a jet of water of cross section area and velocity v impinges normally on a stationary flat plate the mass per unit volume of water ρ
by dimensional analysis determine and expression for the force f exerted by the jet against the plate.​

The tension in the rope B is determined as 10.9 N.

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

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The work done is 8427 J while the energy expended is  80.25 W.

Work done is defined as the product of the force and the distance. We know that the work done in a gravitational field is given as;

W = mgh

Total mass of the water bucket and chain = 13.9 kg +  19.3 kg = 33.2Kg

Distance covered =  25.9 m

W = 33.2Kg * 9.8 m/s^2 * 25.9 m

W = 8427 J

Recall that the work done = Energy expended

Power = Energy expended/ Time

Power = 8427 J/ 1.75 * 60 seconds

Power = 80.25 W

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Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

[tex]\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg[/tex]

We can simplify and rearrange the equation to solve for 'v'.

[tex]\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}[/tex]

Plugging in values:

[tex]v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}[/tex]

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

[tex]\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg[/tex]

Rearranging for 'T":
[tex]T = \frac{mv^2}{r} + mg\\\\[/tex]

Plugging in the appropriate values:
[tex]T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}[/tex]

Answer:

The centripetal acceleration (ac)=314m/s²

Explanation:

look at the attachment ☝️

The magnitude of the tension in the string marked A and B is mathematically given as

A = 52.5 N

Generally, the equation for is mathematically given as

The angle at A is...

[tex]tan\theta = 3/8[/tex]

When below the negative x

B= tan\phi

[tex]tan\phi = 5/4[/tex]

When below the negative x

C=

[tex]tan \rho = 1/6[/tex]

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Hence, considering the Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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The horizontal component of the force  is 336 N

The magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons

The system of the beam, hi_nge and cable are in equilibrium.

The horizontal component of the force exerted by the beam is given below as:

Horizontal component of the force = mg × cos 31°

Horizontal component of the force = 40 × 9.8 × cos 31°

Horizontal component of the force  = 336 Newtons

The magnitude of the force that the beam exerts on the hi_nge is given as:

F = mg × cos 31° + mg × sin 31°

F = 40 × 9.8 × cos 31° + 40 × 9.8 × sin 31°

F = 537.9 Newtons

Hence, the magnitude of the force that the beam exerts on the hi_nge is 537.9 Newtons.

In conclusion, the system of forces acting on the hi_nge and the beam are in equilibrium.

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Answer:

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

(work required to remove bucket)

Center of mass of chain = 25.9 / 2 = 12.95 m   height of CM of chain

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules  

(work required to remove chain)

Total work = 3530 + 2450 = 5980 Joules

P = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

(a) Her distance from the starting location is 21.05 m.

(b) The length of the path she skated is 21.05 m.

The distance around a complete circular path is calculated as 2πr.

The distance for a half circle is calculated as ¹/₂ x 2πr = πr

Distance from the starting location = π x 6.7 m = 21.05 m

The length of the path she skated is the same as her distance from the starting location = 21.05 m.

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A. The efficiency of the ramp is 54.6 %.

V.R = distance moved by effort/distance moved by load = L/h

V.R = (6.12)/(1.53) = 4

M.A = Load/Effort

M.A = (451 x 9.8)/(2025)

M.A = 2.183

E = (M.A / V.R) x 100%

E = (2.183 / 4) x 100%

E = 54.6 %

Thus, the efficiency of the ramp is 54.6 %.

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The radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.

Using the formula:

r = [tex]\frac{mvsin \alpha }{Bq}[/tex]

m = 9.109 × 10-31 kg

α = 35°

B = 0. 040 T

q = 1.6 × 10-19 C

Substitute values into the formula

[tex]r = \frac{9. 109 * 10^-31 * 4.0 * 10^0 * sin 35}{0. 040 * 1.6 * 10^-19 }[/tex]

[tex]r = \frac{2. 089 * 10^-30}{6. 4* 10^-21}[/tex]

[tex]r = 3. 26 * 10^-10[/tex] m

b. [tex]T = \frac{2\pi m}{qB}[/tex]

[tex]T = \frac{2 * 3.142 * 9.109 *10^-31}{1.6* 10^-19* 0.040}[/tex]

[tex]T = \frac{5. 72* 10^-30}{6. 4* 10^-21}[/tex]

[tex]T = 8. 94* 10^-10[/tex]

Therefore, the radius of the helical path and the period of the electron are 3. 26* 10^-10m and 8.94*10^-10m respectively.

