a planet moves around the sun is nearly circular orbit its period of revolution t depends upon
1)radius r of orbit.
2)mass m of the sun.
3) gravitational contact. show dimensionally t^2 proportional to
[tex] {r}^{3} [/tex]


​

Answer:

α = - 0.00148 rad/s^2

Explanation:

Given:-

- The number of blades, n = 7

- The angular speed of blades, w = 0.5 rad/s

- The diameter of golf ball, d = 9.0 x 10^-2 m

- The speed of the ball, v = 0.1 m/s

Find:-

What is the angular acceleration of the windmill?

Solution:-

- We first need to visualize the ball "just passing" through between two successive blades. First we will determine the time taken (t) for the golf-ball to just pass the blades i.e it travelled a distance equal to its diameter (d) with given speed (v) to avoid the blades.

                                      d = v*t

                                      t = d / v

                                      t = 0.09 / 0.1

                                      t = 0.9 s

- It takes t = 0.9s for the golf-ball to just pass the windmill. The same amount of time is taken by the windmill blade to cover an arc distance (s) that is equivalent to the diameter of the ball (d) which is also the width of empty space between two successive blades. The angle (θ) between each blade - denoting empty space can be determined by the seeing that all 7 blades are equally spaced in a circle. So:

                                    θ = 2π / 2*n

                                    θ = π / 7

                                    θ = 0.44879 rads

- So the angular speed of the windmill blade (wf) when the ball passes through can be determined by the formula:

                                    wf = θ / t

                                    wf = 0.44879 / 0.9

                                    wf = 0.49866 rad/s

- Now we will use the first rotational kinematics equation of motion with constant angular acceleration ( α ) as follows:

                                   wf = wi + α*t

- Solve for  ( α ) :

                                    α = ( wf - wi ) / t

- Plug in the values and evaluate ( α ):

                                    α = ( 0.49866 - 0.5 ) / 0.9

                                    α = - 0.00148 rad/s^2

Answer:

X

Explanation:

X represents the location of the image produced by the lens.

A converging lens produced a virtual image when the object is placed in front of the focal point. For such a position, the image is magnified and upright, thus allowing for easier viewing.

Virtual images are always located behind the mirror. Virtual images can be either upright or inverted. Virtual images can be magnified in size, reduced in size, or the same size as the object. Virtual images can be formed by concave, convex, and plane mirrors.

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Answer:

(a) Tension on both side of wire

[tex]T_1=46.706 N[/tex]

[tex]T_2=51.710 N[/tex]

(b) acceleration of the Box

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

(c) The horizontal and vertical components

Horizontal component [tex]T_1=46.706 N[/tex]

Vertical Component  =130.19 N

Explanation:

Refer attached figure for details.

[tex]T_1\ \&\ T_2\ are\ tensions\ in\ the\ string\ and\ a\ is\ the\ acceleration\ of\ the\ masses.[/tex]

Applying Newton's 2 law of motion for 14 kg block in horizontal direction

[tex]T_1 = 14\ a[/tex]-----------(i)

Similarly, applying Newton's 2 law of motion for 8 kg block in vertical direction

[tex]8 g - T_2 =8 a[/tex]-----(ii)

Consider the case of pulley,

[tex]\tau_e_x_t= I\alpha--------(iii)\\\\Where,\\\tau_e_x_t =Torque\ acting\ on\ the\ pulley\\I=moment\ of\ inertia\ of\ pulley\\\alpha= angular\ acceleration[/tex]

where,

[tex]I= \frac{MR^2}{2} (for\ pulley\ disk)[/tex]

[tex]I=\frac{3\cdot0.5^2}{2} =0.375\ kgm^2[/tex] (since mass of the pulley = 3 kg & Radius = 0.5 m)

&[tex]\tau_e_x_t= Net\ force \cdot Distance\ from\ application\ point[/tex]

Hence [tex]\tau_e_x_t = (T_2-T_1) \cdot \frac{1}{2} = 0.375\cdot\alpha[/tex]

[tex]T_2-T_1=0.75\cdot\alpha[/tex]--------(iv)

Relation between linear acceleration (a) and angular acceleration (α) is as follows,

[tex]a = R\alpha=0.5\cdot\alpha \ (R\ is \ radius\ of \ pulley)[/tex]

[tex]\alpha=2a[/tex]--------------------(v)

Putting the value of (v) in to (iv)

