A man exerts a horizontal force of 123 N on a crate with a mass of 40.2 kg.

(A) If the crate doesn't move, what's the magnitude of the static friction force (in N)?
______N

(B) What is the minimum possible value of the coefficient of static friction between the crate and the floor? (Assume the crate remains stationary.)?

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

                      [tex]F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\[/tex]

                     [tex]Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N[/tex]

                [tex]F_V=(29*9.8)-(169.43*sin57)=142.10N[/tex]

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

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The hi.nge will apply a force of 142.10N on the beam in the vertical direction.

We must learn more about the tension in order to find the solution.

                           [tex]F_V=mg-Tsin\alpha[/tex]

                            [tex]Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N[/tex]

                     [tex]F_v=(29*9.8)-(169.43*sin57)=142.10N[/tex]

This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.

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The shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.

We know that proton is positively charge particle while on the other hand, electron is negatively charge particle. Due to opposite charges, these particles attract each other.

So we can conclude that the shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.

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Answer:

electrically charged

Explanation:

got it right

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}[/tex]

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
[tex]B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}[/tex]

Taking out constants from the integral:
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int ds[/tex]

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

[tex]B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds[/tex]

Evaluate:
[tex]B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}[/tex]

Plugging in our givens to solve for the magnetic field strength of one loop:

[tex]B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T[/tex]

Multiply by the number of loops to find the total magnetic field:
[tex]B_T = N B = 0.00631 = \boxed{6.318 mT}[/tex]

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.  

[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}[/tex]

Using the diagram, if 'z' is the point's height from the center:

[tex]r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}[/tex]

Substituting this into our expression:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }[/tex]

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds[/tex]

Evaluate:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]

Multiplying by the number of loops:
[tex]B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]

Plug in the given values:
[tex]B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}[/tex]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

                           [tex]N_x=86.62N[/tex]

                           [tex]N_y=F_V=mg-Tsin59\\[/tex]

                           [tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]

                    [tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]

                 [tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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The hi.nge will be subjected to a force of 261.12N from the beam.

We must understand the tension in order to choose the solution.

                       [tex]N_y=F_v=mg-Tsin59[/tex]

                         [tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]

                  [tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]

                        [tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]

Thus, we can infer that the force the beam applies to the height is 261.12N in size.

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The force exerted per square centimeter is 126 N/cm².

Pressure is the force acting per unit area.

Based on the data given:

volume increase, ΔV/V0 = 6 * 10⁻³

Bulk Modulus, B = 2.1 * 10⁹ N/m²

Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:

Solving for ΔP;

ΔP = B * [(ΔV/V)]

ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)

ΔP = 1.26 * 10⁶ N/m²

Converting to per square centimeter

ΔP = (1.26 * 10⁶ N/m²)/10⁴

ΔP = 126 N/cm²

In conclusion, the force exerted per square centimeter is a measure of the pressure.

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Checking the oil will reduce the possibility of having to rebuild or replaced the engine.

what is an engine ?

A device created to transform one or more sources of energy into mechanical energy is known as an engine or motor. Potential energy, heat energy, chemical energy, electric potential, and nuclear energy are all forms of energy that are readily available.

The lubricating function of engine oil is crucial. It shields and stops all the moving parts from rubbing against one another. Metal-on-metal wear would quickly kill your engine without lubrication.

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(1) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

(2) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

(3) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

v² = u² - 2gh

where;

when the lander's velocity = 0

0 = u² - 2gh

u² = 2gh

h = u²/2g

h = (2.8²)/(2 x 9.8)

h = 0.4 m

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

Thus, the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

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Radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts. This is called differential heating.

Differential heating is the property that causes different surfaces to heat up and cool down at different rates. The earth's surface receives different magnitudes of solar radiation and also the earth's surfaces absorb thermal energy in different magnitudes.

The color, shape, texture, surface and presence of constructions can influence the heating or cooling of the earth.

The earth in the equator line is heated more by solar action than that of the poles, since it receives more amount of radiation per unit area.

In general, dry surfaces heat up and cool down faster than wet ones.

Therefore, we can confirm that when radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts is called differential heating.

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The orbital speed of an ice cube in the rings of Saturn is determined as  355,366.5 m/s.

The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.

