Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b).
By what angle has planet Y rotated through during this time?
Answers
Planet Y has rotated by 135.5° through during this time.
To find the answer, we need to know about the relation between angle and radius of orbit.
What's the expression of angle in terms of radius?Angle= arc/radiusAs arc = orbital velocity × time,angle= (orbital velocity × time)/radius
Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sunSo, angle = (√(GM)× time)/radius^3/2What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet XAs time = 5 years ( a constant)Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂= (4/3)^(3/2) × 88°
= 135.5°
Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.
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Related Questions
A jet of water squirts out horizontally from a hole near the bottom of the tank, as seen in the figure.
If the hole has a diameter of 3.13 mm, what is the height, h, of the water level in the tank? Assume that x = 1.33 m and y = 1.72 m.
Answers
The height, h, of the water level in the tank is mathematically given as
h=1.3675
What is the height, h, of the water level in the tank?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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A car is being tested on a track. The driver approaches the test section at a speed of 28 m s−1. He then accelerates at a uniform rate between two markers separated by 100 m. The car reaches a speed of 41 m s−1.
Answers
The uniform acceleration of the car is 4.485 m/s².
Acceleration of the car
The uniform acceleration of the car is calculated as follows;
v² = u² + 2as
a = (v² - u²)/2s
where;
v is final velocity = 41 m/su is initial velocity = 28 m/ss is distance = 100 ma is acceleration = ?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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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). By what angle has planet Y rotated through during this time?
Answers
According to mathematics, the planet's angle is stated as
dY=704 degrees.
What is the current rotational angle of planet Y?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 circular loop of wire with radius 0.0410 m and resistance 0.169 Ω is in a region of spatially uniform magnetic field, as shown in the following figure (Figure 1). The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 7.78 T and is decreasing at a rate of -0.605 T/s.
a) Is the induced current in the loop clockwise or counterclockwise?
b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?
Answers
(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.
Direction of the currentThe induced current in the loop will be counterclockwise to the direction of magnetic field.
Emf induced in the loopemf = -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
Rate of energy dissipationP = 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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A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, rough surface a distance of 4.95 m.
(a) What is the work done (in J) by the 150 N force?
_________J
(b) What is the coefficient of kinetic friction between the block and the surface?
________
Answers
The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
What is the work done?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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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
Answers
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 figure below shows a motorcycle leaving the end of a ramp with a speed of 39.0 m/s and following the curved path shown. At the peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 37.1 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)
_____m
Answers
The maximum height h of the curved path is 7.38 m.
Maximum height of the curved path
Apply the following kinematic equation;
v² = u² - 2gh
where;
v is the final velocity of the motorcycleu is initial velocity of the motorcycleh is the maximum height(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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A step down transformer with an input voltage of 220V decreases the voltage to half of the input .There is a current flowing of 15A in the primary coil. Find the output current
Answers
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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A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.35 hours. What is density of the planet? Assume that the planet has a uniform density.
Answers
The density of the planet is determined as 1,974.26 kg/m³.
Density of the planet
√(⁴/₃πGρ) = 2π/(2.35 x 3600)
where;
ρ is density of the planetG is universal gravitation constant√(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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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(c) Find the initial velocity of the ball.
______m/s upward
(d) Find the maximum height it reaches.
___________m
Answers
(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.
Acceleration of the ball
The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
Velocity of the ball at maximum heightThe velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.
Initial velocity of the ballv = 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
Maximum height reached by the projectileh = 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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1. Is the image projected on a movie screen real or virtual? What about the image of yourself seen in a bathroom mirror?
2. Hold a shiny spoon in front of you. What differences do you notice about the image of your face seen in the convex and concave sides?
3. Where are the images formed by each side of the spoon? In front or behind the spoon? (Try the parallax method. Look at the image of an overhead light. Hold the tip of a pencil where you think the image is. Move your head from side to side. If the image and pencil tip appear to move relative to each other, adjust the position of the pencil back and forth until they appear as one)
Answers
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.