[tex]T_2 -T_1= 1.5 a[/tex]---------(vi)

adding equation (i),(ii) & (vi) gives

8g =22 a + 1.5 a

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

now putting the value of a in equation (i) & (ii) we get

[tex]T_1=46.706 N[/tex]

[tex]T_2=46.706 +1.5 \cdot 3.336[/tex] = 51.710 N

(a) Hence Tension on both side of wire

[tex]T_1=46.706 N[/tex]

[tex]T_2=51.710 N[/tex]

(b) acceleration of the Box

[tex]a = 3.336 \frac{m}{sec^2}[/tex]

(c) The horizontal and vertical components

Horizontal component [tex]T_1=46.706 N[/tex]

Vertical Component = [tex]T_2+8\cdot g[/tex] =51.710 + 8 x 9.81 =130.19 N

Answer:

Tension, T1 = 46.2 N and T2 = 52 N, where as acceleration = 3.3 ms^-2.

Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical

respectively.

Explanation:

Given:

Mass of the box on rest,[tex]m_1[/tex] = 14 kg

Mass of the attached box,[tex]m_2[/tex] = 8 kg

Mass of the pulley, [tex]m_3[/tex] = 3 kg

Diameter of the pulley, [tex]d[/tex] = 1 m

Radius of the pulley, [tex]r[/tex] = 0.5 m

Here we will be using the concept of net force ([tex]F_n_e_t[/tex]),net torque ([tex]\tau_n_e_t[/tex]) and acceleration of the pulley .

A FBD is attached with.

Lets find the tension on the wire using Fnet.

⇒ [tex]T_1=m_1(a)[/tex]                                                                                       ...for m1

⇒ [tex]m_2g-T_2=m_2(a)[/tex] can be written as [tex]T_2=m_2g-m_2(a)[/tex]                ...for m2

Considering clockwise torque as negative and anticlockwise torque as positive.

Moment of inertia (I) of the disk/pulley = [tex]\frac{m_3r^2}{2}[/tex] and [tex]\alpha=\frac{a}{r}[/tex] .

Now using net torque on the pulley we can say that.

⇒ [tex](T_2-T_1)r=I\alpha[/tex]

⇒ [tex](T_2-T_1)r=\frac{m_3r^2}{2}\times \frac{a}{r}[/tex]

⇒ [tex](T_2-T_1)=\frac{m_3a}{2}[/tex]

⇒ Plugging T1 and T2 .

⇒ [tex]m_2g-m_2a-m_1a=\frac{m_3a}{2}[/tex]

⇒ Isolating a from the rest.

⇒ [tex]m_2g=\frac{m_3a}{2}+m_2a+m_1a[/tex]

⇒ [tex]m_2g=a\ [\frac{m_3}{2}+m_2+m_1][/tex]

⇒ [tex]\frac{m_2g}{\frac{m_3}{2} +m_2+m_1} =a[/tex]

⇒ Plugging the numeric value

⇒ [tex]\frac{(8\times 9.8)}{(\frac{3}{2} +8+14)} =a[/tex]

⇒ [tex]3.3 =a[/tex]

⇒ Acceleration = 3.3 [tex]ms^-^2[/tex]

So,

(a).

Tension in the wire

⇒ [tex]T_1=m_1(a)=14\times 3.3 =46.2\ N[/tex]

⇒ [tex]T_2=m_2g-m_2(a)=8(9.8-3.3)=52\ N[/tex]

(b).

The acceleration of the box is 3.3 ms^-2.

(c).

Forces on the pulley.

Horizontal force, [tex]P_H[/tex] = [tex]T_1[/tex] = [tex]46.2\ N[/tex]

Vertical force,[tex]P_V[/tex] = [tex]T_2+m_3g[/tex] = [tex]52+3(9.8)[/tex] = [tex]81.4\ N[/tex]

The values are as follows:

Tension as T1 = 46.2 N and T2 = 52 N ,where as acceleration =3.3 ms^-2.

Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical

respectively.

Answer:

Mixing a milkshake

Explanation:

Becuse it’s physics becuse your using muscle and moving it and changing it by force.

Answer:

The answer is D.  The electrons will flow from an area of high potential energy to an area of low potential energy.

X== low potential energy

Y = high potential energy

z = flow of electrons

Explanation:

The electrons will flow from an area of high potential energy to an area of low potential energy.

