What are the examples of pulley? Plz tell the answer as fast as possible plz.​

Answer:

Explanation:

The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:

initial velocity is 20 m/s

Here's what we know that we are NOT told:

a = -9.8 m/s/s and

final velocity is 0 at an object's max height in parabolic motion.

We will use the equation:

[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:

0 = 20 + (-9.8)t and

-20 = -9.8t so

t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.

Answer:

Friction always opposes the motion

HAVE A GREAT DAY :)

Explanation:

the car will be brought back

Answer:

store a charge for later use

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Answer:

I = 3.6 kg•m²

Explanation:

Conservation of angular momentum

Let's assume CW is the positive direction

3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)

I(9.3 - 1.8) = 3.4(1.8 + 6.1)

I(7.5) = 3.4(7.9)

I = 3.4(7.9)/(7.5) = 3.5813333333...

The moment of inertia of the second disk will be  [tex]I=3.58\ kg-m^2[/tex]

The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.

here it is given that

MOI of disk one   [tex]I_1=3.4\ kg-m^2[/tex]

Angular velocity  [tex]w_1=6.1\ \frac{rad}{s}[/tex]

Angular velocity of disk two  [tex]w=1.8\ \frac{rad}{s}[/tex]

MOI of the disk two [tex]I=?[/tex]

The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]

Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.

[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]

Now put the values in the formula

[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]

[tex]I_2=3.58\ kg-m^2[/tex]

Thus the moment of inertia of the second disk will be  [tex]I=3.58\ kg-m^2[/tex]

To know more about moment of inertia follow

https://brainly.com/question/18721435

Answer:

b = frequency

Answer:

Because SI system has fundamental units of MKS System

Answer:

Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.

Answer:

The composition of Jupiter is similar to that of the Sun—mostly hydrogen and helium. Deep in the atmosphere, pressure and temperature increase, compressing the hydrogen gas into a liquid. This gives Jupiter the largest ocean in the solar system—an ocean made of hydrogen instead of water.

Answer:

nnnjjdndbsnnshfhhgbfbdbdh

Answer:

Hence the battery's emf E is ε = 11.9 V.

The internal resistance is r = 0.739 ohms.

Explanation:

Now we know that

Voltage V = 11.9 V.

Current I = 16.1 A.

Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.

ε = V + I r

∵ I = 0

ε = V

ε = 11.9 V

Then the ammeter is applied.

Let's take ( r ) to be the total resistance which is equal to internal resistance.

V = I r

r = [tex]\frac{V}{I}[/tex]

 [tex]= \frac{11.9}{16.1}[/tex]

r = 0.739 ohms

The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.

Given the following data:

To determine the battery's emf (E) and internal resistance (r):

For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.

Mathematically, emf (E) is given by this formula:

[tex]E = V + IR[/tex]

Substituting the given parameters into the formula, we have;

[tex]E = 11.9 + 0R\\\\E = 11.9 + 0[/tex]

E = 11.9 Volts.

For the internal resistance (r):

Note: The total resistance is equal to internal resistance.

Applying Ohm's law, we have:

[tex]R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}[/tex]

R = r = 0.739 Ampere.

Read more current here: https://brainly.com/question/25813707

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

Answer:

where are the options

it's not full question

Answer:

The correct solution is:

(a) 1.66

(b) 1.05

Explanation:

Given:

Bending stress,

[tex]\sigma_b = 25 \ kpsi[/tex]

Torsional stress,

[tex]\tau= 15 \ kpsi[/tex]

Yield stress of steel bar,

[tex]\delta_y = 60 \ kpsi[/tex]

As we know,

⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]

        [tex]= \sqrt{(25)^2+3(15)^2}[/tex]

        [tex]=36.055 \ kpsi[/tex]

(a)

The factor of safety against static failure will be:

⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]

By putting the values, we get

        [tex]=\frac{60}{36.055}[/tex]

        [tex]=1.66[/tex]

(b)

According to the Goodman line failure,

[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]

[tex]S_e = 40 \ kpsi[/tex]

[tex]\sigma_m = \sqrt{3} \tau[/tex]

     [tex]=\sqrt{3}\times 15[/tex]

     [tex]=26 \ kpsi[/tex]

[tex]Sut = 80 \ kpsi[/tex]

⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]

      [tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]

              [tex]\eta_y = 1.05[/tex]

Answer:

5572.8 N

Explanation:

Applying,

F  = ma.............. Equation 1

Where F = Force, m = mass of the car, a = acceleration.