The orbital speed of ice cube in the rings of Saturn is calculated as follows;

v = √GM/r

where;

v = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁵)

v = 355,366.5 m/s

Thus, the orbital speed of an ice cube in the rings of Saturn is determined as  355,366.5 m/s.

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The lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t) is

"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.

This is further explained below.

Generally, the equation for the modulated wave containing frequency components  is  mathematically given as

[tex]Fc,Fc+w,Fc-W = > 1.5k,2.5k,0.5kHz[/tex]

[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]

"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.

b)

In conclusion,, then the modulated wave contains frequency components as

[tex]Fc,Fc+W,Fc-W = > 0.75k,1.75k,-0.25kHz[/tex]

[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]

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The angle of refraction of the light beam is determined as  16.46 ⁰.

n = sin i / sin r

where;'

Angle between the ray and the normal = incident angle = 24.8⁰

1.48 = sin (24.8) / (sin r)

sin r = sin (24.8) / (1.48)

sin r = 0.283

r = sin ⁻¹(0.283)

r = 16.46 ⁰

Thus, the angle of refraction of the light beam is determined as  16.46 ⁰.

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The solution to the questions are given as

Given, $B(t)=(1.4 T) e^{-0.057 t}$

[tex]$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}[/tex]

[tex]\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$[/tex]

[tex]\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}[/tex]

[tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]

(b) [tex]Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$[/tex]

[tex]\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}[/tex]

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

A2 = 0.05 m x 0.02 m = 0.001 m²

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

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The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

                                [tex]dU=dQ-dW[/tex]

                                  [tex]dU=0[/tex],

                             [tex]dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2[/tex]

                          [tex]T_1=T_2=100^0C=373K[/tex]

                               [tex]W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J[/tex]

Where, R is the universal gas constant.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

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The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.

We need to understand the thermodynamic processes in order to locate the solution.

                    [tex]dU=dQ-dW[/tex]     , It is 0 here.

                  [tex]dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K[/tex]

As a result, the system's final temperature will be equal to its starting temperature.

                  [tex]W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules[/tex]

R is the gaseous universal constant.

Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.

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The vertical weight carried by the builder at the rear is 240.89N. The weight he must now carry is 352.26N

We have w = 1.8

Then we have L as 3.30

θ = 24.0

FC = 147

We have to find FB

147 (3.3 + 1.8 tan24)/(3.3 - 1.8 tan24)

= 240.896

The vertical weight carried by the builder is 240.896

2. 240.896 + 147

= 387.896

387.896/[1 + (1.8 + 3.3 tan24) /(1.8 - 3.3 tan24)]

= 387.896/10.885

= 35.64

387.896 - 35.64

= 352.26N

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The mechanical energy lost due to friction is 360 J.

The distance the runner slides is, s = 0.655 m

The mechanical energy lost due to friction acting on the runner is equal to the change in kinetic energy.

Change in Kinetic energy = 1/2m(v -u)²

Change in Kinetic energy = 80 * (0 - 3.0)²/2

Change in Kinetic energy = 360 J

Mechanical energy lost = 360 J

Distance he slide is determined using the formula below as follows:

Acceleration of runner = coefficient of friction * acceleration due to gravity

acceleration, a = 0.7 * 9.8 = 6.86

v² = u² - 2as

s = v² + u²/2a

s = 0 + 3³/2 * 6.86

s = 0.655 m

In conclusion, the mechanical energy lost due to friction is the loss in kinetic energy.

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The position of the object is = -68cm

The magnification of the mirror= 0.3

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

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Answer:

11,890

Explanation:

First we need to know what is considered a significant figure.

A significant figure is a value that is not a zero at the start OR end of a value.

Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.

The 0 in the value of 3056 is considered a significant figure.

So from the table, we can deduce:

0.275 has 3 significant figures

750 has 2 significant figures

[tex]10.4 \times {10}^{5} = 1040000[/tex]

has 3 significant figures.

11,890 has 4 significant figures.

320,050 has 5 significant figures.

So from the above, we can already see the answer.