Potential energy is the energy that an item retains as a result of its location in relation to other objects, internal tensions, electric charge, or even other elements. Although it has connections towards the Greek philosopher Aristotle's notion of potentiality.

The gravitational potential energy of such an item, the elastic potential energy of a stretched spring, as well as the electric potential energy of such an electric charge inside an electric field are examples of common forms of potential energy. The electrons will flow from an area of high potential energy to an area of low potential energy.

X== low potential energy

Y = high potential energy

z = flow of electrons

Therefore, the electrons will flow from an area of high potential energy to an area of low potential energy.

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Answer:

Common stock

Explanation:

Common stock is also regarded to as shares in which dividends are paid to the holders according to how profitable the organization is.

Common stock is a type of equity security and it is also the least considered in the case of bankruptcy due to its variability and the probability of stockholders not getting anything because of the debtor being insolvent.

Answer:

A heavyweight suspended within a moving box needs to overcome inertia which leads to a slight delay in the motion of the weight as the box moves.

Explanation:

A seismograph is an instrument used to record earthquake waves

A heavyweight suspended within a moving box needs to overcome inertia which leads to a slight delay in the motion of the weight as the box moves.

The first earthquake waves arrive at a seismograph station, a short time after the earthquake occurs.

Answer:

Weights, vibrating rod, pendulum : sensitive to vibrations.

Explanation:

Seismograph is an instrument used to measure earthquakes by recording seismic waves. It provides us all details about earthquake - centre, time, depth, energy.

The device is sensitive to vibrations. It consists of a vibrating rod connected to a pendulum, that vibrates due to earthquake shaking. The weight is also complementary attached with rotating drum & pen, to record ground motion. The seismograph output is then recorded & processed on paper.

Answer:

57.16

Explanation:

The minimum weight of the ballast required to bring the CG of the aircraft within the given range is approximately 15.2 lbs. This is calculated using the weight of the aircraft, the distances the CG needs to move and the distance from the CG to the location where the ballast is placed.

The subject matter of this problem is physics, specifically dealing with the concept of Centres of Gravity (CG). In order to find the minimum weight of the ballast required to bring the CG within the given CG range, we first need to establish how far the current CG is from the lower limit of the range. That is 32.0 - 30.5 = 1.5 inches. We can assume that the weight required to move the CG 1 inch can be found using the formula: (Weight of aircraft X Distance CG needs to move) / Distance from CG to ballast. So, in this case, the minimum ballast weight will be: (4954 lbs * 1.5 inches) / (162 inches - 30.5 inches). Carrying out the calculations, this gives us a ballast weight of approximately 15.2 lbs.

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Answer:

C. 1500.

Explanation:

750 / .5 = 1500.

Hope this helps & best of luck!

Feel free to message me if you need more help! :)

The hydrostatic force on one end of the trough can be found using the formula F = pghA, where p is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid, and A is the area of the end of the trough. In this case, the ends of the trough are equilateral triangles with sides 4m long.

To find the hydrostatic force on one end of the trough, we can use the formula F = pghA, where F is the force, p is the density of the liquid, g is the acceleration due to gravity, h is the height of the liquid, and A is the area of the end of the trough. Since the ends of the trough are equilateral triangles, each side has a length of 4m. The area of an equilateral triangle is given by A = sqrt(3)/4 * s^2, where s is the length of the side. In this case, A = sqrt(3)/4 * (4m)^2 = 4sqrt(3)m^2. Plugging in the values, we have F = (835kg/m^3)*(9.8m/s^2)*h*(4sqrt(3)m^2).

Answer:

Magnetic compass

Explanation:

A magnetic field is a field that describes the magnetic effect of electric charges in a relative motion.

Magnetic field allows magnets to interact without contact. The principle involved in the function of the compass is that a magnetic field exerts a force on any moving charge and can be measured and detected by this effect.

Answer:

Magnetic compass

Explanation:

The magnetic compass is the most familiar compass type. It functions as a pointer to "magnetic north", the local magnetic meridian, because the magnetized needle at its heart aligns itself with the horizontal component of the Earth's magnetic field

Compasses are mainly used in navigation to find direction on the earth. This works because the Earth itself has a magnetic field which is similar to that of a bar magnet (see the picture below). The compass needle aligns with the Earth's magnetic field direction and points north-south.

Answer:

For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases

The height of the image stays the same and the image distance increases.)

