In a simple machine, the energy input is 120 J. If the efficiency of the machine is 80%, calculate the energy output

Answer:

a. T= 0

b. T = 1286N

c. T= 2058N

Explanation:

Answer:

(a) V = 0.75 m/s

(b) V = 0.125 m/s

Explanation:

The speed of the flow of the river can be given by following formula:

V = Q/A

V = Q/w d

where,

V = Speed of Flow of River

Q = Volume Flow Rate of River

w = width of river

d = depth of river

A = Area of Cross-Section of River = w d

(a)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 20 m

d = 20 m

Therefore,

V = (300 m³/s)/(20 m)(20 m)

V = 0.75 m/s

(b)

Here,

Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s

w = 60 m

d = 40 m

Therefore,

V = (300 m³/s)/(60 m)(40 m)

V = 0.125 m/s

1750N/m

According to Hooke's law, the spring constant (k) of an elastic material is the ratio of the force (F) applied to the material to the extension (x) of the material caused by this force. i.e

k = F / x         --------------(i)

From the question, the elastic material (spring) of 25 coils has a spring constant of 350N/m. This means that for every 350N force applied to the spring, the spring extends by 1m.

Now if the spring is cut into five equal parts each with five coils, imagine they are attached together such that the same force of 350N is applied to still cause a total extension of 1m. Each spring contributes 1/5 of this extension.

Therefore, the extension caused by each spring is 1/5 of 1m = 0.2m

Since the same force of 350N is applied, substitute F = 350N and x =  0.2m into equation (i) as follows;

k = 350 / 0.2

k = 1750N/m

Therefore, the spring constant of each of the 5-coil spring is 1750N/m

The spring constant of each of the 5-coil springs is 1,750 N/m.

The given parameters;

The total spring constant of the 25 coils is calculated as follows;

[tex]K_t = 25 \times 350 \ N/m\\\\K_t = 8,750 \ N/m[/tex]

The spring constant of each of the given 5 coils is calculated as follows;

[tex]K = \frac{8,750 }{5} \\\\K = 1,750 \ N/m[/tex]

Thus, the spring constant of each of the 5-coil springs is 1,750 N/m.

Learn more here:https://brainly.com/question/22499689

Answer:

272.89g

Explanation:

Find the diagram to the question in the attachment below;.

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.

Moment = Force * Perpendicular distance

Taking the moment of force about the pivot.

Anticlockwise moment:

The 85g mass will move in the anticlockwise

Moment of 85g mass = 85×36.6

= 3111gcm

Clockwise moment.

The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.

CW moment = 11.4×M = 11.4M

Equating CW moment to the ACW moment we will have;

11.4M = 3111

M = 3111/11.4

M = 272.89g

The mass of the metre stick is 272.89g

Speed = (distance) / (time)

Speed = (1,233 mile) / (2.4 hour)

Speed = 513.75 mile/hour

Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)

Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)

Speed = 229.7 meter/second

Answer:

Explanation:

The force of the lake bottom on the boat = force the boat exerts on the bottom

force the boat exerts on the bottom  = weight of the boat - buoyant force on the boat

so  

The force of the lake bottom on the boat  = weight of the boat - buoyant force on the boat

So

The force of the lake bottom on the boat will be less than its weight .

Answer:

Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.

Explanation:

Answer:

heat required = mass × specific heat × change in temperature

Explanation:

0.25 × 4200 × 40

=42000 J

Answer:

The answer to this question can be defined as follows:

Explanation:

In the given question, an equation is missing which can be defined as follows:

[tex]I \omega =\sqrt{J(J+1)}h[/tex]

solution:

Angular momentum:

[tex]L=I \omega =\sqrt{J(J+1)}h[/tex]

Convert Angular momentum in terms of kinetic energy:

[tex]K = \frac{L^2}{2I}[/tex]

    [tex]= \frac{h^2(J(J+1))}{2I}[/tex]

Answer:

B

Explanation:

Solution:-

- Take a coordinate system as follows:

                   x: Directed along the ground

                   y: Directed vertical to ground

- We will assume that the initial vertical and horizontal non-zeroes velocities are given as follows:

                          [tex]v_x_i = v_o_x\\\\v_y_i = v_o_y\\[/tex]

- After a ball is thrown it continues a path of parabolic projectile. The motion of the ball can be analyzed in each coordinate system. We will assume that effects of air-resistance are negligible.

