Please do in degrees. An 14kg block is pulled by a string up a 30* incline by a force of 112N where mu = .15 find the acceleration of the block

Answer:

It doubles as well.

Explanation:

Assuming that the area on which the force acts remains constant, and knowing that by definition:

P = F/A

We see that if we have a double force:

p = 2 F/A

Then the pressure is doubled.

Answer:

pressure doubles if area is constant

Explanation:

we know,

p=F/A

when force is double by keeping area constant

F=2F

Then,

Change in pressure (p') will be

P'=2F/A

or,p'=2×F/A

or,p'=2p(since, P=F/A)

therefore, when force is double by keeping area constant the pressure will increase by 2 times

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Answer:

The density of this object is approximately [tex]1.36\; {\rm kg \cdot L^{-1}}[/tex].

The density of the oil in this question is approximately [tex]0.600\; {\rm kg \cdot L^{-1}}[/tex].

(Assumption: the gravitational field strength is [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex])

Explanation:

When the gravitational field strength is [tex]g[/tex], the weight [tex](\text{weight})[/tex] of an object of mass [tex]m[/tex] would be [tex]m\, g[/tex].

Conversely, if the weight of an object is [tex](\text{weight})[/tex] in a gravitational field of strength [tex]g[/tex], the mass [tex]m[/tex] of that object would be [tex]m = (\text{weight}) / g[/tex].

Assuming that [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex]. The mass of this [tex]63.8\; {\rm N}[/tex]-object would be:

[tex]\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}[/tex].

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was [tex](63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}[/tex]. Hence, the weight of water that this object displaced would be [tex]47.0 \; {\rm N}[/tex].

The mass of water displaced would be:

[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}[/tex].

The volume of that much water (which this object had displaced) would be:

[tex]\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}[/tex].

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately [tex]4.793\; {\rm L}[/tex].

The mass of this object is [tex]6.50\; {\rm kg}[/tex]. Hence, the density of this object would be:

[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].

(Rounded to [tex]\text{$3$ sig. fig.}[/tex])

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately [tex]4.793\; {\rm L}[/tex].

The weight of oil displaced would be equal to the magnitude of the buoyancy force: [tex]63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}[/tex].

The mass of that much oil would be:

[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}[/tex].

Hence, the density of the oil in this question would be:

[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].

(Rounded to [tex]\text{$3$ sig. fig.}[/tex])

There is no A in the vector, so I assume you mean λ.

The magnitude of any unit vector is 1, so

[tex]\|0.4\,\vec\imath + 0.8\,\vec\jmath + \lambda \,\vec k\| = \sqrt{0.4^2 + 0.8^2 + \lambda^2} = 1[/tex]

Square both sides and solve.

[tex]0.4^2 + 0.8^2 + \lambda^2 = 1^2 \implies \lambda = \boxed{\pm \sqrt{0.2}}[/tex]

The total electric potential energy is [tex]\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }[/tex].

Electric Potential Energy of a System of Charges :

The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.

Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.

To bring  q₁ no work is done,

[tex]V_{p} = \frac{kq_{1} }{r_{1} }[/tex]

where, V = electric potential energy.

            q = point charge.

            r = distance between any point around the charge to the point charge.

           k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

[tex]V_{2} = \frac{kq_{2} }{r_{2} }[/tex]

Work done by q₁ ;

[tex]W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }[/tex]

Now bring  q₃,

[tex]V_{3} = \frac{kq_{3} }{r_{3} }[/tex]

Work done on q₃ by q₁ and q₂

[tex]W= q_{3} [ V_{1} + V_{2} ][/tex]

    [tex]=\frac{kq_{1} q_{3} }{r_{13} }[/tex][tex]+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }[/tex]

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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The total electric potential energy is [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]

Electric Potential Energy of a System of Charges

The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.

Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.

No work has been done to bring q1,

[tex]V_{1} = \frac{kq1}{r1}[/tex]

where, V = electric potential energy.

           q = point charge.

           r = distance between any point around the charge to the point charge.

          k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

[tex]V_{2} = \frac{kq2}{r2}[/tex]

Work done by q₁ ;

W1 = [tex]V_{p} q2[/tex] = [tex]\frac{kq1q2}{r12}[/tex]

Now bring  q₃,

[tex]V_{3} = \frac{kq3}{r3}[/tex]

Work done on q₃ by q₁ and q₂

W= q3{[tex]V_{1} + V_{2}[/tex]}

W = [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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The vertical component of the force being applied to the beam by the high-tension cable is 36.37 N.

Calculation of the cable's tension-

Utilizing the moment principle, determine the cable's tension:

Torque clockwise = TL sin∅

Counterclockwise torque = 1/2WL

TL sin∅ = 1/2WL

⇒T sin∅ = 1/2W

⇒T = W/2sin∅

⇒T = (29* 9.8)/ (2*sin57)

⇒T = 169.43 N

Calculation of the vertical component of the force-

Forces that operate perpendicular to the surface vertical plane are called the vertical force. Gravity always pulls objects straight down to the earth's core.

