sovle it please help me​

Answer:

BELTS

Explanation:

why would you need belts for running and stuff

Belts are not an important part of exercise clothing.

An exercise is a physical activity performed to increase strength and agility of the muscles as well as to enhance fitness.

During an exercise, it is important to wear clothes that will not hinder movement or cause injury.

Some items of clothing needed in exercise include:

Therefore, belts are not important during exercise.

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Explanation:

Given,

Position vector, [tex]c(t)=6ti+((3t)^2)j+((t)^3)k[/tex]

We need to find the velocity vector of the given path

[tex]v=\dfrac{d(c(t))}{dt}\\\\v=\dfrac{d(6ti+((3t)^2)j+((t)^3)k)}{dt}\\\\v=\dfrac{d(6ti)}{dt}+\dfrac{d((3t)^2j}{dt}+\dfrac{d(t^3)k}{dt}\\\\v=6i+18tj+3t^2k[/tex]

Hence, this is the velocity vector of the given path.

Answer:

57.0 m

Explanation:

Assume the time it takes for the sound to reach the top of the well is negligible.

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.41 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (3.41 s) + ½ (9.8 m/s²) (3.41 s)²

Δy = 57.0 m

Answer:

57 m

Explanation:

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, [tex]F_1[/tex] = 140 N

let the second force = [tex]F_2[/tex]

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

[tex]F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N[/tex]

[tex]F_2 = 280 \ East[/tex]

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

[tex]F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N[/tex]

[tex]F_2 = 560 \ West[/tex]

Answer:

Subliminal simulation

Explanation:

When an individual registers sensory input without being aware of the process consciously, Subliminal simulation occurs.

Subliminal stimulation defined as sensory stimulation which is below a person's threshold for perception.

It can't be seen by the bare eye or consciously heard. An example is visual stimuli which is flashed so quickly on a screen that a person can't process it. So, they are unaware they have seen anything.

Subliminal stimuli helps facilitate conscious processing of related information, changes the current mood, boost our motivation. It can even change a person's political attitudes and voting intentions.

When an individual registers sensory input without being aware of the process consciously, Subliminal simulation occurs.

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Answer:

The. Machine must detect a shift of

1 Hz

Explanation:

Frequency shift is given as

={ ( Vsound +V/ V sound -V) -1}f emitted

So by substitution we have

= { 1540+4E-4/1540-4E)-1) 2*10^6

= 1Hz

Explanation:

Distance = speed × time

d = (27 m/s) (2.0 s)

d = 54 m

Answer:

Approximately [tex]3.1\; \rm m \cdot s^{-1}[/tex].

Explanation:

The content of this pail is in a centripetal motion because its path forms part of a vertical circle. Let [tex]m[/tex] denote the mass of the contents of this pail, let [tex]v[/tex] denote the (linear) velocity of the content, and let [tex]r[/tex] denote the radius of this circle. The net force on the contents of this pail will thus be:

[tex]\displaystyle F(\text{net}) = \frac{m\, v^2}{r}[/tex] towards the center of the circle.

Assume that there is no friction between the content and walls of the pail. The only two possible forces on the contents pail towards the center would be:

Note that at the top of the circle, both the gravitational pull and the normal force from the bottom point towards the center of the circle. On the other hand, the normal force from the walls of the pail would be perpendicular to the line towards the center of the circle. At that point in the circle, there's no upward force to support the content of the pail. The uniform rotation will be sufficiently fast if it could allow the content to stay in the pail at the top of the circle.

Let [tex]g[/tex] denote the gravitational field strength at the top of this circle. The size of the gravitational pull on the content would be [tex]m\cdot g[/tex]. Let [tex]F(\text{normal})[/tex] denote the normal force from the bottom of the pail on the contents. The sum of these two forces should be equal to the vertical net force on the contents of this pail. That is:

[tex]F(\text{net}) = m\cdot g + F(\text{normal})[/tex].

From the centripetal motion of the content:

[tex]\displaystyle \frac{m\, v^2}{r} = m\cdot g + F(\text{normal})[/tex].

Rearrange to obtain an expression for the normal force:

[tex]\displaystyle F(\text{normal}) = \frac{m\, v^2}{r} - m\cdot g[/tex].

Note, that the normal force the bottom of the pail exerts on the contents should be greater than or equal to zero. While the pail is at the top of the circle, the normal force from the bottom of the pail cannot pull the contents upwards. Hence:

[tex]\displaystyle \frac{m\, v^2}{r} - m\cdot g = F(\text{normal}) \ge 0[/tex].

