Answer:
v = 2.75 10⁴ m / s
Explanation:
For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system
F = ma
where force is the force of gravity
F = [tex]G \frac{m M}{r^2}[/tex]
acceleration is centripetal
a = [tex]\frac{v^2}{r}[/tex]
we substitute
G m M / r² = m v² / r
[tex]\frac{GM}{r}[/tex] = v²
v = [tex]\sqrt{GM/r}[/tex]
indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system
r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m
let's calculate
v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11
v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8
v = 2.75 10⁴ m / s
a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?
What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, [tex]$\omega = \frac{1}{8}$[/tex] rev/sec
[tex]$=\frac{2 \pi \times 7.5}{8}$[/tex] rad/sec
= 5.89 rad/sec
Therefore, force required,
[tex]$F=m.\omega^2.r$[/tex]
[tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
when blueshift occurs,the preceived frequency of the wave would be?
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
if 145kl of energy is added to water, what mass of water can be heated from 35C to 100C then vaporized at 100C
Answer:
m = 0.057 kg = 57 g
Explanation:
Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water
[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]
where,
E = Energy added to water = 145 KJ
m = mass of water = ?
C = specific heat capacity of water = 4.2 KJ/kg.°C
ΔT = change in temperature = 100°C - 35°C = 65°C
H = Latent heat of vaporization of water = 2260 KJ/kg
Therefore,
[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]
m = 0.057 kg = 57 g
The mass of water that can be heated is equal to 0.527 kilograms.
Given the following data:
Quantity of energy = 145 kJ = 145,000 Joules.Initial temperature = 35.0°CFinal temperature = 100.0°CScientific data:
Specific heat capacity of water = 4200 J/kg°CLatent heat of vaporization of water = 2260 KJ/kgTo calculate the mass of water that can be heated:
The quantity of energy and heat.Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.
Mathematically, this is given by this expression:
[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]
Making m the subject of formula, we have:
[tex]m=\frac{E}{c\theta + H}[/tex]
Substituting the parameters into the formula, we have;
[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]
Mass, m = 0.527 kilograms.
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A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.
(a) How much mechanical energy was lost during the collision with the floor?
(b) A basketball player dribbles the ball from a height of 1.37 m by exerting a constant downward force on it for a distance of 0.132 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.37 m, what is the magnitude of the force?
Answer:
a)[tex]|\Delta E|=4.58\: J[/tex]
b)[tex]F=61.90\: N[/tex]
Explanation:
a)
We can use conservation of energy between these heights.
[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]
[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]
Therefore, the lost energy is:
[tex]|\Delta E|=4.58\: J[/tex]
b)
The force acting along the distance create a work, these work is equal to the potential energy.
[tex]W=\Delta E[/tex]
[tex]F*d=mgh[/tex]
Let's solve it for F.
[tex]F=\frac{mgh}{d}[/tex]
[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]
Therefore, the force is:
[tex]F=61.90\: N[/tex]
I hope is helps you!
A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position
Answer:
Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2
x1 = [-m / (m + M)] * L / 2 is the original position of the CM
x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right
[-m x - m L / 2 + m x - M x] / (M + m) * L/2
= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm
Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
Help me with my physics, please
Two sinusoidal waves have the same frequency and wavelength. The wavelength is 20 cm. The two waves travel from their respective sources and reach the same point in space at the same time, resulting in interference. One wave travels a larger distance than the other. For each of the possible values of that extra distance listed below, identify whether the extra distance results in maximum constructive interference, maximum destructive interference, or something in-between.
a. 10 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
b. 15 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
c. 20 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
d. 30 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
e. 35 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
f. 40 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
Answer:
Explanation:
When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.
For constructive interference , path diff = n λ
For destructive interference path diff = ( 2n+ 1 ) λ /2
where λ is wave length of wave , n is an integer.
a )
path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.
b )
path diff = 15 cm which is neither half the wavelength nor full wavelength , so in between is the right option.
c )
path diff = 20 cm which is equal to the wavelength , so maximum constructive interference will occur.
d)
path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.
e)
path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.
f )
path diff = 40 cm which is 2 times the wavelength , so maximum constructive interference will occur
•. What is called the error due to the procedure and used apparatuses?
a. Random error
b. Index error
c. Systematic error
d. Parallax error.
