A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)

Answer:

The  length is  [tex]l = 8.6 \ m[/tex]

Explanation:

From the question we are told that

   The frequencies of the two successive harmonics are [tex]f_1 = 220 \ Hz[/tex] ,  [tex]f_2 = 240 \ Hz[/tex]

   The speed of sound in the air is  [tex]v_s = 343 \ m/s[/tex]

Generally the frequency of a given harmonic is mathematically represented as

     [tex]f_n = \frac{n v }{2l}[/tex]

Here  n defines  the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

    [tex]220 = \frac{n v}{2l}[/tex]

and  

     [tex]240 = \frac{(n+1) v}{2l}[/tex]

So

   [tex]\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220[/tex]

=>  [tex]\frac{v}{2l} = 20[/tex]

=>   [tex]l = 8.6 \ m[/tex]

Answer:

3.4093

Explanation:

NPSHa = hatm + hel + hf +hva

the elevation head is the hel

friction loss head is hf

NPSHa is the head of vapour pressure of fluid

atmospheric pressure head is hatm

log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]

[tex]A, B, C are fixed[/tex]

log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]

= 4.07827 - 1343.943/333.377

=4.07827 - 4.0313009

= 0.0469691

we take the log

p* = 1.114218

we convert this value to get 111421.8

hvap = 111421.8 * 1/776.14 * 1/9.81

= 14.63

hatm = 1.1 *101325/1 * 1/9.81 *1/776.14

=14.64

hf = 7000/1 * 1/776.14 * 1/9.81

= 0.9193

NPSHa = 2.5

hel = 0.9193 + 2.5 + 14.63 - 14.64

hel = 3.4093

The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.

The vapor pressure can be calculated by,

[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]

Where,

[tex]\mathrm {NPSH_A}[/tex] = available NPSH

[tex]p_i[/tex]     = absolute pressure at the inlet = 1.1 atm

[tex]V_i[/tex]     = average velocity at the inlet = 10, 000 kg/h

[tex]\rho[/tex] = fluid density = 886 kg/m3.  

g = acceleration of gravity = 9.8 m/s²

[tex]p_v[/tex] = vapor pressure of the fluid = ?

Put the values in the equation, we get

[tex]p_v = 1.114\ atm[/tex]

Therefore, the vapor pressure of the liquid is 1.114 atm.

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Answer:

a lower speed

Explanation:

Let us look closely at the Einstein's photoelectric equation;

KE= E-Wo

Where;

KE= kinetic energy of the emitted photoelectron

E= energy of the incident photon

Wo= work function of the metal

Hence,where Wo for metal 1 > Wo for metal 2, it follows that KE for metal 1 must also be less than KE for metal 2.

This is because the difference between E and Wo for metal 1 is smaller than the same difference for metal 2 hence the answer.

Answer:

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

[tex]U*m*g=\frac{mv^2}{r}[/tex]

[tex]U*g=\frac{v^2}{r}[/tex]

substituting our given data in to expression we can solve for the speed V

[tex]0.5*9.81=\frac{v^2}{94}[/tex]

making v the subject of formula we have

[tex]0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47[/tex]

v= 21.47m/s

hence the maximum velocity of the car is 21.47m/s

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]

Putting the values

[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

By using lens formula

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

Putting the values

[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]

v = 210 mm .

B ) Magnification is the ratio of the size of the image to the size of the an object.

[tex]\rm magnification = \dfrac{v} { u}[/tex]

[tex]M= \dfrac{210} { 15}[/tex]

M= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]

D = 25 cm , f_e = focal length of eye piece

[tex]= 14 \times \dfrac{ 250} { 21}[/tex]

= 166.67

= 170 ( in two significant figures )

Hence all the answers are:

(a) The distance of the image v=220mm

(b) SIze of the image 15 mm

(c) Distance of lens be from the image produced by the objective lens 21 mm

(d) overall magnification of the microscope 170

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Answer:

24.34 m/s

Explanation:

recall that one of the equations of motions takes the form:

v = u + at

where,

v = final velocity

u = initial velocity (given as 22.2 m/s)

a = acceleration (given as 2.68m/s²)

t = time elapsed during acceleration (given as 0.80s)

since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:

v = u + at

v = 22.2 + (2.68) (0.8)

v = 24.34 m/s

Answer:

[tex]\Large \boxed{\sf kilo}[/tex]

Explanation:

kilo is a prefix that means [tex]1000[/tex] of the base unit.