[tex]\displaystyle \frac{m\, v^2}{r} - m\cdot g \ge 0[/tex].

Rearrange and simplify to obtain:

[tex]\displaystyle \frac{v^2}{r} - g \ge 0[/tex].

[tex]v^2 \ge g\cdot r[/tex].

[tex]v \ge \sqrt{g \cdot r}[/tex].

In other words, if the gravitational field strength is [tex]g[/tex] and the radius of the circle is [tex]r[/tex], the minimum linear velocity required to keep the content in the pail at the top of the circle is [tex]\sqrt{g \cdot r}[/tex].

If [tex]g = 9.81\; \rm N \cdot kg^{-1} = 9.81\; \rm m \cdot s^{-2}[/tex] and [tex]r = 0.82 \; \rm m + 0.185\; \rm m \approx 1.005\; \rm m[/tex], then the minimum value of [tex]v[/tex] would be approximately:

[tex]\sqrt{9.81 \; \rm m \cdot s^{-1} \times 1.005\; \rm m} \approx 3.1\; \rm m \cdot s^{-1}[/tex].

Answer:

The value is [tex]W_N = 11849 \ J [/tex]

Explanation:

From the question we are told that

The mass of the external weight is [tex]m = 51 \ kg[/tex]

The height through which the mass drops is [tex]h = 5.6 \ m[/tex]

     The  number of revolution made is   [tex]N  = 10100 \  kg[/tex]

    The  diameter of the free moving piston is     [tex]d  = 0.51 \  m \  kg[/tex]

     The distance moved by the free moving piston is   [tex]s  = 0.71 \  m \  kg[/tex]

     The  atmospheric pressure is  [tex]P  = 101 \ kPa = 101*10^{3}\ Pa [/tex]

Generally the workdone by the  external weight is mathematically represented as

      [tex]W_w  =m *  g   *  h[/tex]

=     [tex]  51 *  9.8  *  5.6 [/tex]

=     [tex]2799 N  [/tex]

Generally the workdone by the free moving piston  is mathematically represented as  

        [tex]W_p  =P *  A  *  s[/tex]

Here   A is the cross-sectional area with value  

        [tex] A  =  \pi *  \frac{d^2}{4}[/tex]

        [tex] A  =  3.142 *  \frac{0.51^2}{4}[/tex]

So

       [tex]W_p  =101*10^{3} * 3.142 *  \frac{0.51^2}{4}  *  0.71[/tex]

=>       [tex]W_p  = 14651 [/tex]

So

   The net workdone is mathematically evaluated as

        [tex]W_N   = -W_w+W_p  [/tex]

The negative sign shows that it is acting in opposite direction to [tex]W_N[/tex]

So

       [tex]W_N   = -2799+14651  [/tex]

      [tex]W_N   = 11849 \ J   [/tex]

Answer:

360000

Explanation:

Answer:

360000 milliseconds

Explanation:

I have this remembered dont worry homie i got chu

Answer: 12.5 km/h

Explanation: 20/1.6 = 12.5 km/h

The average velocity of an individual will be "0.08 km/min". To understand the calculation, check below.

According to the question,

Time, t = 20 minutes

Distance, d = 1.6 km

We know the formula,

→ Distance = Speed × Time,

Or,

Speed, v = [tex]\frac{Distance}{Time}[/tex]

By substituting the values,

               = [tex]\frac{1.6}{20}[/tex]

               = 0.08 km/min

Thus the avg. velocity will be "0.08 km.min" is correct.

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Answer:

48.9°, 29°

Explanation:

We kno that snell's law States that

( sin i / sin r ) = ( n 2 / n 1 )

And here

i = 90 - α

r = 90

n 2 = n a = 1.00029

n 1 = n = 1.52

So

( sin ( 90 - α ) / sin 90 ) = (1.00029 / 1.52 )

= 0.658

sin ( 90 - α ) = 0.658

90 - α = sin -1 ( 0.658 )

= 41.15

α = 90 - 41.15

= 48.9°

( b ) to find angle when prism immerssed in water

Using snell's law again

( sin i / sin r ) = ( n 2 / n 1 )

here i = 90 - α

r = 90

n 2 = n w = 1.33

n 1 = n = 1.52

So

( sin ( 90 - α ) / sin 90 ) = (1.33 / 1.52 )

= 0.875

sin ( 90 - α ) = 0.875

90 - α = sin -1 ( 0.875 )

= 61

α = 90 - 61

= 29°

Answer:

You are putting more force on the ball no the racket so the racket wouldnt swing back but if you hit a racket on the ground it will swing backits just the way the force reaction works

Explanation:

^^^^^^^^^^^^^^^

The racket wouldn't swing back because you were applying more power to the ball than to the racket, so the racquet swings backward when the ball hits it.

Any action that aims to maintain, modify, or deform a body's motion is considered a force in mechanics. Isaac Newton's three laws of motion, which are outlined in his Principia Mathematica, are frequently used to illustrate the concept of force (1687).

According to Newton's third law, whenever one body applies force to another, the second body also applies an equal amount of force to the first body. The action-reaction concept explains why a force tends to cause a body to deform, or change shape, whether it moves the body.

The racket wouldn't swing back because you were applying more power to the ball than to the racket, but it would if you struck the racket on the ground; that's how the force reaction operates.

Therefore, the racquet swings backward when the ball hits it.

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Answer:

Explanation:

W=?

Mass=1800kg

g=(9.8m/s^2 constant)

W=mg

W=1800*9.8

W=17649N

Answer:

Nearly all frequencies and wavelengths of electromagnetic radiation can be used for spectroscopy. Radio waves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays are all types of electromagnetic radiation. Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.

Explanation:

Answer:

hello your question lacks the required diagram attached below is the complete question with the required diagram

answer : Qtotal = 807.4 Mw

Explanation:

Given Data :

disk properties :

∈ = 0.65

D = 200 mm

Ts = 400⁰c

attached below is the detailed solution

The total rate of Heat transferred from the disk

Qtotal = 807.4 Mw

Answer:

The answer is

Explanation:

The speed of an object given it's kinetic energy and mass can be found by using the formula

[tex]v = \sqrt{ \frac{2KE}{m} } [/tex]

where

KE is the kinetic energy

m is the mass

v is the speed

From the question

KE = 31,500 J

m = 7000 kg

The speed is

[tex]v = \sqrt{ \frac{2(31500)}{7000} } \\ = \sqrt{ \frac{63000}{7000} } \\ = \sqrt{ \frac{63}{7} } \\ = \sqrt{9} \: \: \: \\ = 3 \: \: \: \: \: \: \: [/tex]

We have the final answer as

Hope this helps you

Answer:

The car does not hit the deer.

Explanation:

In order to find out, whether the car stops before hitting the dear or not, we will use 3rd equation of motion.

2as = Vf² - Vi²

where,

s = distance covered by car before stopping = ?

a = deceleration of car = - 4.2 m/s²

Vf = Final Velocity of the Car = 0 m/s (Since, the car finally stops)

Vi = Initial Velocity of the Car = 25 m/s

Therefore,

2(- 4.2 m/s²)s = (0 m/s)² - (25 m/s)²

s = (- 625 m²/s²)/(-8.4 m/s²)

s = 74.4 m

So, the car stops in 74.4 m, while the deer is at a distance of 75 m.

Hence, the car does not hit the deer.

Using third equation of kinematics

[tex]\boxed{\sf v^2-u^2=2as}[/tex]

[tex]\\ \sf\longmapsto (0)^2-(25)^2=2(-4.2)s[/tex]

[tex]\\ \sf\longmapsto -625=-8.4s[/tex]

[tex]\\ \sf\longmapsto 8.4s=625[/tex]

[tex]\\ \sf\longmapsto s=\dfrac{625}{8.4}[/tex]

[tex]\\ \sf\longmapsto s=74.4m[/tex]

Answer:

Mirage is an optical illusion due to the refraction of light as it travels through different layers of air at different temperature.

Explanation:

On a clear hot afternoon, the sun heats up the ground, and the ground heats up the air immediately above it. The result is that the air directly above the ground is hotter than the air above it, leaving the air layers with different refractive indies. Since refractive index decreases with increase in temperature, the air immediately above the ground bends the light that travels through it from the sky, directly into our eyes. The image seen is perceived to be reflected from the ground, but in an actual sense is a direct image of the sky, giving one the impression of seeing a pool of water on the ground.