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The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

The average power produced by the rocket is calculated as follows;

P = FV

where;

Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.

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(a) The ball will accelerate at a speed of 9.81 m/s² while it is in flight.

(b) The ball will be moving at a speed of 0 m/sec when it reaches its highest point.

(c) The ball's initial upward velocity is 20.30 m/s.

(d) It will reach a maximum height of 21.0 m.

Speed is defined as the change in distance with regard to time. A scalar quantity, speed. It is a temporal element, m/sec is its unit.

The information provided in the issue is;

u is the fall's beginning speed in m/sec;

h is the fall's distance

g is the fall's acceleration or 9.81 m/sec²

a. The ball's speed while in flight is equal to the acceleration caused by gravity, which is 9.81 m/s².

b. When the ball reaches its highest point, its velocity will be zero.

c. The ball's starting velocity is;

v=u-gt

0=u-gt

u=gt

u=9.81 m/s² × 2.07 sec

u=20.30 m/s

d. The formula is used to determine the highest height it can reach.

[tex]\rm u^2 = - 2gh \\\\ (20.30)^2 = -2 \times (-9.8) \times h \\\\ 412.09 = 19.6 \times h\\\\ h = 412.09/19.6\\\\ h = 21.00 m[/tex]

As a result, the ball's greatest height, beginning velocity, and acceleration are all 9.81 meters per second, zero meters per second, 20.30 meters per second, and 21 meters accordingly.

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Answer:

when switch is off no electricity will flow and then the circuit is called an open circuit

Explanation:

Electricity will not flow in open circuit

Answer:

C

Explanation:

Mechanical energy =   kintetic energy + grav potential energy

ME                            = KE    + GPE           <====  re-arrange

GPE = ME - KE               <==== KE = 1/2 m v^2

GPE = ME - 1/2 mv^2

Answer:

uduehdhehshdhdhdhdegsgshe

Explanation:

a

Answer:

13671

Explanation:

The impact will last for 3.3 milliseconds, the average force applied to the nail will be [m (v² - u²)]/2s, and the average force applied to the nail will be 883.2 N, respectively.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Average force,F = ?

Mass,m = 0.46 kg

Acceleration,a

Final velocity,v = 0 m/sec

Initial velocity,u = 6.5 m/sec

The duration of the impact, assuming the acceleration is constant is found with the help of Newton's third equation of motion as;

v²=u²+2as

0  = (6.5)² + 2 ×a × 1.1 × 10⁻²

- 42.25 = 2 × a × 1.1 × 10⁻²

a = - 1920 m/s²

a = (v-u)/t

-1920 = (0-6.5)/t

t = 3 × 10⁻³ sec

t = 3.3 milli second

From Newton's third equation of motion;

v² = u² +2as

a = (v² - u² ) /2s

The average force exerted on the nail, in terms of the mass, initial velocity, and distance traveled is;

F = ma

F = [m (v² - u² )] /2s

The average force, in newtons, exerted on the nail is;

F = [m (v² - u² )] /2s

F= ma

Substitute the given value;

F = 0.469 × 1920

F = 883.2 N

Hence the duration of the impact, the average force exerted on the nail and the average force, in newtons, exerted on the nail will be 3.3 milliseconds, F = [m (v² - u² )] /2s and 883.2 N respectively.