Answer:

kilo is the correct answer

Explanation:

because my exam says sooo....

Answer:

Explanation:

To solve for the power consumed by the trains motor we have to employ the formula for power which is

Power= current * voltage

Given that

voltage V= 800 V

current I= 2130 A

Substituting in the formula for power we have

Power= 2130*800=  1704000 watt

Power = 1704 kW

This is the amount of energy consumed, transferred or converted per unit of time

Hence the power consumed  by the trains motor is 1704 kW

Explanation:

Given that,

Frequency of LCR circuit is 120 Hz

RMS voltage, [tex]V_{rms}=82\ V[/tex]

Resistance of circuit, R = 71 Ω

Impedance, Z = 107 Ω

We need to find the average power is delivered to the circuit by the source. Firstly, finding the rms value of current,

[tex]I_{rms}=\dfrac{V_{rms}}{Z}\\\\I_{rms}=\dfrac{82}{105}\\\\I_{rms}=0.78\ A[/tex]

Power is given by :

[tex]P=I_{rms}V_{rms}\cos\phi[/tex]

[tex]\cos\phi = \dfrac{R}{Z}\\\\\cos\phi = \dfrac{71}{105}\\\\\cos\phi =0.676[/tex]

Now, power,

[tex]P=0.78\times 82\times 0.676\\\\P=43.23\ W[/tex]

So, the average power of 43.23 watts is delivered to the circuit by the source.

Answer:

analyze data without drawing conclusions

Explanation:

Research reports are written in order to communicate clearly, information obtained primarily from research and analysis of data.

Typical reports of scientific research endeavours are written in such a way that they convey the research process succinctly without excessive extraneous information. A report is typically made up of; summary of the contents, introduction/ background, methods, results, discussion, conclusion and recommendations.

Hence a report does not really make inferences from the research findings.

Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop.  There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

Answer:

only in or near star-forming clouds

Explanation:

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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Answer:

Repulsion

Explanation:

Answer:

Number of turns of wire(N) = 3,036 turns (Approx)

Explanation:

Given:

Diameter = 13 Cm

emf = 5.6 v

Note:

The given question is incomplete, unknown information is as follow.

Magnetic field increases = 0.25 T in 1.8 (Second)

Find:

Number of turns of wire(N)

Computation:

radius (r) = 13 / 2 = 6.5 cm = 0.065 m

Area = πr²

Area = (22/7)(0.065)(0.065)

Area = 0.013278 m²

So,

emf = (N)(A)(dB / dt)

5.6 = (N)(0.013278)(0.25 / 1.8)

5.6 = (N)(0.013278)(0.1389)

N = 3,036.35899

Number of turns of wire(N) = 3,036 turns (Approx)

Answer:

-105J

Explanation:

See attached file

Answer:

Stay the same

Explanation:

Since, friction is negligible:

Initial Momentum = Final Momentum

Initial KE = Final KE

m1 * v1 = m2 * v2

When m increases v decreases.

The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.

Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.

Given:

The paperclips into an open cart rolling along a straight horizontal track with negligible friction,

Calculate the momentum, Since friction is negligible,

Initial Momentum = Final Momentum

Initial Kinetic Energy = Final Kinetic Energy

m₁ × v₁ = m₁  × v₂

When m increases, v decreases,

Thus, momentum will remain the same.

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Answer:

All electromagnetic waves travel at the same speed in a vacuum

Explanation:

All the wave listed in the question are electromagnetic waves. The speed of electromagnetic waves (collectively called light) in a vacuum is fixed. Its value is 3×10^8 ms^-1. This is a constant for all electromagnetic waves irrespective of their frequency.

Hence for any electromagnetic wave, its speed is 3×10^8 ms^-1, this will be the common velocity of all the electromagnetic waves listed in the question in a vacuum thus we can not rank them according to speed.

