a large crane has a mass of 8500kg calculate the weight of the crane
​

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

[tex]F = \frac{k|q_1||q_2|}{r^2}[/tex]

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

[tex]F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N[/tex]

Therefore, the mutual force between the two point charges is 319.64 N

Answer:

Tension, T = 0.556 N

Explanation:

It is given that,

Length of vocal cords, l = 22 mm = 0.022 m

Mass per unit length, [tex]\mu=0.0042\ kg/m[/tex]

We need to find the tension is required in the vocal cords in order to produce a tone of middle C of frequency 261.62 Hz. The frequency in terms if tension is given by :

[tex]f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/tex]

T = tension in the vocal cords

[tex]f^2=\dfrac{1}{4l^2}\times \dfrac{T}{\mu}\\\\T=4l^2\mu f^2\\\\T=4\times (0.022)^2\times 0.0042 \times (261.62 )^2\\\\T=0.556\ N[/tex]

So, the tension in the vocal cords is 0.556 N.

Answer:

2. 500 R

3. 16.3 kPa, 2.36 psi

Explanation:

2. Convert Celsius to Fahrenheit.

1.8 (4°C) + 32 = 39.2°F

Convert Fahrenheit to Rankine

39.2°F + 459.67 = 498.87 R

Rounding to one significant figure, the temperature is 500 R.

3. Absolute pressure = gauge pressure + atmospheric pressure

P = Pg + Pa

First, convert mmHg to kPa (remember that a vacuum is negative gauge pressure).

-638 mmHg × (101.3 kPa / 760 mmHg) = -85.0 kPa

So the absolute pressure is:

P = -85.0 kPa + 101.3 kPa

P = 16.3 kPa

Converting to psi:

P = 16.3 kPa × (14.7 psi / 101.3 kPa)

P = 2.36 psi

Answer:

The wait time is [tex]t_w = 3.4723 \ s[/tex]

Explanation:

From the question we are told that

    The distance of the hot air balloon above the ground is  [tex]z = 50 \ m[/tex]

    The distance of the hot air  balloon from the target is  [tex]k = 100 \ m[/tex]

    The  speed of the wind is  [tex]v = 15 \ m/s[/tex]

Generally the time it will take the balloon to hit the ground  is  

           [tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]

where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]

   substituting values  

                  [tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]

                 [tex]t = 3.194 \ s[/tex]

Now at the velocity the distance it will travel before it hit the ground is mathematically represented as

               [tex]d = v * t[/tex]

   substituting values

              [tex]d = 15 * 3.194[/tex]

             [tex]d = 47.916 \ m[/tex]

Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as  

         [tex]b = k - d[/tex]

   substituting values

        [tex]b =100 -47.916[/tex]

         [tex]b = 52.084 \ m[/tex]

Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as

        [tex]t_w = \frac{b}{v}[/tex]

substituting values

       [tex]t_w = \frac{52.084}{15}[/tex]

       [tex]t_w = 3.4723 \ s[/tex]

Answer:

Explanation:

he moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is:  m L²/ 3  where m is mass and L is length of rod

If the axis of rotation passes through the center of the rod, then the moment of inertia is:   m L² / 12

So for the former case , moment of inertia is higher that that in the later case .

In the former case , the axis is at one extreme end . Hence range of distance of any point on the rod from axis is from zero to L .

In the second case , as axis passes through middle point , this range of distance of any point on the rod from axis is from zero to L / 2 .

Since range of distance from axis is less , moment of inertia too will be less because

Moment of inertia = Σ m r² where r is distance of mass m from axis .

Answer:

The average induced emf in the loop is 0.3 V

Explanation:

Given:

d = 22 cm

Magnetic field  is  B = 1.5 T

Change in time  Δt = 0.20 sec.

Radius of loop  = r = d/2 = 11 x [tex]10^{-2}[/tex] m

According to the faraday's law,  Induced emf is given by   e = - (ΔФ / Δt)

where Ф = magnetic flux

Ф = BAcos0   (in this occasion, Ф=0)

where area A = pi * r²

note that we have to neglect negative sign due to its lenz law...so

       B * pi * r

e = ----------------

          Δt

       1.5 ( 3.1416) 11 x [tex]10^{-2}[/tex] )²

e =  ------------------------------------

                        0.20

e = 0.3 V

Therefore, the average induced emf in the loop is 0.3 V

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.

[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.

Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:

[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]

And final speed is now calculated after clearing it:

[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]

[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Answer:

The longer the cord, the lower the illumination

Explanation:

The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.

Long wires have more electrical resistance than shorter ones.

Let us consider this formula:

Resistance =[tex]\frac{\rho L}{A}[/tex]

From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.

The velocity and force are required.