Explanation:

For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image

1. 3 significant figures

2. 6 significant figures

3. 4 significant figures

The answer for the following question is explained below.

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

  W = m × g × h

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

  g = 10 m/ s²

h represents the height of the cliff

From the above formula;

  W = 65 × 10 × 2000

 W = 130,000 J

  W = 130 Kilo Joules

Therefore the work done is 130 kilo Joules.

Answer:

As you learned, scientists standardized the short-hand way we represent elements, by their chemical symbol. One of the main reasons this was developed was because using letters was the easiest way to represent the elements. Another reason that we use chemical symbols is to allow us to write chemical formulas easily

Or

One set of chemical symbols means that every science student and scientist -- no matter what languages they speak -- can identify these chemicals.

I hope this helps

As you taught, scientists standardized the use of an element's chemical symbol as a shorthand for representing it. This was created in part because representing the elements with letters was the simplest method. Utilizing chemical symbols also makes it easier to formulate chemical formulas.

All identified chemical elements are arranged in rows (referred to as periods) and columns (referred to as groups) in the periodic table of chemical elements, also known as the periodic table, in ascending order of atomic number.

The periodic table is used by scientists to quickly refer to details about an element, such as its atomic mass and chemical symbol. Scientists can identify trends in element properties like electronegativity, ionization energy, and atomic radius thanks to the periodic table's arrangement.

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Answer:

[tex]T = 12910.5\,N[/tex] for a craft with a mass of 1500 kg.

Explanation:

Let consider that craft has a mass of 1500 kg. The submersible craft is modelled after the Newton's Laws, whose equation of equilibrium is:

[tex]\Sigma F = T - W +F_{D} = 0[/tex]

The tension experimented by the cable while the craft is lowering to the seafloor is:

[tex]T = W - F_{D}[/tex]

[tex]T = (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)-1800\,N[/tex]

[tex]T = 12910.5\,N[/tex]

Answer:

Explanation:

Given that,

Mass of counterweight m= 4kg

Radius of spool cylinder

R = 8cm = 0.08m

Mass of spool

M = 2kg

The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:

Then we have,

τ(net) = R~ × T~

τ(net) = R~•i × mg•j

τ(net) = Rmg• k

τ(net) = 0.08 ×4 × 9.81

τ(net) = 3.139 Nm •k

The magnitude of the net torque is 3.139Nm

b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:

L~ = R~ × m~v + I~ω

L = mRv + MR v

L = (m + M)Rv

L = (4 + 2) × 0.08

L = 0.48 Kg.m

C. τ =dL/dt

mgR = (M + m)R dv/ dt

mgR = (M + m)R • a

a =mg/(m + M)

a =(4 × 9.81)/(4+2)

a = 6.54 m/s

Answer:

a) τnet = 3.1392 N-m

b) L = (0.48 Kg-m)*v

c) a = 6.54 m/s²

Explanation:

Given

m = 4 Kg

R = 8 cm = 0.08 m

M = 2 Kg

a) τnet = ?

b) L = ?

c)  a = ?

Solution

a) We use the formula

τnet = R*m*g*Sin 90°

τnet = 0.08 m*4 Kg*9.81 m/s²*(1)

τnet = 3.1392 N-m

b) We apply the equation

L= R*m*v + R*M*v = R*(m + M)*v

then

L = (0.08 m)*(4 Kg + 2 Kg)*v = (0.48 Kg-m)*v

c) We use the relation

τ = dL/dt = d((0.48 Kg-m)*v)/dt = (0.48 Kg-m)*dv/dt

τ = (0.48 Kg-m)*a

then

τ/(0.48 Kg-m) = a

⇒   a = 3.1392 N-m/((0.48 Kg-m)

a = 6.54 m/s²

Answer:

The current is flowing To The Left. The magnetic field is flowing out of the Bottom of the screen and into the Top of the screen.

Answer:

The two word term is "ICE SHOVES"

Explanation:

An ice shove also known as ice surge, ice heave, ivu, or shoreline ice pileup is simply a surge of ice which occurs from an ocean or large lake and drops on the shore. These ice shoves are usually caused by ocean currents, strong winds, or temperature differences which tend to push the ice to the shore. The ice shoves can create ice piles up to 12 metres high and more.

Ice shoves has been recorded in at least three lakes in the past years. They are Lake Dauphin (Manitoba, Canada), Mille Lacs Lake (Minnesota), Lake Winnebago (Wisconsin), etc.