What is virtual screen?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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Two long, straight, parallel wires, 10.0 cm apart carry equal 4.00-A currents in the same direction, as shown in (Figure 1).
a) Find the magnitude of the magnetic field at point P1 , midway between the wires.
b) What is its direction?
c) Find the magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 .
d) What is its direction?
e) Find the magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 .
f) What is its direction?
Answers
(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.
Magnetic field midway between the wiresB = 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;
I is current in the wiresr is the distance between the wiresB = (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.
Magnetic field at 25 cm right of P1B = 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.
Magnetic field at 20 above P1B = 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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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
Answers
The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
Speed of the satelliteThe speed of the satellite is calculated as follows;
v = √GM/r
where;
M is mass of Earthr is radius of satellitev = √[(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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D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the it to density of the gold.
Answers
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
An object of height 5 cm is kept in front of a
lens. An inverted image of height 5 is formed. identity the lens and position of the
object and image.
Answers
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:
What affects fuel consumption in automobiles?
A. Drag
B. Nothing
C. Air resistance
D. Time of day
Answers
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 decibel level of the sound from a certain hair dryer is measured at 60 dB. Find the intensity of the sound.
Answers
Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².
How to determine intensity of the sound?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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an 8 kilogram ball is fired horizontally form a 1 times 10^3 kilogram cannon initially at rest after having been fired the momentum of the ball is 2.4 time 10^3 kilogram meters per second east calculate the magnitude of the cannons velocity after the ball is fired
Answers
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.
How to find the magnitude of the cannons velocity after the ball is fired?The law of conservation of linear momentum states that, the initial momentum of a system will be equal to the final momentum.Here, then initial momentum of the system will be equal to zero, since the initial velocity of the cannon is equal to zero.Given that,[tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]
When the ball is fired from the canon, then the cannon will have a recoil velocity in the opposite direction of motion of the ball.Since it has a final momentum towards east, the recoil momentum will be in the west.Thus, the velocity of the cannon after when the ball is fired will be,[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?
According to the law of conservation of linear momentum, a system's starting and ultimate velocities are equal.Since the cannon's initial velocity is 0 in this case, the system's initial momentum will be equal to zero as well.We have,[tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]
When a cannon fires a ball, the cannon will recoil quickly and in the opposite direction of the ball's motion.Given that it is now moving eastward, the recoil momentum is towards the west.As a result, when the ball is fired, the cannon's velocity will be,[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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Part B
A roller coaster ride starts with the roller coaster car being pulled to the top of the first hill with pulley system. The car is
released from the top with an initial velocity close to zero, then accelerates downward. From that first hill, the roller coaster just
coasts; there is no driving force, other than gravity, to keep It going. Assuming no friction, what can you say about the height of
the other hills in the roller coaster ride?
Answers
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.
How to find the answer?Since it influences the mechanical energy of the system, the first hill must be the highest.One of the fundamental tenets of physics is that, in the absence of friction, mechanical energy must be conserved. Mechanical energy is the product of kinetic energy and potential energy.When the vehicles ascend the first hill on the roller coaster, mechanical energy is provided to the system because the speed is zero at this point.Mechanical energy = U = mgh
Where m represents the car mass, g represents gravity, and h represents height
If the system is to continue moving, the other hills on the mountain must be lower than the first hill. When the vehicles are released, this energy is converted into kinetic and potential energy when it lowers and ascends, but the sum of these two cannot be larger than the starting energy.Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.
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A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on the same track at a speed of 1 m/s. What is the initial momentum of the yellow and orange trains combined?
A. 200 kgm/s
B. 1000 kgm/s
C. 800 kgm/s
D. 600 kgm/s
Answers
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.
Given: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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Why is air resistance friction not useful for an airplane?
A. Causes turbulence
B. Speeds it up
C. Slows it down
D. Creates heat
Answers
The correct option is C. Air resistance friction is not useful for an airplane because it slows it down.
What is air resistance?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.
Effect of air resistance on airplaneFriction 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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please help me with this physics question ASAP
Answers
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
A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.845 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.351 and he exerts a constant horizontal force of 280 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude_____N
What is the direction?