T+F = W

⇒F = W-T

⇒F = (21*9.8)-169.43

⇒F = 36.37 N

So, the force on the beam has a vertical component of 36.37 N.

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The law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.

Pascal's law states that when an object is immersed in a fluid, it experiences equal pressure on all surfaces.

P = Patm + pgh

where;

Thus, the law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.

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The orbital speed of an ice cube in the rings of Saturn is determined as  11,237.7 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 10⁸)

v = 11,237.7 m/s

Thus, the orbital speed of an ice cube in the rings of Saturn is determined as  11,237.7 m/s.

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Answer: A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale. Then, the magnitude of the force on the left-hand pole will be, 0.167N.

Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                                   [tex]tan\alpha =\frac{30}{30}[/tex]

                                   [tex]\alpha =tan^{-1}(1)=45[/tex] degree.

                                  [tex]y- direction,\\T_2sin\alpha =mg\\T_2=\frac{mg}{sin \alpha } =\frac{17.1*10^{-3}kg*9.8m/s^2}{sin 45}=0.236 N[/tex]

                                  [tex]x- direction\\T_1-T_2cos\alpha =0\\T_1=0.236*cos45=0.167N[/tex]

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

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Answer:

  0.1426 N

Explanation:

A chain of uniform mass density is suspended between two poles 30 cm apart. The geometry of the problem is such that the left support only supplies a horizontal force on the chain. The right support must both balance that horizontal force and supply a vertical force that balances the weight of the chain.

For some tension T in the chain at the right support, the vertical force will be ...

  vertical force = T·sin(α) = W . . . . . matches the weight (W) of the chain

for some angle α between the horizontal and the chain at the right pole.

The corresponding horizontal force is ...

  horizontal force = T·cos(α)

This force balances the horizontal force at the left support pole. In terms of W, this force is ...

  horizontal force = W/sin(α)·cos(α) = W/tan(α)

The curve assumed by a chain of uniform mass density can be demonstrated to be a catenary. For supports 30 cm apart, its equation can be described by ...

  y = 30·cosh(x/30)

The diagram shows that y=4 for x=0, so we need to subtract 26 cm from this:

  y = 30·cosh(x/30) -26

The slope of the curve at any point is the derivative of this function:

  y' = 30(1/30)(sinh(x/30)) = sinh(x/30)

At the right support, the slope of the curve is ...

  y' = sinh(30/30) = sinh(1) ≈ 1.1752012

This is the tangent of the angle that the curve makes with the horizontal at the right support.

  tan(α) = 1.1752012

Note, you can see from the grid squares on the graph that the slope at the right support is slightly more than 1.

The weight of the chain is the product of its mass and the acceleration due to gravity:

  W = ma = (0.0171 kg)(9.8 m/s²) = 0.16758 N

Then the force on the left-side pole is ...

  horizontal force = W/tan(α) = (0.16758 N)/1.1752012

  horizontal force ≈ 0.1426 N

__

Additional comment

The attached graph is a plot of the catenary curve we have assumed for the gold chain. We have attempted to match the vertical height on the left side, but we note that there seems to be a small discrepancy at the right side. The graph in the problem statement seems to show the right attach point at about y=21, not 20.3.

Answer:

You would weigh the same.

Explanation:

At the moment, since Earth is not a perfect sphere, the Earth "bulges out" at the equator, so you're further from the centre of the Earth. Since gravity acts through a body's center of mass, the further you are from the centre the weaker the gravitational acceleration you will feel, because gravity weakens over distance.

So, you're actually lighter at the equator than you'd be at the poles.

However, if the Earth was a perfect sphere, this "bulge" at the equator would not happen, and so you would weigh the same at the poles and at the equator.

Hope this makes sense.

The ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.

An emf is induced in a coil placed in a magnetic field when a current carrying conductor moves in the field.

emf = NdФ/dt

where;

Thus, the ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.

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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

K is cation by losing of electron whereas Br is anion due to accepting of electrons.

Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.

So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.

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The initial speed of the sled at the given height is 4.85 m/s.

Apply the principle of conservation of energy;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v = √(2 x 9.8 x 1.2)

v = 4.85 m/s

Thus, the initial speed of the sled at the given height is 4.85 m/s.

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The maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.

v(max) = √μrg

where;

v(max) = √(0.55 x 90 x 9.8)

v(max) = 22.02 m/s

Thus, the maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.

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A. During long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

Inertia is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

When an athlete runs before taking the jump, he is trying to increase his inertia, that is his reluctance to stop, thereby increasing his forward motion or jump.