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a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,[tex]\rm v_{SA} = 9 \ m/sec[/tex]

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

[tex]\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec[/tex]

Air velocity with reference to the ground is;

[tex]\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec[/tex]

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

[tex]\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec[/tex]

The time the bird takes;

[tex]\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec[/tex]

c)\

The total round-trip time compared to what it would be with no wind. is;

[tex]\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec[/tex]

The time for the round trip is;

[tex]\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec[/tex]

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

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Objective observations do not include bias or opinions.

Observations can capture a child's strengths and weaknesses

This refers to the types of observation that is done through the senses through what we see, touch, feel, hear, smell, and taste.

Hence, we can see that objective observation is important because they provide insight into the strengths and needs of a child.

One of the ways you can capture your observations is through note-taking.

The federal law which ensures that all children and family education information is confidential is called The Family Educational Rights and Privacy Act (FERPA)

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The sound of a lamb is grave due to its high pitch depending on its low frequency. Option B is correct.

A sound pressure wave's frequency is the number of times it repeats itself every second.

The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.

Pitch and the frequency of sound are inverse to each other. A lamb's sound is grave because of its high pitch and low frequency.

The sound of a lamb is grave due to its high pitch depending on its low frequency.

Hence, option B is correct.

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A) 20 m

The magnitude of displacement is defined as the shortest distance between the initial and final position of the object.

For a particle in motion, the magnitude is either less than or equal to the distance travelled.

Also,

Velocity is simply defined as the rate of change of displacement with time.

Mathematically, it can be expressed as:

Velocity = displacement / time

According to the question,

The following data were obtained:

Time = 5 s

Velocity = 4 m/s

Displacement =?

Velocity = displacement / time

4 = d / 5

D = 4 × 5

D= 20 m

Hence,

20 m is the magnitude of the displacement of the object after it travelled for five seconds.

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Answer:

speed and velocity

Explanation:

A 520 Hz tone is sounded at the same time as a 516 Hz tone now the beat frequency is 4 Hz.

The pace of direction changes in current per second is known as frequency. It is expressed in hertz (Hz), a unit of measurement that is used internationally. One hertz is equal to one cycle per second. One hertz (Hz) is equivalent to one cycle per second. A complete alternating current or voltage wave is referred to as a cycle.

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Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

[tex]v=at+v_o[/tex] where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

[tex]v=at+v_o[/tex]

[tex](1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})[/tex]

[tex]1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t[/tex]

[tex]\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}[/tex]

[tex]10\text{ [s]}=t[/tex]

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

Verification:

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is [tex]a=\dfrac{100 [\frac{m}{s}]}{1[s]}[/tex], the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

Answer:

it's 6a - 18

Explanation:

because 6 is multiplied witk a and -3

Answer: 0.025 cm

Explanation:

The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm

The magnification of the lens is given by

m = v/u = h'/h

v/u = h'/ h

h' = ( v/u ) × h

h' = ( 2cm /-120cm ) × 1.5cm

h' = - 0.025cm

Therefore, the height of the image is 0.025 cm

Answer:

The purpose of an experiment is to test out your hypothesis. If your hypothesis is correct, then it is a theory that could work every single time the experiment has been performed by scientists.

Answer:

[tex]1000\; {\rm W}[/tex].

Explanation:

Joule is a unit for work, while watt is a unit for power- the rate at which work is being done. Divide work by time to find power.

By definition, one watt is equivalent to one joule-per-second. In other words: [tex]1\; {\rm W} = 1\; {\rm J \cdot s^{-1}}[/tex]. Note that the unit of time in this question is "minute", not second.

Apply unit conversion and ensure that the unit of time is "second":

[tex]\begin{aligned} 1\; \text{minute} &= 1\; \text{minute} \times \frac{60\; {\rm s}}{1\; \text{minute}} = 60\; {\rm s}\end{aligned}[/tex].

The power of this generator would thus be:

[tex]\begin{aligned} \text{power} &= \frac{\text{work}}{\text{time}} \\ &= \frac{60\, 000\; {\rm J}}{60\; {\rm s}} \\ &= 1\, 000\; {\rm W} \end{aligned}[/tex].