- Therefore, only gravity acts on the ball in the vertical direction. We can use kinematic equation of motions to determine the velocity of ball in either ( x-y ) direction at any instant of time ( t ).

- Use first kinematic equation of motion in both x and y directions.

                         [tex]v_x_f = v_o_x + a_x*t\\\\v_y_f = v_o_y + a_y*t\\[/tex]

- The accelerations ( ax and ay ) in the direction of each axis are to be determined. We know that the gravity acceleration ( g ) acts in vertical direction or along y-axis ( ay ) and always directed downwards while velocity is directed up. Since, we neglected the effects of air-resistance there is no acceleration in the x-direction ( ax = 0 ) .

                        [tex]v_x_f = v_o_x\\\\v_y_f = v_o_y - gt\\[/tex]

- We see that the horizontal velocity of the ball ( vxf ) at any point in time remains equal to the initial horizontal velocity; hence, it is constant throughout the journey.

- However, the velocity of the ball in vertical direction( vyf ) is changing for every unit of time ( t ) under the influence of gravitational acceleration. Hence, it is not constant throughout the journey

Answer:

A)    θ = 28.1º , B)         v = 2.47 m / s

Explanation:

A) The angle of the ramp can be found using trigonometry

         sin θ = y / L

         Φ = sin⁻¹ y / L

         θ = sin⁻¹ (115/244)

         θ = 28.1º

B) For this pate we can use the relationship between work and kinetic energy

       W =ΔK

where the work is

       W = -fr x

the negative sign is due to the fact that the friction force closes against the movement

Lavariacion of energy cineta is

         ΔEm = ½ m v² - mgh

        -fr x = ½ m v² - m gh

the friction force has the equation

         fr = very N

at the highest part there is no speed and we take the origin from the lowest part of the ramp

To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular

Axis y . perpendicular

            N- Wy = 0

            cos tea = Wy / W

            Wy = W cos treaa

             N = mg cos tea

we substitute

- (very mg cos tea) x = ½ m v²2 - mgh

            v2 = m (gh- very g cos tea x)

   let's calculate

           v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)

           v = RA (7.84 -1.729)

           v = 2.47 m / s

Answer:

Δf = 73.72Hz

Explanation:

In order to calculate the difference in frequency between the direct waves and the reflected waves, you first take into account the Doppler's effect for an observer getting closer to the source:

[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]         (1)

You can assume that the reflected waves come from a source "the whale". Then you have:

f': frequency of the reflected waves = ?

f: frequency of the source = 22.0*kHz = 22.0*10^3 Hz

v: speed of the sound in water = 1482m/s

vs: speed of the source = 4.95m/s

vo: speed of the observer = 0m/s

You replace the values of the parameters in the equation (1):

[tex]f'=(22.0*10^3Hz)(\frac{1482m/s}{1482m/s-4.95m/s})=22073.72Hz[/tex]

Then, the difference in frequency is:

[tex]\Delta f = f'-f=22000Hz-22073.72Hz=73.72Hz[/tex]

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

Answer:

h = 8.48*10^-3m

Explanation:

In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.

The total momentum of the system before the collision is equal to the total momentum after:

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]        (1)

Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.

m1: mass of the marble = 0.0215kg

m2: mass of the pendulum concentrated at its end = 0.250kg

v1: horizontal speed of the arble before the collision = 5.15m/s

v2: horizontal speed of the pendulum before the collision = 0m/s

v: horizontal speed of both marble and pendulum after the collision = ?

You solve the equation (1) for v, and replace the values of the other parameters:

[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]

Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:

[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]

v = 0.40m/s

g: gravitational acceleration = 9,8m/s^2

[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]

Then, the height reached by marble and pendulum is 8.48*10^-3m

Answer:

The width is  [tex]Z = 0.0424 \ m[/tex]

Explanation:

From the question we are told that

    The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]

    The wavelength of the light is  [tex]\lambda = 721 \ nm[/tex]

      The position of the screen is  [tex]D = 2.83 \ m[/tex]

Generally angle at which the first minimum  of the interference pattern the  light occurs  is mathematically  represented as

        [tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]

Where m which is the order of the interference is 1

substituting values

       [tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]

      [tex]\theta = 0.5317 ^o[/tex]

 Now the width of first minimum  of the interference pattern is mathematically evaluated as

       [tex]Y = D sin \theta[/tex]

substituting values

       [tex]Y = 2.283 * sin (0.5317)[/tex]

       [tex]Y = 0.02 12 \ m[/tex]

 Now the width of  the  pattern's central maximum is mathematically evaluated as

        [tex]Z = 2 * Y[/tex]

substituting values

      [tex]Z = 2 * 0.0212[/tex]

     [tex]Z = 0.0424 \ m[/tex]

Answer:

nvbnncbmkghbbbvvvvvvbvbhgggghhhhb

Answer:

a)[tex]h_{max}=14536.16 m[/tex]

b)[tex]h = 15687.9 m[/tex]

c)[tex]PD=7.62\%[/tex] The estimate is low.

Explanation:

a) Using the energy conservation we have:

[tex]E_{initial}=E_{final}[/tex]

we have kinetic energy intially and gravitational potential energy at the maximum height.

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

[tex]h_{max}=\frac{v^{2}}{2g}[/tex]

[tex]h_{max}=\frac{43^{2}}{2*0.0636}[/tex]

[tex]h_{max}=14536.16 m[/tex]  

b)  We can use the equation of the gravitational force

[tex]F=G\frac{mM}{R^{2}}[/tex]   (1)

We have that:

[tex] F = ma [/tex]    (2)

at the surface G will be:

[tex]G=\frac{gR^{2}}{M}[/tex]

Now the equation of an object at a distance x from the surface.

is:

[tex]F=\frac{mgR^{2}}{(R+x)^{2}}[/tex]

[tex]m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}[/tex]

Using that dv/dt is vdx/dt and integrating in both sides we have:

[tex]v_{0}=\sqrt{\frac{2gRh}{R+h}}[/tex]

[tex]h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}[/tex]

[tex]h=15687.9[/tex]

c) The difference is:

So the percent difference will be:

[tex]PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%[/tex]

[tex]PD=7.62\%[/tex]

The estimate is low.

I hope it helps you!

Answer:

a) [tex]\Delta x=56.25 m[/tex]

b) imagen adjunta

Explanation:

a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.

Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:

[tex]v_{f}^{2}=v_{0}^{2}+2a \Delta x [/tex]

Queremos encontrar la posición hasta detenerse, osea vf = 0.

[tex]\Delta x=\frac{-v_{0}^{2}}{2a}[/tex]

[tex]\Delta x=\frac{-22.5^{2}}{-2*4.5}[/tex]

[tex]\Delta x=56.25 m[/tex]

b) Para este caso el gráfico se encuentra adjunto.                                      

Espero que te sirva de ayuda!                                                                                                                                                                          

Explanation:

The collision forces are equal and opposite.  Therefore, the magnitudes are equal.

Given that,

radius = 1.24 m

According to question,

The rope cannot push outwards. It must always have some slight tension or the bucket will fall.

We need to calculate the tension in the rope

At the top the force of gravity is

[tex]F=mg[/tex]

The force needed to move the bucket in a circle is centripetal force.

So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.

The rope have a tension of less than zero.

We need to calculate the velocity of swing bucket

Using centripetal force

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]mg=\dfrac{mv^2}{r}[/tex]

[tex]g=\dfrac{v^2}{r}[/tex]

[tex]v^2=gr[/tex]

[tex]v=\sqrt{gr}[/tex]

Put the value into the formula

[tex]v=\sqrt{9.8\times1.24}[/tex]

[tex]v=3.49\ m/s[/tex]

Hence, The minimum tension in the rope is less than zero .

The bucket swings with the velocity of 3.49 m/s.

Answer:

fggdfddvdghyhhhhggghh

Answer:

sphere, disk, hoop

Explanation:

See attached file

Answer:

a) virtual

Explanation:

This is a lens problem, in this case the equation that describes the process is the constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, weights the distance to the object and qq the distance to the image

In a projection system the image that we see is q, which can have several characteristics

a) virtual It can never be projected, since this image is formed by the extensions of the light rays, therefore it is not real but a construction of the brain that interprets where the rays must come from.

b) Real. This image can be projected since light rays pass through the image

c) Inverted. Inverted images are real so they can be projected, so rays pass through the image

d) expanded. In this case the image is greater than the object, this occurs when the object the distance to the image is greater than the distance to the object, therefore the distance q is negative, therefore this image is straight and is formed by the extensions from the rays and you can't project

Answer:

option A) virtual

Explanation:

hope this helps :)

Answer:

hand saws

power saws

Circular Saw

Explanation:

that is all that i know

Answer:

E = 326.17 N/C

Explanation:

(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]         (1)

vo: initial speed of the proton = 0m/s

t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s

a: acceleration of the proton = ?

x: distance traveled by the proton = 1.60cm = 0.016m

You solve the equation (1) for a, and replace the values of all parameters:

[tex]a=\frac{2x}{t^2}=\frac{2(0.016m)}{(3.20*10^{-6}s)^2}=3.125*10^{10}\frac{m}{s^2}[/tex]

Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:

[tex]F_e=qE=ma[/tex]           (2)

q: charge of the proton = 1.6*10^-19C

m: mass of the proton = 1.77*10^-27kg

You solve the equation (2) for E:

[tex]E=\frac{ma}{q}=\frac{(1.67*10^{-27}kg)(3.125*10^{10}m/s^2)}{1.6*10^{-19}C}\\\\E=326.17\frac{N}{C}[/tex]

The magnitude of the electric field in between the parallel plates is 326.17N/C

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

Answer:

the length that would produce a sound tone under the same experimental contditions must be increased by  Δl = [tex]\frac{v}{2f}[/tex]

Explanation:

Recall

V = f ×λ

where λ is ⁴/₃l₂ for second resonance

f = [tex]\frac{3v}{4l_{2} }[/tex]

l₂ = [tex]\frac{3v}{4f}[/tex]

where λ is 4l₁ for 1st resonance

f = [tex]\frac{v}{4l_{1} }[/tex]

l₁ = [tex]\frac{v}{4f}[/tex]

∴ Δl = l₂ - l₁ =  [tex]\frac{3v}{4f}[/tex] ⁻  [tex]\frac{v}{4f}[/tex]

Δl=  [tex]\frac{2v}{4f}[/tex]

Δl = [tex]\frac{v}{2f}[/tex]

Therefore, the length should increase by [tex]\frac{v}{2f}[/tex]

Plz check attachment for answer.

Hope it's helpful

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

(b) Find the angle in radians through which it rotates in this time interval.

Now we convert rad to angle

The angle in radians = 27.9°

Answer:

Explanation:

In interferometer , when the movable mirror is moved away by distance d , there is fringe shift on the screen . If n be number of fringes shifted

2 d = n λ

where λ is wavelength of light

Applying this theory for first case when no of fringes shifted is 146

2 d₁ = 146 x 656.3 nm

d₁ = 47909.9 x 10⁻⁹ m

= .048 x 10⁻³ m

= .048 mm

For second case  n = 122

2d₂ = 122 x 434 x 10⁻⁹

d₂ = 26474 x 10⁻⁹ m

= .026 mm

So in the second case , mirror must have been moved towards the beam splitter by .048 - .026 = .022 mm

So movement =  -  0 .022 mm ( negative displacement )