The speed of the racket is 8.7 m/s

The required force is 471.43 N.

[tex]m_1[/tex] = Mass of racket = 1000 g

[tex]m_2[/tex] = Mass of ball = 60 g

[tex]u_1[/tex] = Initial velocity of racket = 12 m/s

[tex]u_2[/tex] = Initial velocity of ball = -15 m/s

[tex]v_1[/tex] = Final velocity of racket

[tex]v_2[/tex] = Final velocity of ball = 40 m/s

[tex]\Delta t[/tex] = Time = 7 ms

The equation of the momentum will be

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]

Force is given by

[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]

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Answer:

D(t)= 11x2.09t cos( 20pi,t)

Explanation:

Pls attached file

Answer:

D

Explanation:

Gravity is the force of attraction between two objects.

Each object creates a gravitational field in wich every other object is affected by it.

Answer:

x  =  0.27 cm

Explanation:

given data

mass = 150 g

velocity = 10 cm/s = 0.1 m/s

spring constnat = 200 N/m

solution

as we know that here ball is moving with constant speed

so

0.5 × m × v² = 0.5 × k × x²      .......................1

here x is compression in spring

so put here value and we get

0.5 × 150 × (0.1)² = 0.5 × 200 × x²

solve it we get

x  =  0.27 cm

✴ Case - I

⟶ Force = F

⟶ Mass = m

⟶ Acceleration = 12m/s²

✴ Case - II

⟶ Force = 2F

⟶ Mass = 3m

➳ Acceleration in second case.

⇒ This question is completely based on the concept of newton's second law of motion.

⇒ As per this law, Force is defined as the product of mass and acceleration.

Mathematically, F = ma

[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]

New acceleration would be 12 m/s²

Given that;

Acceleration of object = 12 m/s²

New net force = 2f

New mass = 3m

Find:

New acceleration

Computation:

[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]

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Answer:

The magnetic  flux is  [tex]\phi = 15 \ Nm^2 /C[/tex]

Explanation:

From the question we are told that

    The value of the charge is  [tex]q = 0.80 \ nC = 0.80 *10^{-9} \ C[/tex]

     The length of each side of  the cube is [tex]d = 4.0 \ m[/tex]

Generally the magnetic flux through a closed surface is  mathematically represented as

          [tex]\phi = \frac{q}{\epsilon_o} * D[/tex]

Where  D is the area enclosing the charge

  Now  a  cube is made up of six faces but in this question we are considering only one face which is mathematically represented as

      [tex]D = \frac{1}{6}[/tex]

So  the electric flux through one face of the cube  is  mathematically represented as

         [tex]\phi = \frac{q}{6 * \epsilon _o }[/tex]

where  [tex]\epsilon _o[/tex] is the permitivity of free space with value  

           [tex]\epsilon_o = 8.85 *10^{-12} F/m[/tex]

substituting value  

      [tex]\phi = \frac{0.80 *10^{-9}}{6 * 8.85 *10^{-12} }[/tex]

     [tex]\phi = 15 \ Nm^2 /C[/tex]

Answer:

The options are

A. He should aim it at the monkey

B. He should aim it below the monkey

C. He should aim it above the monkey

D. None of the above

The answer is A. He should aim it at the monkey

This is because the monkey has a large surface area and a bigger body mass. This will make aiming the food at the monkey feasible in it getting it as it could use other parts of the body to get the food aimed at it. The monkey won’t reach the food when falling if it is aimed above it. It also won’t get to the it when it is shot at below it.

Answer:

The quantities to measure are:

* the distance to the screen

* The distance from the central maximum to each interference

* in order of interference

* wavelength

Explanation:

To determine the gap spacing we must use the constructive interference equation

            d sin θ = m λ

as the angles are small

          tan θ = sin θ / cos θ

          tan θ = sin θ

and the definition of tangent is

          tan θ = y / L

Thus

         sin θ = y / L

when replacing

          d y / L = m λ

          d = m λ L / y

with this equation we can know what parameter should be measured.

The quantities to measure are:

* the distance to the screen

* The distance from the central maximum to each interference

* in order of interference

* wavelength

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved  mirror than the centre of curvature.

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Answer:

ω = 3.61 rad/sec

Explanation:

Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.

μmg = mv^2/r = mω^2r

Thus;

μg = ω^2r

ω^2 = μg/r

ω = √(μg/r)

ω = √(0.321 * 9.8)/0.241

ω = √(13.05)

= 3.61 rad/sec

Answer:

1.03A

Explanation:

For computing the magnitude of the current in the circuit we need to do the following calculations

LCR circuit impedance

[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]

= 148.7Ω

Now the phase angle is

[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]

Now the rms current flowing in the circuit is

[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]

= 0.98 A

The current flowing in the circuit is

[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]

= 1.39 A

And, finally, the current across the generator is

[tex]I'= I cos \phi[/tex]

[tex]= (1.39) cos 42.3^{\circ}[/tex]

= 1.03A

Hence, the magnitude of the circuit current is 1.03A

Answer:

P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.