Answer:

ICE SHOVE

Explanation:

An Ice Shove is a situation or phenomenon that occurs when the wind carries or pushes melting ice from frozen lakes down to the lake shore.

Agents involved in the pushing of melting ice down to the lake shore includes: currents in the ocean, very strong winds as well as variations and fluctuations in the temperature of the lake.

Ice shoves can result in a pile up and accumulation of ice on the lake shore to be at about 30-40 feet high.

Ice shoves can result in negative effects in the lives of living organisms. Such negative effects includes:

a. Damages to the shoreline of the lakes

b. Damages to the habitats of any living organisms by the lake.

c. Damages to structures (buildings) that can be found at the lake shore.

Answer: higher temperature increases reaction rates for both endothermic and exothermic reactions

Explanation: i took the test

The wrong statement with the student's reasoning is Higher temperature increases reaction rates for both endothermic and exothermic reactions. Hence option A is correct.

When the reactant absorbs heat energy from the surroundings to form products is called an endothermic reaction. When the heat is absorbed from the surroundings, the system remains cooler and the enthalpy of the system increases.

When the energy is released in the form of heat to form the product is called an exothermic reaction. In this reaction, the temperature of the substance increases due to the release of heat, and the enthalpy of the system decreases.

If the temperature increases the reaction rates increase in the exothermic reaction and it decreases in the endothermic reaction. Hence, increases in temperature increases the reaction rates for both endothermic and exothermic reactions is wrong.

Hence, the ideal solution is A.

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Answer:

0.0176m

Explanation:

Given that,

railroad track has a length of 40 m

temperature is T₁ −5 ◦C

temperature is T₂ 35 ◦C

linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1

Lo = 40 m

T₁ = -5° C

T₂ = 35° C

dT = T₂ - T₁

    = 35 - (-5)

     = 40°C

L = Lo*(1+alpha*dT)

dL = Lo*alpha*dT

dT =  40°C

alpah = 11 x 10⁻⁶

Lo = 40 m

dL = 40 × 11 x 10⁻⁶  × 40

 = 0.0176m

Answer:

ΔL = 0.0176m

Explanation:

We are given;

Length of railroad track; L = 40 m

First Temperature; T1 = −5 ◦C

Second temperature; T2 = 35 ◦C

linear expansion coefficient of steel; α = 11 × 10^(−6) (◦C)^(-1)

The increase in length is given by the equation;

ΔL = α•L•ΔT

Where,

α is linear coefficient

L is length

ΔT is change in temperature.

ΔT = first temperature - Second Temperature

Thus, ΔT = 35 - (-5) = 35 + 5 = 40°C

Thus,plugging in relevant values,

ΔL = α•L•ΔT = 11 × 10^(−6)•40•40

ΔL = 0.0176m

Answer:

D. Alaska is in the half of Earth that's illuminated longer by sunlight.

Explanation:

Alaska is closer to the north poll than Florida causing longer day during the summer because Alaska is facing the sun (roughly) 20 hours out of the day.

Train has the most momentum if they were all traveling 50 mph

Explanation:

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of [tex]m_2[/tex] would be four times that of [tex]m_1[/tex].

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

[tex]\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2[/tex].

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is [tex]T[/tex], then the angular velocity of the SHM would be

[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].

Assume that the mass starts with a zero displacement and a positive velocity. If [tex]A[/tex] represent the amplitude of the SHM, then the displacement of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{x}(t) = A\sin(\omega\cdot t)[/tex].

The velocity of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)[/tex].

The acceleration of the mass at time [tex]t[/tex] would be:

[tex]\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)[/tex].

Let [tex]m[/tex] represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time [tex]t[/tex] would be:

[tex]\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)[/tex],

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant [tex]k[/tex] will be equal to:

[tex]\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}[/tex].

Since [tex]\displaystyle \omega = \frac{2\pi}{T}[/tex], it can be concluded that:

[tex]\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}[/tex].

For the first mass [tex]m_1[/tex], if the time period is [tex]T_1[/tex], then the spring constant would be:

[tex]\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2[/tex].

Similarly, for the second mass [tex]m_2[/tex], if the time period is [tex]T_2[/tex], then the spring constant would be:

[tex]\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2[/tex].

Since the two springs are the same, the two spring constants should be equal to each other. That is:

[tex]\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2[/tex].

