹ × 1/4) / ( 1/2)

[tex]Q_{2} =[/tex] 5 Js⁻¹

Therefore, the at which the bar conducts now is 5 Js⁻¹

Answer:

 v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]

Explanation:

For this exercise we must use the projectile launch ratios, let's start by finding the time it takes to reach the bottom of the cliff, the initial vertical velocity is zero

          y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

at the bottom of the cliff y = 0 and as the body is thrown horizontally the initial vertical velocity is zero

          0 = y₀ + 0 - ½ g t²

          t = [tex]\sqrt{2y_o/g}[/tex]

this time is the same as the horizontal movement.

Let's use Newton's second law to find the acceleration on this x-axis due to the force of the air

           F = m aₓ

they tell us that force is equal to the weight of the body

           -mg = maₓ

           aₓ = -g

the sign indicates that the acceleration is to the left

we write the kinematics equation

          x = x₀ + v₀ₓ t + ½ aₓ t²

They indicate that the final position is the foot of the cliff (x = 0), when it leaves the top it is at x₀ = 0 and has a velocity v₀ₓ = v

we substitute

          0 = 0 + v t + ½ (-g) t²

          v = ½ g t

we use the drop time

          v = ½ g [tex]\sqrt{\frac{2yo}{g} }[/tex]

          v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]

Answer:

The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Explanation:

Given;

first net force, F₁ = 100 N

second net force, F₂ = 50 N

If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Let a₁ be the acceleration produced by the first net force

then, a₂ be the acceleration produced by the second net force

Thus, a₁ = 2a₂

Answer: B

Explanation:

The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.

[tex]V_y=48*sin(53)=38.3m/s[/tex]

Explanation:

There are two answers

m/(s/s)=m

or

(m/s)/s=m/s²

Answer:

If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.

Explanation:

Answer:

Explanation: