Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?

a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

D) The bicycle is not in motion.

Explanation:

Study the position-time graph for a bicycle.

Which statement is supported by the graph?

A) The bicycle has speed but not velocity.

B) The bicycle is moving at a constant velocity.

C) The bicycle has a displacement of 3 m.

D) The bicycle is not in motion.

Solution:

Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.

Speed is the time rate of change of distance. It is the ratio of distance to time taken.

From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.

From the position-time data provided, it can concluded that the bicycle is not in motion.

Motion of a body involves a change in the position of that body with time.

A body in motion is constantly changing position or orientation as time passes.

The body may move with constant velocity/speed or changes in its velocity.

A position-time graph provides information about the motion of a body.

From the data provided:

The position of the bicycle remains the same for all time intervals.

Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.

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Answer:

[tex]\triangle P=1.95*10^{-4}[/tex]

Explanation:

Mass [tex]m=0.001[/tex]

Diameter [tex]d=1.2m[/tex]

Length [tex]l=10m[/tex]

Generally the equation for Volume flow rate is mathematically given by

 [tex]Q=AV[/tex]

 [tex]V=\frac{Q}{\pi/4D^2}[/tex]

 [tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]

 [tex]V=8.84*10^{-4}[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]F=\frac{64}{Re}[/tex]

Where Re

Re=Reynolds Number

 [tex]Re=\frac{pVD}{\mu}[/tex]

 [tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]

 [tex]Re=1040[/tex]

Therefore

 [tex]F=\frac{64}{Re}[/tex]

 [tex]F=\frac{64}{1040}[/tex]

 [tex]F=0.06[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]Head loss=\frac{fLv^2}{2dg}[/tex]

 [tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]

 [tex]H=19.9*10^{-9}[/tex]

Where

[tex]H=\frac{\triangle P}{\rho g}[/tex]

[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]

[tex]\triangle P=H*\rho g[/tex]

[tex]\triangle P=1.95*10^{-4}[/tex]

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

I don't know

because I don't know

Answer:

Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

Answer:

Things you should do for your family

things you shouldn't

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

Power = work done/time

Put all the values,

[tex]P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W[/tex]

So, the required power is 70 W.

Answer:

43994

Explanation:

Hope this helps!

Answer: 31.89seconds

Explanation:

Based on the information given, we are meant to calculate deceleration which will be:

t = V/a

where, a = mg

Therefore, t = V/mg

t = 25/0.08 × 9.8

t = 25/0.784

t = 31.89seconds

Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.

Answer:

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),

Explanation:

Answer:

W = 16.4 kJ

Explanation:

Given that,

There are 135 steps from the ground floor to the sixth floor.

Each step is 16.6 cm tall.

The mass of a person, m = 73.5 kg

We need to find the work done by the person. We know that,

Work done = Fd

Where

d is the displacement, d = 135 × 0.166 = 22.41

So,

W = 73.5 × 10 × 22.41

= 16471.35  J

or

W = 16.4 kJ

So, 16.4 kJ is the work done by the person.

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

Answer:

The angle is 4.1 rad.

Explanation:

The centripetal acceleration (α) is given by:

[tex] \alpha = \omega^{2} r [/tex]    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

[tex] a = \alpha r [/tex]      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

[tex] \omega^{2} r = 8.2\alpha r   [/tex]

[tex] \omega^{2} = 8.2\alpha [/tex]    (3)      

Now, we can find the angle with the following equation:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]

Where:

[tex] \omega_{f}[/tex]: is the final angular velocity                                                                              [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)

[tex]\Delta \theta[/tex]: is the angle

[tex] \omega^{2} = 2\alpha \Delta \theta [/tex]     (4)    

By entering equation (3) into (4) we can calculate the angle:

[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]

[tex] \Delta \theta = 4.1 rad [/tex]

Therefore, the angle is 4.1 rad.

I hope it helps you!                  

Explanation:

V= IR

35=I×7

I=35/7

I=5amperes

pls give brainliest

The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.

The correct answer is option E.

To calculate the energy of a photon, we can use the equation:

E = (hc) / λ

where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.

Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:

E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)

E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)

E = 3.3695 x [tex]10^-^1[/tex] eV

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The question probable may be:

The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)

A) 2.81 eV

B) 3.89 eV

C) 2.10 eV

D) 2.78 eV

E)  0.337 eV

Answer:

a. 59 ev. helpful answer

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]

The maximum magnetic field is calculated using the following intensity formula;

[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

Answer:

because it is helpful to human beings I think

a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]

b. The constant force exerted on the pole vaulter is 6799.52 N

a. We use Newton's equation of motion,

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

Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.

Given that,  s = 5.1 m , t = 0.29s, u = 0

Substitute in above equation.

            [tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]

the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]

b. The constant force exerted on the pole vaulter due to the collision is given as,             [tex]Force=mass*acceleration[/tex]

             [tex]Force=56*121.42=6799.52N[/tex]

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Answer:

No of proton is 13 and nucleus is 13

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

Answer:

12

Explanation:

I'm a beginner so am not sureeeeee