The angle of the red light is mathematically given as

[tex]\theta = 7.24 \textdegree[/tex]

Generally, the equation for the grating element is  mathematically given as

d= 1 / N

Therefore

d= 1/1965

d= 5.089 * 10^{-6} m

Generally, the equation for the  differential formula is  mathematically given as

[tex]d sin \theta = m\lambda[/tex]

Therefore

[tex]sin \theta = \lambda / d[/tex]

[tex]sin \theta= (640 * 10 ^ {-9} m)/(5.089 * 10 ^ {-6} m)[/tex]

[tex]\theta = 7.24 \textdegree[/tex]

In conclusion, The angle of the red light

[tex]\theta = 7.24 \textdegree[/tex]

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Answer:

Explanation:

An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa

Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. [tex]density=\frac{mass}{volume}[/tex]

[tex]1000=\frac{mass}{2 * 1 * 0.2}[/tex]

[tex]1000*0.4=mass[/tex]

[tex]400kg = mass[/tex]

3. [tex]density=\frac{mass}{volume}[/tex]

[tex]0.6=\frac{120}{volume}[/tex]

[tex]volume=\frac{120}{0.6}[/tex]

[tex]volume= 200cm[/tex]

The change in momentum in time interval, given the data will be F × Δt

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

This is defined as the change in momentum of an object.

Impulse = change in momentum

But

Impulse = force × time

Therefore

Force × time = change in momentum

Force × time = change in momentum

F × Δt = change in momentum

Change in momentum = F × Δt

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The total resistance in the circuit, R is 4 Ω .

The meters connected to the 12 Ω resistance are:

If the switch is opened, only the 4 Ω and 2 Ω resistances will stop working.

The resistances in the circuit are connected both in series and in parallel

The resistances in series are the 4 Ω and the 2 Ω resistances.

Equivalent resistance = 4 + 2 = 6 Ω

The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.

Total resistance, R is calculated as follows:

1/R = 1/12 + 1/6

1/R = 3/12

R = 12/3

R = 4Ω

The meters connected to the 12 Ω resistance are:

If the switch is opened, only the 4 Ω and 2 Ω resistances will stop working because the circuit connecting them to the cell is broken whereas the circuit to the 12 Ω resistance is continuous.

In conclusion, resistances can be connected in parallel or in series.

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The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

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The original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

The energy possessed by a body in motion is known as Kinetic Energy. The S. I unit is Joule.

Given that an 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler. The given parameters are;

The original kinetic energy of the player can be calculated by using the formula K.E = 1/2m[tex]v^{2}[/tex]

Substitute all the parameters into the formula

K.E = 1/2 x 87 x [tex]5.2^{2}[/tex]

K.E = 1176.24

K.E = 1200 J

Power is the rate at which work is done.

Work done = energy

The average power is required to stop him can be calculated by using the formula P = E/t

Substitute all the parameters into the formula

P = 1200/1

P = 1200 W

Therefore, the original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

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Force necessary to support the object on piston 2 is 24× 10⁴ N.

To find the answer, we need to know about the force and pressure on piston 1 and piston 2.

= 991 Kg × 9.8 = 9414.5N

= 9414.5N × π(9.46² cm² /4)

= 9414.5N × π(9.46²× 10^(-4)m²/4)

= 66.2 N/m²

= 66.2/ π(1.87² cm² /4)

= 66.2/ π(1.87²×10^(-4)m² /4)

= 24× 10⁴ N

Thus, we can conclude that force necessary to support the object on piston 2 is 24× 10⁴ N.

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Three moons can fit inside the volume of the sun.

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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The speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

Apply the principle of conservation of energy,

E(up) - E(friction) = E(bottom)

mg sin(15) + ¹/₂(M + m)u² - μ(M + m)cos 15 = ¹/₂(M + m)v²

[tex]v = \sqrt{2[\frac{mgd \ sin15 \ + \frac{1}{2}(M + m)u^2 \ -\mu (M + m)g cos\ 15 }{M + m}] }[/tex]

where;

u = √2gh

u = √(2gL sin15)

u = √(2 x 9.8 x 28 x sin 15)

u = 11.92 m/s

[tex]v = \sqrt{2[\frac{(30)(9.8)(70) \ sin15 \ + \frac{1}{2}(30 + 50)(11.92)^2 \ - 0.12 (30 + 50)9.8 cos\ 15 }{30 + 50}] }\\\\v = 16.52 \ m/s[/tex]

Thus, the speed of the boy and his friend at the bottom of the slope is 16.52 m/s.

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The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

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