1. Opposite as the motion of the crate
2. Same as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
______J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_______m/s
Answers
(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.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
Net work done on the crateW = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
Acceleration of the cratea = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
Speed of the cratev² = 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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A rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
Answers
According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is -6048 N
What does Newton's third law say ?The law state that in every action, there will be equal and opposite reaction.
Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.
Let us first calculate the force involved in the acceleration of the rocket by using the formula
F = ma
Where mass m = 87.4 kg, acceleration a = 69.2 m/s2
Substitute the two parameters into the formula
F = 87.4 x 69.2
F = 6048.08 N
According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.
Therefore, the normal force acting on the astronaut is -6048 N approximately
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Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as
Answers
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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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
Answers
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.
What's the expression of escape velocity of a satellite?Mathematically, escape velocity= √(2GM/R)G = gravitational constant, M = mass of planet, R= radius of the planetHere, M = 4.74×10²⁴kg, R = 5870 kmEscape velocity= √(6.67×10^(-11)×4.74×10²⁴/5.870×10⁶)= 7.3 Km/s
What's the expression of time period of a circularly orbiting satellite?T= {2π×r^(3/2)}/√(GM)r= (T/2π)⅔× (GM)^(1/3)r is the radius of the orbitWhat's the radius of the circular orbit, if the time period of the satellite is 16.6 hours?T = 16.6 hours = 16.6×3600 second = 59760sr = (59760/2π)^⅔× (6.67×10^(-11)×4.74×10²⁴)^(1/3)= 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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*Look at attachment for photo of object**
An object, whose mass is 0.520 kg, is attached to a spring with a force constant of 106 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).
The object is pulled to the right a distance A = 0.150 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
__________ N
(b) At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
________ m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
- Toward the equilibrium position (i.e., to the left in the figure).
- Away from the equilibrium position (i.e., to the right in the figure).
- You cannot tell without more information.
Answers
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).
A. How to determine the force Extension (e) = 0.150 mSpring constant (K) = 106 N/mForce (F) = ?F = Ke
F = 106 × 0.15
F = 15.9 N
B. How to determine the accelerationMass (m) = 0.52 KgForce (F) = 15. 9 NAcceleration (a) =?F = ma
Divide both sides by m
a = F / m
a = 15.9 / 0.52
a = 30.58 m/s²
C. How to determine the direction of the acceleration vectorConsidering 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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describe one similarity and one difference between the velocity on the reference circle and the velocity on the pendulum
Answers
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.
What is the velocity at the bottom of a pendulum?
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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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
Answers
(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.
Acceleration due to gravity of the planet
g = GM/R²
where;
M is mass of the planetR is radius of the planetg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²
g = 0.13 m/s²
Initial speed of the rockv² = u² - 2gh
where;
v is final velocityu is initial velocityat maximum height, v = 0
u² = 2gh
u = √2gh
u = √(2 x 0.13 x 1,440)
u = 19.35 m/s
Speed of the satellitev = √GM/r
M is mass of the planet Globr is the total distance from the center of the planet Globr = 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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By how many newtons does the weight of a 100-kg person decrease when he goes from sea level to mountain top at an altitude of 5000 m? The mean radius of the earth is 6.38 × 106 m.
Answers
The weight of a 100-kg person decreases by 4 N when he goes from sea level to mountain top at an altitude of 5000 m.
What determines the weight of a person?The weight of a person is determined by the mass of the body and the acceleration due to gravity.
The acceleration due to gravity, g is dependent on the mass of the earth, M the radius of the earth and the gravitational force constant , G.
Mathematically, the acceleration due to gravity at the mountain top is determined using the formula:
g = GM/r²where:
G = 6.67 × 10⁻¹¹ Nm²/kg²
M = 5.9736 x 10²⁴ kg
r = 638000 + 5000 = 6385000
g = (6.67 × 10⁻¹¹ * 5.9736 x 10²⁴ )(6385000)²
g = 9.77 m/s²
His weight at the mountain top will be:
weight = 100 * 9.77
weight = 977 N
Weight at sea level = 100 * 9.81 = 981 N
Decrease in weight = 981 - 977
Decrease in weight = 4 N
In conclusion, the weight of the man varies according to his distance from the earth.
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