Thus, during long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

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288.51 N is  the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

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

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The  horizontal component of force + the vertical component of force  

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51  N.

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Answer:

Explanation:

——»To measure centimeters, we can use ruler.

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]

Answer: 1. The movie one is virtual and the bathroom mirror is real

2. The image is distorted in a way

3. Behind the spoon

Explanation:

The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.

In virtual desktops, the desktop environment is segregated from the physical device being used to access it. They are preloaded images of operating systems and apps. Over a network, users can remotely view their virtual desktops.

A virtual desktop can be accessed from any endpoint device, including a laptop, smartphone, or tablet. The user interacts with the client software that was installed on the endpoint device by the virtual desktop provider.

A virtual desktop mimics the appearance and feel of a real workstation. Because robust resources, like storage and back-end databases, are easily accessible, the user experience is frequently even better than that of a physical workstation.

Therefore, The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.

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Answer:

Tension in the vertical rope: approximately [tex]613 \; {\rm N}[/tex].

Tension in the horizontal rope: approximately [tex]3.74 \times 10^{3}\; {\rm N}[/tex].

Assumption: [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].

Explanation:

Since the system is not moving, the tension in the vertical rope would be equal to the weight of the crate:

[tex]\begin{aligned}\text{weight of crate} &= m\, g \\ &= 62.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 613.25 \; {\rm N}\end{aligned}[/tex].

Note that [tex]\theta = 57.7^{\circ}[/tex] is the angle between the beam (the lever) and the vertical rope. The torque that this vertical rope exert on the beam would be:

[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &=(7.65\; {\rm m}) \, (613.25\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.965 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

This torque is in the clockwise direction.

The weight of the beam ([tex]m = 116\; {\rm kg}[/tex]) would be:

[tex]\begin{aligned}\text{weight of beam} &= m\, g \\ &= 116 \; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 1.138 \times 10^{3}\; {\rm N}\end{aligned}[/tex].

Note that [tex]\theta = 57.7^{\circ}[/tex] is also the angle between the beam and the direction of the (downward) gravitational pull on this. Since this beam is uniform, it would appear as if the weight of this beam is applied at the center of this beam (with a distance of [tex](7.65\; {\rm m}) / 2[/tex] from the pivot.) Thus, the torque gravitational pull exerts on this beam would be:

[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &= \genfrac{(}{)}{}{}{7.65\; {\rm m}}{2} \, (1.138 \times 10^{3}\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.679 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

This torque is also in the clockwise direction.

The tension in the horizontal rope would need to supply a torque in the counterclockwise direction. The magnitude of that torque would be approximately:

[tex]\begin{aligned} & 3.965\times 10^{3}\; {\rm N \cdot m} + 3.679\times 10^{3}\; {\rm N \cdot m} \\ \approx \; & 7.645 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

Note the angle between the direction of this tension and the beam is [tex](90^{\circ} - \theta) = 32.3^{\circ}[/tex]. This force is applied  [tex](7.65\; {\rm m}) / 2[/tex] from the pivot. Hence, achieving that torque of [tex]7.645 \times 10^{3}\; {\rm N \cdot m}[/tex] would require:

[tex]\begin{aligned} F &= \frac{\tau}{r\, \sin(90^{\circ} - \theta)} \\ &\approx \frac{7.645\times 10^{3}\; {\rm N \cdot m}}{((7.65\; {\rm m}) / 2) \times \sin(32.3^{\circ})} \\ &\approx 3.74 \times 10^{3}\; {\rm N} \end{aligned}[/tex].

(a) The speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

(c) The radius of the synchronous orbit of a satellite is 69,801 km .

v = √GM/r

where;

v = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 7,338.93 m/s

v = √2GM/r

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

v = 2πr/T

r = vT/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Thus, the speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

The radius of the synchronous orbit of a satellite is 69,801 km .

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The behavior of water when subjected to temperature change is that the volume of water will increase if cooled from 3° to 2°C.

The chemical compound water, which can exist in the gaseous, liquid, and solid phases, is made up of the elements hydrogen and oxygen in the ratio 2: 1 i.e. 2 atoms of hydrogen and 1 atom of oxygen.

In general

Volume of water depends on the temperature and is directly proportional to it.

Thus, as the temperature rises, the molecules of water gain energy and move more quickly, which causes the molecules to spread apart and increase the volume of the liquid.

When water cools, it initially contracts (decreases in volume) until a temperature of about four degrees Celsius (4°C).

But at temperatures below 4.0° C, water undergoes an abnormal expansion that causes its volume to start to rise.

This ability is related to the formation of hexagonal structures, which take up a lot of room and increase the volume of the water, as a result of strong hydrogen bonding between water molecules at a lower temperature.

Hence, the volume of water will increase if cooled from 3° to 2°C

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The stretching force acting on the second wire, given the data is 588 N

F₁ / r₁ = F₂ / r₂

450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³

cross multiply

3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³

Divide both side by 3.9×10⁻³

F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³

F₂ = 588 N

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Answer: See below

Explanation:

Given:

The potential between plates, V = 240 V

Distance between plates, d = 0.02 m

The mass of drop, m = 2x10^-11

Charge on electron, e = 1.6x10^-19

Part (a)

The free-body diagram is attached below

Part (b)

The electric field is given by,

[tex]E=\frac{V}{d}[/tex]

On applying force balance, the force on oil drop is equal to the weight of the oil,

[tex]$$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$$[/tex]

Substituting the given values in the above equation,

[tex]\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}[/tex]

Therefore, the charge on the oil drop is 1.63x10^-14 C

Part (c)

There will be an excess of electrons on the oil drop.

The number of electrons on oil drop can be calculated as,

[tex]\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}[/tex]

Therefore, the number of excess electrons is 1.01x10^5

(a) The normal force exerted by the road on the car is 8,506.2 N.

(b) The normal force exerted by the car on the driver is 394.9 N.

(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.

The normal force exerted by the road on the car is calculated as follows;

Normal force = weight of the car - centripetal force of car

Weight of the car = (1400 x 9.8) = 13,720 N

Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N

Normal force = 13,720 N -  5,213.8 N

Normal force = 8,506.2 N

Normal force = weight of driver - centripetal force of driver

Weight of driver = (65 x 9.8) = 637 N

Centripetal force of driver = (65² x 18²)/(87) = 242.07 N

Normal force =  637 N - 242.07 N = 394.9 N

N = mg - mv²/r

0 = mg - mv²/r

mv²/r = mg

v²/r = g

v² = rg

v = √rg

v = √(87 x 9.8)

v = 29.2 m/s

Thus, the normal force exerted by the road on the car is 8,506.2 N.

The normal force exerted by the car on the driver is 394.9 N.

The speed of the car at which the normal force on driver is zero is 29.2 m/s.

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The results of the calculation are;

a) The feet spends 0.41 s in air

b) The highest point above board is 2.62 m

c) The velocity when her feet hit the water is  7.2 m/s

From the data presented;

v = u + at

But v = 0 m/s at the maximum height thus;

0 = 4 - (9.8 * t)

4 = 9.8 * t

t = 4/9.8

t = 0.41 s

b) from;

h = ut - 1/2gt^2

h = (4 * 0.41) - (0.5 * 9.8 * (0.41)^2)

h = 1.64 - 0.82

h = 0.82 m

The total height above board = 0.82 m + 1.8 m = 2.62 m

c) The total time in air is obtained from;

h = ut + 1/2gt^2

u = 0m/s because she dropped off the board

h = 1/2gt^2

2.62 = 0.5  * 9.8 * t^2

t = √2.62/0.5  * 9.8

t = 0.73 seconds

Hence, the velocity when her feet hit the water is obtained from;

v = u + gt

when u = 0 m/s

v = gt

v = 9.8 *  0.73 s

v = 7.2 m/s

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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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(a) The time taken by the ball to reach its greatest height is 2 seconds.

(b) The time taken by the ball to travel from A to B, C is 4.08 m/s.

(c) The speed with which the ball hits the hoarding is 69.3 m/s.

H = u²sin²θ/2g

H = (40² x (sin 30)²) / (2 x 9.8)

H = 20.4 m

h = ut - ¹/₂gt²

20.4 = (40 x sin 30)t - (0.5)(9.8)t²

20.4 = 20t - 4.9t²

4.9t² - 20t + 20.4 = 0

solve the quadratic equation using formula method;

t = 2 seconds

This is the time of motion of the ball;

T = 2usinθ/g

T = (2 x 40 sin 30)/9.8

T = 4.08 s

v = u + gt

vy = 40 sin30  + (9.8 x 4.08)

vy = 59.98 ms

vx = 40 x cos30

vx = 34.64 m/s

vf = √(vy² + vx²)

vf = √(59.98² + 34.64²)

vf = 69.3 m/s

Thus, the time taken by the ball to reach its greatest height is 2 seconds.

The time taken by the ball to travel from A to B, C is 4.08 m/s.

The speed with which the ball hits the hoarding is 69.3 m/s.

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The minimum value of the coefficient of static friction is equal to 0.92.

First of all, we would derive an expression for the horizontal and vertical component of forces acting acting on the car.

For the vertical component, we have:

∑Fy = 0

Fn + Fg = 0

Fn - mg = 0

Fn = mg     .....equation 1.

For the horizontal component, we have:

∑Fx = mAc

uFn = m(V²/r)    .....equation 2.

Substituting eqn. 1 into eqn. 2, we have:

umg = m(V²/r)

u = 1/g(V²/r)

u = (V²/gr)

u = (30²/9.8 × 100)

u = 900/980

u = 0.92.

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