Explanation:

The force, F on a conductor of length, L in a magnetic field of magnetic field strength, B and current, I flowing through it is given by

F = BIL.

Now, if the conductor has a velocity, v, the energy dissipated by this force is P

P = Fv = BIL × v = BILv.

Now, we know that the induced e.m.f due to the motion of the conductor is given by ε = BLv.

From P above, P = BILv = I(BLv)

substituting ε = BLv into P, we have

P = Iε

Thus, P is the electrical energy dissipated due to the motion of the conductor.

Now since P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.

Answer:

0.46N

Explanation:

See attached file

Explanation:

in order for the cable to not break the tension force on the cable must be equal to the weight it supports . apparent weight of the 3 passengers plus the elevator itself must be equal to the tension force. mathematicaly:

Total apparent weight = Tension force

(a) tension of the cable if it is going up at an acceleration of 1m/s2

elevator App weight=m(a + g)

elevator App weight=100kg(10m/s2 + 1m/s2)

elevator App weight=1100N

pass. App weight=m(a + g)

pass. App weight=200kg(10m/s2)+ 1m/s2)

pass. App weight=2200N

tension F= total apparent weight

tension F=2200N+1100N

tension F=3300N

(b) tension of the cable if it is going down with the same acceleration of 1m/s.

in this case ,due to the downward movement of the elevator the acceleration is assigned a negative sign(-1m/s2)

elevator App weight=m(a + g)

elevator App weight=100kg(10m/s2 +( - 1m/s2))

elevator App weight=900N

pass. App weight=m(a + g)

pass. App weight=200kg(10m/s2)+(- 1m/s2))

pass. App weight=1800N

tension F= total apparent weight

tension F=1800N+900N

tension F=2700N

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K  , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

Answer:

0.45 mm

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = [tex]\frac{\pi d^{2} }{4}[/tex]

where d is the diameter of the wire

1.613 x 10^-7 = [tex]\frac{3.142* d^{2} }{4}[/tex]

6.448 x 10^-7 = 3.142 x [tex]d^{2}[/tex]

[tex]d^{2}[/tex] =[tex]\sqrt{ 2.05*10^-7}[/tex]

d = 4.5 x 10^-4 m = 0.45 mm

The value of the diameter of the wire is 0.45 mm.

Given that:

Current in the fuse = 1 Ampere

Current density = 620 A/cm²

Area of the wire = [tex]\dfrac{\text I}{\rho}[/tex]

Area = [tex]\dfrac{1}{620}[/tex]

Area = 1.613 x 10⁻³ cm²

Also, we know that:

10000 cm² = 1 m²

1.613 x 10⁻³ cm² = 1.613 x 10⁻⁷ m²

The area of a wire can be calculated as:

Area = [tex]\dfrac{\pi \text d^2}{4}[/tex]

where, d = diameter

Substituting the values in the above equation, we get:

1.613 x 10⁻⁷ = [tex]\dfrac{3.14 \times \text d^2}{4}[/tex]

6.448 x 10⁻⁷ = 3.142 x d²

Hence, the value of d will be:

d² = 2.05 x 10⁻⁴

d = [tex]\sqrt{2.05 \times 10^{-7}}[/tex]

d = 4.5 x 10⁻⁷

Thus, the value of the diameter of the wire is 0.45 mm.

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Answer:

use a hammer to hit it

Explanation:

if u hit it u will be able to hear the shattered noise

Answer:

The rate at which power is generated in the coil is 10.24 Watts

Explanation:

Given;

number of turns of the coil, N = 160

area of the coil, A = 0.2 m²

magnitude of the magnetic field, B = 0.4 T

time for field change = 2 s

resistance of the coil, R =  16 Ω

The induced emf in the coil is calculated as;

emf = dΦ/dt

where;

Φ is magnetic flux = BA

emf = N (BA/dt)

emf = 160 (0.4T x 0.2 m²)/dt

emf = 12.8 V/s

The rate power is generated in the coil is calculated as;

P = V²/ R

P = (12.8²) / 16

P = 10.24 Watts

Therefore, the rate at which power is generated in the coil is 10.24 Watts

Answer:

Explanation:

When the path difference is half the wave length or λ /2 , destructive interference takes place which results in reduced or zero intensity in case equal intensity waves interfere as in Young's double slit experiment

Hence dark fringe is formed at that place where intensity is zero .

Hence the right option is D