Answer:

24W

Explanation:

The series connection has a resistance of 2R

The parallel connection has a resistance of R/2 .. the resistance has decreased by a factor 4

Assuming the battery still provides the same pd .. the current increases by a factor of 4 .. increasing the power output by a factor of 4 also (P = V x A)

Power output = 4 x 8W .. .. So P = 24 W

Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

[tex]T_1[/tex] = 27°C = (273 + 27) K = 300 K

[tex]T_2[/tex]= 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

[tex]k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg[/tex]

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = [tex]\dfrac{\rho \times v \times x }{\mu}[/tex]

reynolds number = [tex]\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}[/tex]

reynolds number = [tex]\dfrac{0.98}{18.6 \times 10^{-6}}[/tex]

reynolds number = 52688.11204

Prandtl number = [tex]\dfrac{c_p \mu}{k}[/tex]

Prandtl number = [tex]\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}[/tex]

Prandtl number = [tex]\dfrac{0.018693}{0.02476}[/tex]

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate  can be computed as follows:

Nusselt no = [tex]\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}[/tex]

[tex]\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}[/tex]

[tex]\dfrac{h \times 0.4}{0.02476} =161.4252008}[/tex]

[tex]h =\dfrac{161.4252008 \times 0.02476}{ 0.4}[/tex]

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = [tex]h\times A \times (T_2-T_1)[/tex]

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

Answer:

The velocity is  [tex]v = 8.743 \ m/s[/tex]

Explanation:

From the question we are told that

    The frequency of the signal sent out  is  [tex]f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz[/tex]

    The frequency of the signal received is  [tex]f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz[/tex]

     The  speed of sound is  [tex]v_s = 341 \ m/s[/tex]

Generally the frequency of the sound received is  mathematically represented as

         [tex]f_r = f_s [\frac{v_s + v}{v_s - v} ][/tex]

where v is the velocity of the object

       =>      [tex]40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ][/tex]

       =>      [tex]1.05263 = \frac{341+v }{341-v}[/tex]

       =>   [tex]358.94 - 1.05263v = 341 + v[/tex]

      =>    [tex]17.947 = 2.05263 v[/tex]

      =>    [tex]v = 8.743 \ m/s[/tex]

Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

Hope this Helps!! Sorry its late

Answer:

f = 0.4 Hz

Explanation:

The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = radial acceleration = 34.1 m/s²

r = radius = length of beam = 5.55 m

Therefore,

f = (1/2π)√[(34.1 m/s²)/(5.55 m)]

f = 0.4 Hz

Answer:

E = 2.48 eV

Explanation:

The energy of a photon is given by the following formula:

E = hυ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

υ = frequency of photon = c/λ

Therefore,

E = hc/λ

where,

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)

E = (3.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 2.48 eV

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

We can calculate the energy (E) of a photon with a wavelength (λ) of 500 nm using the Planck's-Einstein relation.

[tex]E = \frac{h \times c}{\lambda } = \frac{(6.63 \times 10^{-34}J.s ) \times (3.00 \times 10^{8}m/s )}{500 \times 10^{-9}m } = 3.98 \times 10^{-19} J[/tex]

where,

We can convert 3.98 × 10⁻¹⁹ J to eV using the conversion factor 1 J = 6.24 × 10¹⁸ eV.

[tex]3.98 \times 10^{-19} J \times \frac{6.24 \times 10^{18} eV }{1J} = 2.48 eV[/tex]

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

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Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

6.13 seconds

Explanation:

At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:

v² = u² + 2gh

0 = 40² - (2)(9.81)(h)

0 = 1600 - 19.62h

19.62h = 1600

h = 1600/19.62

h = 81.55 m

Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.

We are told the first piece lands in a time of 2.71 s,

Using 3rd equation of motion, we have;

h = ut + ½gt²

81.55 = u(2.71) + ½(9.81 × 2.71²)

81.55 = 2.71u + 36.0228

2.71u = 81.55 - 36.0228

2.71u = 45.5272

u = 45.5272/2.71

u = 16.8 m/s

The time it takes a projectile to return back to its original launch point assuming the projectile was launched

vertically with speed u = 16.8 m/s is;

t = 2u/g

t = (2 × 16.8)/9.81

t = 3.43 s

Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds

Answer:

Option (A) : Nuclear Fusion and Beta Decay (electron emission)

Answer:

A : Fusion followed by beta decay (electron emission)

Explanation:

Ap3x

Answer:

c. 1.11 m/s down

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Assuming the balloon and projectile are originally at rest:

(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)

0 kg m/s = (90 kg) v + 100 kg m/s

v = -1.11 m/s