Simplify to obtain:

[tex]\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2[/tex].

the minimum speed the asteroid should have to just graze the surface of the Earth is approximately 9426 m/s.

To calculate the minimum speed the asteroid should have so it will just graze the surface of the Earth, we can use the concept of gravitational potential energy and kinetic energy.

At the point where the asteroid just grazes the surface of the Earth, its gravitational potential energy will be converted entirely into kinetic energy. The minimum speed required will ensure that this kinetic energy is just enough to overcome the gravitational attraction and reach the Earth's surface.

The gravitational potential energy (U) of an object at a distance (r) from the center of the Earth is given by the formula:

[tex]\[ U = -\frac{GMm}{r} \][/tex]

Where:

- ( U ) is the gravitational potential energy,

- ( G ) is the gravitational constant [tex](\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]

- ( M ) is the mass of the Earth[tex](\( 5.972 \times 10^{24} \, \text{kg} \)),[/tex]

- ( m ) is the mass of the asteroid (which cancels out in this context), and

- ( r ) is the distance from the center of the Earth.

The kinetic energy (K) of an object moving at a speed (v) is given by the formula:

[tex]\[ K = \frac{1}{2} mv^2 \][/tex]

At the point of grazing, the gravitational potential energy is equal to the kinetic energy:

[tex]\[ -\frac{GMm}{r} = \frac{1}{2} mv^2 \][/tex]

The mass of the asteroid (m) cancels out from both sides of the equation. We can rearrange this equation to solve for the minimum speed (v):

[tex]\[ v = \sqrt{\frac{2GM}{r}} \][/tex]

Substitute the known values:

[tex]- \( G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),\\- \( M = 5.972 \times 10^{24} \, \text{kg} \), and\\- \( r = 24000 \, \text{km} = 24000 \times 10^3 \, \text{m} \).[/tex]

[tex]\[ v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{24000 \times 10^3}} \][/tex]

Calculate the minimum speed (v):

[tex]\[ v \approx \sqrt{8.89 \times 10^8} \]\[ v \approx 9426 \, \text{m/s} \][/tex]

So, the minimum speed the asteroid should have to just graze the surface of the Earth is approximately 9426 m/s.

Answer:

65 years

Explanation:

The arc that ends at 4000 Hz intersects the 20 db hearing intensity at 65 years.

Answer:

65

Explanation:

the answer is -183 degree celcius

Answer:

90= C+273

C=273-90= - 183

testing the model and analyzing the results of the tests

Explanation:

Kepler's Third Law helps calculate the period of satellites orbiting Earth based on their radius, aiding in various applications.

Kepler's Third Law: The period of an orbiting object is related to the radius of its orbit. Using the formula n²a³ = GM (where n is the mean motion, a is the semi-major axis, and GM is the product of the gravitational constant and the Earth's mass), we can calculate the period of the satellite's circular orbit.

Calculations: For a satellite at an altitude of 20,200 km above Earth's surface (a = 26,578 km), with GM = 3.986005 x 10¹⁴ m³/s², the period T comes out to be 43,122 seconds.

Application: Understanding Kepler's laws helps in determining crucial parameters for artificial satellites orbiting Earth, aiding in satellite communication, weather observation, and other scientific endeavors.

Answer:

6 m/s²

Explanation:

We know that the force acting on an object varies directly with the acceleration. The force, F, and acceleration, a, are related by:

F = m * a

Where m is the mass of the object

We then need to find the mass of the object before we can find its acceleration when the force is 30 m/s².

Therefore, when the force is 20 N and the acceleration is 4 m/s²:

20 = m * 4

=> m = 20/4 = 5 kg

The mass of the object is 5 kg. Hence, we can find the acceleration of the object when the force is 30 N

30 = 5 * a

=> a = 30/5 = 6 m/s²

The acceleration of the object is 6 m/s²

Answer:

6m/s²

Explanation:

According to Newton's second law

Force F acting on an object varies directly with the object's acceleration a. Mathematically it is expressed as

F = ma where;

m is the mass of the object

F is the applied force

a is the acceleration

Let m be constant

m = F/a

If a force of 20 N acts on a certain object with an acceleration of 4m/s²

m = 20/4

m = 5kg

If the force is changed to 30 N, its new acceleration can be gotten using the previous formula.

F = ma

a = F/m

a = 30/5 (since mass is kept constant)

a = 6m/s²