We can find a by applying,

v² = u²+2as............. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration,  = distance.

From the question,

Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m

Substitute these value into equation 2

0² = 1.7²+2×0.21×a

a = -1.7²/(2×0.21)

a = -2.89/0.42

a = -6.88 m/s²

Also given: m = 810 kg

Substitute these value into equation 1

F = 810(-6.88)

F = -5572.8 N

Hence the force on the bumber is 5572.8 N

Answer:

Newton's First Law of Motion states that a body will stay at rest or continue its path with constant velocity unless an external force acts upon it. Newton's Second Law of Motion states that the net force that acts upon a body is equal to the mass of the body multiplied by the acceleration due to the net force.

Answer: [tex]3.92\ m/s[/tex]

Explanation:

Given

Mass of the attached object is [tex]m=2.3\ kg[/tex]

Spring is stretched by [tex]A=0.5\ m[/tex]

Speed reaches zero after [tex]t=0.4\ s[/tex]

Speed is zero at the extremities of the S.H.M motion that is

[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]

Time period of motion is [tex]0.8\ s[/tex] which can also be given by

[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]

Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]

[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]

Answer:

a)   # _photon = 2.5 10¹⁸ photons / s,   b) E = 10⁻² N / C,  c)     B = 3 10⁻¹¹ T

d)  r=  2 10⁹ m

Explanation:

a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation

          E₀ = h f

          c = λ f

          E₀ = h c /λ

          E₀ = 6.63 10⁻³³⁴   3 10⁸/500 10⁻⁹

          E₀ = 3.978 10⁻⁻¹⁹ J

Let's use a direct ratio rule to find the number of photons

         #_foton = E / Eo

         #_fototn = 1 / 3.978 10⁻¹⁹

         # _photon = 2.5 10¹⁸ photons / s

b) The intensity received by the detector is related to the electric field

          I = E²

Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space

          I = P / A

          P = I A

Let's use index 1 for the point on the bulb and index 2 for the point on the detector.

The area of ​​a sphere is

          A = 4π r²

         P = I₁ A₁ = I₂ A₂

         I₁ r₁² = I₂ r₂²

         I₂ = I₁  r₁²/r₂²

         I₂ = I₁    1 / 100²

         I₂ = I₁ 10⁻⁴

we must know the intensity at the output of the bulb suppose that I₁ = 1 J

          I₂ = 10⁻⁴ J

let's look for the electric field

         E =√I

         E = √10⁻⁴

         E = 10⁻² N / C

c) for the calculation of the magnetic field we use that the field is in phase

               E / B = c

               B = E / c

               B = 10⁻² / 3 10⁸

               B = 3 10⁻¹¹ T

d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area  A = 4π R² where R = 100 m how many photons are there in the area of ​​the detector r = 1 cm,   A’= 10⁻⁴ m²

             #_photons = 2.5 10¹⁸ A_detector / A_sphere

             #_photons = 2.5 1018 10-4 / 4π 10⁴

             #_photons = 2 10⁹ photons in the detector area

for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m

Solution :

Let the positive [tex]x-axis[/tex] is along the East and the positive [tex]y[/tex] direction is along the north.

Given :

Mass of the Tesla car, [tex]m_1[/tex] = [tex]750 \ kg[/tex]

Mass of the Ford car, [tex]m_2 = 1250 \ kg[/tex]

Now let the initial velocity of Tesla car in the south direction be = [tex]-v_1j[/tex]

The initial momentum of Tesla car, [tex]p_1 = -750 \ v_1[/tex]

Let the initial velocity of Ford car in the east direction be = [tex]v_2 \ i[/tex]

So the initial momentum of the Ford car is [tex]p_2=1250\ v_2 \ i[/tex]

Therefore, the initial velocity of both the cars is [tex]p_i = p_1+p_2[/tex]

                                                                  [tex]=1250 \ v_2 \ i - 750\ v_1 \ j[/tex]

Now the final velocity of both the cars is [tex]v = 18 \ m/s[/tex]

So the vector form is :

[tex]v = 18\cos 32\ i-18 \sin 32 \ j[/tex]

  [tex]= 15.26 \ i - 9.54 \ j[/tex]

Therefore the momentum after the accident is

[tex]p_f=(m_1+m_2) \times v[/tex]

    [tex]=(750+1250) \times (15.26 \ i - 9.54 \ j)[/tex]

    [tex]= 30520\ i -19080\ j[/tex]

According to the law of conservation of momentum, we know

[tex]p_i = p_f[/tex]

[tex]1250 \ v_2 \ i - 750\ v_1 \ j[/tex]  [tex]= 30520\ i -19080\ j[/tex]

[tex]1250 \ v_2 = 30520[/tex]

[tex]v_2=24.4 \ m/s[/tex]

From, [tex]750\ v_1 = 19080[/tex]

We get, [tex]v_1=25.4 \ m/s[/tex]

Therefore the speed of Tesla car before collision = 25.4 m/s

The speed of ford car before collision = 24.4 m/s

Answer:

12

Explanation:

i agree the answer is 12 because the acceleration is given by the difference in the velocity divided by the time taken

a=v-u/t

60-0/5

=12m/s²

I hope this helps

Answer:

Explanation:capacitor

Answer:

Ammeter

Explanation:

pls mark me as a brainlist

Answer:

Part A - 4.084 mJ

Part B - 0.908 mJ

Part C - 8.168 mJ

Explanation:

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',

1/C' = 1/C₂ + 1/C₃      (Since C₁ = C₂ = C₃ = C)

1/C' = 1/C + 1/C

1/C' = 2/C

C' = C/2

Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2

C" = 3C/2

The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V

W = 1/2C"V²

W = 1/2(3C/2)V²

W = 3CV²/4

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/4

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4

W = 16335/4 × 10⁻⁶ FV²

W = 4083.75 × 10⁻⁶ J

W = 4.08375 × 10⁻³ J

W = 4.08375 mJ

W ≅ 4.084 mJ

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

If the capacitors are connected in series, their equivalent resistance is C'

and 1/C' = 1/C₁ + 1/C₂ + 1/C₃

Since C₁ = C₂ = C₃ = C

1/C' = 1/C + 1/C + 1/C

1/C' = 3/C

C' = C/3

The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(C/3)V²

W = CV²/6

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = CV²/6

W = 24.2 × 10⁻⁶ F (15.0 V)²/6

W = 24.2 × 10⁻⁶ F × 225 V²/6

W = 5445/6 × 10⁻⁶ FV²

W = 907.5 × 10⁻⁶ J

W = 0.9075 × 10⁻³ J

W = 0.9075 mJ

W ≅ 0.908 mJ

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

If the capacitors are connected in parallel, their equivalent resistance is C'

and C' = C₁ + C₂ + C₃

Since C₁ = C₂ = C₃ = C

C' = C + C + C

C' = 3C

The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(3C)V²

W = 3CV²/2

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/2

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2

W = 16335/2 × 10⁻⁶ FV²

W = 8167.5 × 10⁻⁶ J

W = 8.1675 × 10⁻³ J

W = 8.1675 mJ

W ≅ 8.168 mJ

Answer:

"the minimum frequency of radiation that will produce a photoelectric effect."

Explanation:

That answer was derived from gogle cuz my explanations was harder to explain but good luck

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

[tex] F = qv \times B = qvBsin(\theta) [/tex]     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

[tex] K = \frac{1}{2}mv^{2} [/tex]

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

Answer:

if you want to drop into the water at the greatest possible distance from the edge, you must hang for 1.662s.

Explanation:

The time period of the oscillation is,

[tex]T = 2\pi \sqrt{ \frac{I} {g }[/tex]

[tex]T = 2\pi \sqrt{\frac{11}{9.8} } \\\\T= 6.65 s[/tex]

This would be the time taken for the person to move from.

The duration of time he hangs over the river be one-fourth of the time period.

Here,

[tex]t= \frac{T}{4} \\\\t=\frac{6.65}{4}\\\\t = 1.662 s[/tex]

Answer:

English please?

Explanation:

Explain in english?

Answer:

The answer is "Option B".

Explanation:

[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]

[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]

Explanation:

the process of conversion of energy from one form to another is called transformation of energy.