100m with a constant speed 200km/hr the pilot drops abomb from the plane. determine (neplect air resistance of friction) X Q​

The minimum speed required for a satellite in order to break free permanently from the planet and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.

To find the answer, we need to know about the escape velocity and time period of revolving satellite.

          = 7.3 Km/s

        = 3.1 × 10⁴ km

Thus, we can conclude that the escape velocity and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.

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(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

(b) The speed of the satellite is 50.24 m/s.

g = GM/R²

where;

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²

g = 0.13 m/s²

v² = u² - 2gh

where;

at maximum height, v = 0

u² = 2gh

u = √2gh

u = √(2 x 0.13 x 1,440)

u = 19.35 m/s

v = √GM/r

r = radius of planet Glob + radius of the satellite

r = 63200 m + 145,000 m = 208,200 m

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]

v = 50.24 m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

The speed of the satellite is 50.24 m/s.

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(a) The induced current in the loop will be counterclockwise.

(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

The induced current in the loop will be counterclockwise to the direction of magnetic field.

emf = -NdФ/dt

emf = -NBA/dt

where;

A is area of the loop

A = πr² = π(0.041)² = 5.28 x 10⁻³ m²

emf = -(-0.605 - 7.78) x 5.28 x 10⁻³

emf = 8.385 x 5.28 x 10⁻³

emf = 0.0442 V

P = emf²/R

P = (0.0442)²/0.169

P = 0.012 W

Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

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The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

The work done is given by the use of the formula;

W = F * x

Where;

F = force applied

x = distance covered

W = 150 N *  4.95 m = 742.5 J

Now;

The coefficient of kinetic friction is given by;

μ = F/mg

μ = 150/ 33 * 9.8

μ = 0.46

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The correct option is C.   Air resistance friction is not useful for an airplane because it slows it down.

Air resistance is the opposition to motion of an object caused by air flow.

High air resistance may cause turbulence during the motion of an air plane. This can lead to engine failure or some mishap during flight.

Air resistance is also a type of friction between air and another material such as airplane.

Friction between the air and the plane slows the airplane down. This is known as air resistance, or drag.

The faster an object travels through the air, the more it has to fight against drag.

Thus, air resistance friction is not useful for an airplane because it slows it down.

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The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.

The speed of the satellite is calculated as follows;

v = √GM/r

where;

v = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]

v = 4,188 m/s

Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.

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The uniform acceleration of the car is 4.485 m/s².

The uniform acceleration of the car is calculated as follows;

v² = u² + 2as

a = (v² - u²)/2s

where;

a = (41² - 28²)/(2 x 100)

a = 4.485 m/s²

Thus, the uniform acceleration of the car is 4.485 m/s².

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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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(a) The  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

(b) The velocity of the ball when it reaches its maximum height is zero.

(c) The initial velocity of the ball is 17.36 m/s.

(d) The maximum height it reaches is 15.36 m.

The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.

v = u - gt

at maximum height, final velocity, v = 0

0 = u - gt

u = gt

u = 9.81 x 1.77

u = 17.36 m/s

h = ut - ¹/₂gt

h = 17.36(1.77) - ¹/₂(9.81)(1.77²)

h = 15.36 m

Thus, the  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball when it reaches its maximum height is zero.

The initial velocity of the ball is 17.36 m/s.

The maximum height it reaches is 15.36 m.

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(a) The magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

B = μ/2π[I₁/0.5r + I₂/0.5r]

B = (μ/2π) x (I/0.5r + I/0.5r)

B = (μ/2π) x (2I/0.5r)

B = μI/0.5r

B = 2μI/r

where;

B = (2 x 4π x 10⁻⁷ x 4)/(0.1)

B = 1.005 x 10⁻⁴ T

The direction of the magnetic field is out of the page.

B = μI/2πd

d = 5 cm + 25 cm = 30 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.3)

B = 2.67 x 10⁻⁶ T

The direction of the magnetic field is into the page towards P1.

B = μI/2πd

d = √(20² + 5²)

d = 20.62 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.2062)

B = 3.88 x 10⁻⁶ T

The direction of the magnetic field is downwards towards P1.

Thus, the magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

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Answer:

Wc = 7.84    weight of crown

Ww = 7.84 - 6.86 = .98       weight of water displaced

Density = 7.84 / .98 = 8     crown is 8 X that of water

Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold  

According to mathematics, the planet's angle is stated as

dY=704 degrees.

We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.

The equation for the period is often expressed numerically as

[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]

pX / pY=8

In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is 23.7 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

F(net) = F - μFf

F(net) = 280 - 0.351(92 x 9.8)

F(net) = -36.46 N

W = F(net) x L

W = -36.46 x 0.65

W = - 23.7 J

a = F(net)/m

a = -36.46/92

a = - 0.396 m/s²

v² = u² + 2as

v² = 0.845² + 2(-0.396)(0.65)

v² = 0.199

v = √0.199

v = 0.45 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is 23.7 J.