Answer:

The  value is   [tex]F_t =  76024 \  N[/tex]  

Explanation:

From the question we are told that

  The  flow rate is  [tex]v_1 =  5 \  m/s[/tex]

   The  entrance diameter is  [tex]d =  0.7 \  m[/tex]

  The  exit diameter is evaluated as [tex]d_e  =  0.6  *  0.7 =  0.42  \  m[/tex]

   The entrance gauge pressure is  [tex]P_g =  350 \  kPa =  350*10^{3} \  Pa[/tex]

    The  droped gauge  pressure is  [tex]P_d =  50 \  kPa =  50*10^{3} \  Pa[/tex]

 The presure at the exist is evaluated as  [tex]P_e =(350  -    50 ) \  kPa =  300*10^{3} \  Pa[/tex]

Generally the entrance cross-sectional area is mathematically represented as

     [tex]A =  \pi \frac{d^2}{4}[/tex]

     [tex]A = 3.142 \frac{0.7^2}{4}[/tex]

      [tex]A =0.385 \ m^2[/tex]

Generally the exit  cross-sectional area is mathematically represented as

        [tex]A_e =  \pi \frac{d_e^2}{4}[/tex]

     [tex]A_e = 3.142 \frac{0.42^2}{4}[/tex]

      [tex]A_e =0.139\ m^2[/tex]

Generally from the continuity equation

    [tex]v_1 * A  =  v_2 * A_e[/tex]

=>  [tex]v_2  =  \frac{v_1 * A}{A_e}[/tex]

=>  [tex]v_2  =  \frac{5 * 0.385}{0.139}[/tex]

=>  [tex]v_2  =  13.8 m/s [/tex]

Generally the net force in horizontal axis is equivalent to the net momentum change in the horizontal direction

So

     [tex]P_g *  A +  F_t - P_e *  A_e  =  P_d [ A_e * v_2^2 -  A* v_1^2][/tex]

Here  [tex]F_t[/tex] is the horizontal thrust on the fitting

So  

        [tex]350*10^{3} * 0.385  +  F_t -0.385  *  0.139  =  50*10^{3} [ 0.385* 13.8^2 -  0.385* 5^2][/tex]

Answer:

-Suzy flies up into the sky because of the upward force on the jetpack.

-Newton's third law of motion can be ascribed to this.

Explanation:

We know that in momentum, Force is described as the rate of change of momentum. Thus, for us to understand the forces, we will need to establish how the momentum of the water changes.

Suppose the mass flow rate of the water is denoted as M kg/sec and the upward velocity is denoted as v_u.

Now, since water is lying stationary at the sea, the initial momentum of the water will be zero. If every second the boat pumps upwards with mass flow rate M at velocity vu, the momentum per second would be;

Δp = F_water = Mv_u

Now, from Newton's third law the boat/pump would experience a downwards force with the formula:

F_boat = −Mv_u

Furthermore at the jetpack, the water does a reverse turn in direction and is now again directed back down at a different velocity v_d. Therefore the final momentum per second is Mv_d and therefore change in momentum is now;

Δp = M(v_d - v_u)

The velocity v_d will be negative since the boat/pump would experience a downwards force. Thus, the force on the water will be downwards and F_water ≤ 0 .

We will finally use Newton's third law to conclude that the upward force on the jetpack will be;

F_jetpack = M(v_u - v_d)

Thus upward force is what makes Suzy to fly up in the sky.

Answer:

We know that the atomic number of sodium is 11. This tells us that sodium has 11 protons and because it is neutral it has 11 electrons. The mass number of an element tells us the number of protons AND neutrons in an atom (the two particles that have a measurable mass)

Answer:

a) Magnitude = 1.03 m/s², Direction: south

b) [tex]V_{f} = 8.16 m/s [/tex]

Explanation:

a) The magnitude and direction of the acceleration can be calculated using the following equation:

[tex] V_{f} = V_{0} + at [/tex]     (1)

Where:

[tex]V_{f}[/tex]: is the final speed = 9.40 m/s  

[tex]V_{0}[/tex]: is the initial speed = 13.0 m/s

t: is the time = 3.50 s

Solving equation (1) for a, we have:

[tex] a = \frac{V_{f} - V_{0}}{t} = \frac{9.40 m/s - 13.0 m/s}{3.50 s} = -1.03 m/s^{2} [/tex]

Hence, the magnitude of the acceleration is 1.03 m/s² and the direction of the bird's acceleration is the opposite of the initial velocity direction, which means that the bird is decelerating.  

b) The final velocity of the bird can be found using the same equation 1:

[tex] V_{f} = V_{0} + at [/tex]

[tex] V_{f} = 13.0 m/s + (-1.03 m/s^{2})*(3.50 s + 1.20 s) = 8.16 m/s [/tex]

Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.

I hope it helps you!