The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

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The height, h, of the water level in the tank is  mathematically given as

h=1.3675

Generally, the equation for kinematics is  mathematically given as

S=v_o t+(1/2)at^2

Therefore

y=0+(1/2)gt^2

Where

t=(2y/g)^{1/2}

t=(2*1.72/9.8)^{1/2}

t=0.5925s

The horizontal exit velocity will be given as

[tex]Vx=x/t[/tex]

Therefore

[tex]Vx=\frac{0.3.13}{0.5925}[/tex]

Vx=0.5283

In conclusion, applying Bernoulli's Law the tank bottom and tank surface

P+(1/2)pv0^2+pgh=P+(1/2)pvx^2+pgh

(1/2)(p)(0m/s)^2+pgh=(1/2)pvx^2+pg(0m)

gh=(1/2)vx^2

h=(0.5283m/s)^2/(2)(9.8)

h=1.3675

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according to definition of acceleration

a=v-u/t

t=v-u/a(equation 1)

according to the formula of average velocity

v+u/2*s/t

s=v+u/2*t(equation 2)

now putting the value of t in equation 2

s=v+u/2*v-u/a

s=v^2-u^2/2a

v^2=u^2+2as

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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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The maximum height h of the curved path is 7.38 m.

Apply the following kinematic equation;

v² = u² - 2gh

where;

(u² - v²)/2g = h

(39² - 37.1²)/(2 x 9.8) = h

7.38 m = h

Thus, the maximum height h of the curved path is 7.38 m.

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Solutions for the three problems are is mathematically given as

(a) First cosmic speed (arbitral velocily)

[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]

v=7338.9349

(b) Second cosmic speed (escape velo.)

$$

\begin{gathered}

[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]

[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]

(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force

[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]

[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]

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Answer:

The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.

According to the question, a real, inverted, and same-sized image of the object is formed.

A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.

When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.

The distance of the image from the lens is equal to the distance of the object from the lens.

For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.

That's why the convex lens and center of curvature are the correct answer to this question.

Explanation:

The initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

                            ∴    P = mv

                                  P = 200 x 1

                                  P = 200 kgm/s

∴   The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

                            ∴    P = mv

                                  P = 100 x 8

                                  P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

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The output current for the given step down transformer is 30A.

What is the step down transformer?

A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the  input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.

The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:

Input voltage / Output voltage = Output current / Input current

220 / 110 = Outout current / 15

Output current = 30A

Hence, the output current (30A) obtained is higher than the given input current (15A).

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SHM can be acquired by perpendicular projection of uniform circular motion of a particle on its diameter such a particle is called reference particle and its circular path is called reference circle.

A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the term for conservation of energy to determine the maximum height of the block.

As the pendulum swings downward, gravity converts this potential energy into kinetic energy, so that at the bottom of the swing, the pendulum bob has zero potential energy, and its kinetic power, (1/2)mv2, equals the initial potential energy (mgh). (So the velocity, v, equals √(2gh).)

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Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.

Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.

                 [tex]F_g=\frac{GMm}{r^2}[/tex]

                 [tex]F_c=\frac{mv^2}{r} \\[/tex]

                    [tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]

                  [tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]

                  [tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]

Thus, we can conclude that the speed of satellite will be 4188.11 m/s.

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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.

In order to get the solution, we must understand the satellite's planetary motion equation.

                          [tex]F_g=\frac{GMm}{r^2}[/tex]

                         [tex]F_c=\frac{MV^2}{r}[/tex]

                          [tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]

As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.

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Answer:

A and C

Explanation:

drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).

The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.

To find the answer, we have to know more about the law of conservation of linear momentum.

                      [tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]

                   [tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]

Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.

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After the ball is shot, the cannon will move at a speed of 2.4 m/s.

We must learn more about the rule of conservation of linear momentum in order to locate the solution.

How can I determine the cannon's post-ball velocity magnitude?

                          [tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]

                               [tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]

Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.

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Answer:

See below

Explanation:

With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings

V = IR

I = V/R = 5 / (10+5+5) = .25 A

b) With S1 closed   5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm

 then the series circuit current becomes  

     5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps

            ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3        .273 *   10 / 15 =.182 amps

The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.

To find the answer, we have to know more about the mechanical energy of a system.

                  Mechanical energy = U = mgh

Where m represents the car mass, g represents gravity, and h represents height

Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.

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A. The magnitude of the spring force (in N) acting upon the object is 15.9 N

B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²

C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).

F = Ke

F = 106 × 0.15

F = 15.9 N

F = ma

Divide both sides by m

a = F / m

a = 15.9 / 0.52

a = 30.58 m/s²

Considering the diagram, we can see that the spring was pulled away from the equilibrium point.

Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.

Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.

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The density of the planet is determined as 1,974.26 kg/m³.

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

where;

√(⁴/₃πGρ) = 2π/(2.35 x 3600)

√(⁴/₃πGρ) = 2π/8460

(⁴/₃πGρ)  = (2π/8460)²

⁴/₃πGρ = 4π²/(8460)²

ρ = 12π/(8460² x 4G)

ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)

ρ = 1,974.26 kg/m³

Thus, the density of the planet is determined as 1,974.26 kg/m³.

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Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².

Mathematically, sound intensity level can be calculated by using this formula:

[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]

Where:

I is the intensity of the sound.

Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².

Rewriting the formula, we have:

β/10 = logI - logIo

Substituting the parameters into the formula, we have;

60/10 = logI - log(1.0 × 10⁻¹²)

6 = logI + 12

logI = 6 - 12

logI = -6

I = 1.0 × 10⁻⁵ W/m².

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