Answer:
[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]
Your dog is running around the grass in your back yard. He undergoes successive displacements 3.20 m south, 8.16 m northeast, and 15.6 m west. What is the resultant displacement
Answer:
D1 = 3.50 m, south; D2 = 8.20 m, northeast; D3 = 15.0 m, west. Converting all these displacements from east where zero degrees is at east or + x-axis, the converted displacements are: D1 = 3.50 m 270°; D2 = 8.20 m 45° and D3 = 15.0 m 180°. We then tabulate these vectors including there x and y components. The x-components are solved by magnitudes * cos of direction angle while the y-components of the three vectors are solved by magnitudes * sin of direction angle.
The resultant is computed by summing the components algebraically. The direction in degrees is the arc tangent of the sum of all y divided by the sum of all x.
Explanation:
A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West
A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?
Answer:
3.1m
Explanation:
Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.
use vf²=v²+2a(y)
At the maximum height the velocity is 0 and since the object is in freefall, a=-g
Plug in all values
0=15.1875-2*9.8(y)
solve for y
-15.1875*2/-9.8=y
y=3.1m
Answer:
0.774m
Explanation:
The formula for maximum height is given by:
hmax = ∨₀² ₓ Sin (α)² / 2 × g
where;
∨₀ = initial velocity
Sin (α) = angle of launch
g = gravitational acceleration which is equal to 9.8m/s²
Plugging in our values, we will have:
hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²
hmax= 20.25m/s × 0.75 / 19.8m/s²
hmax = 15.1875 / 19.8
hmax = 0.774m
Which is a mixture?
'a' sodium metal
'b' chlorine gas
'c' sodium metal and chlorine gas
'd' sodium chloride (salt) and water
Answer:
d. Sodium chloride (salt) + water
Explanation:
A mixture is made up of two or more substance combined together (combined chemically).NaCl (salt) can completely dissolve in water and sodium chlorine (aqueous) is a homogeneous mixture.sodium metal when extracted is a soft, silvery white solid.chlorine gas is a pure gas.sodium metal and chlorine gas are at pure state hence they are not mixture.learn more: https://brainly.com/question/2331419
Answer:D. Sodium chloride (salt) and water
Explanation:
I got it right on edge 2023
hope this is helpful!
What would the separation between two identical objects, one carrying 4 C of positive charge and the other 4 C of negative charge, have to be if the electrical force on each was precisely 8 N
Answer:
7.46×10⁻⁶ m
Explanation:
Applying,
F = kqq'/r²............ Equation 1
make r the subject of the equation
r = √(F/kqq').......... Equation 2
From the question,
Given: F = 8 N, q' = q= 4 C
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 2
r = √[8/(4×4×8.98×10⁹)]
r = √(55.7×10⁻¹²)
r = 7.46×10⁻⁶ m
Give reason why a man getting out of moving bus must run in the same direction for a certain distance.
Explanation:
Explanation: It's because when he stop down from a moving bus his feet come at rest while the upper portion of his body is still in motion and he falls in the forward direction.
Define measurements.
Answer:
act or process of measuring
Explanation:
Explanation:
the comparison of an unknown quantity with a known quantity.
1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)
Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).
At point A, the block has total energy
E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²
E (A) = 686 J + 1/2 (10.0 kg) v₀²
At point B, the block's potential energy is converted into kinetic energy, so that its total energy is
E (B) = 1/2 (10.0 kg) v₁²
The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,
E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J
Throughout this whole process, energy is conserved, so
E (A) = E (B) = E (C) = E (D)
(a) Solve for v₀ :
686 J + 1/2 (10.0 kg) v₀² = 2548 J
==> v₀ ≈ 19.3 m/s
(b) Solve for v₁ :
1/2 (10.0 kg) v₁² = 2548 J
==> v₁ ≈ 22.6 m/s
Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:
• net horizontal force:
∑ F = -f = ma
• net vertical force:
∑ F = n - mg = 0
where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :
n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N
f = µn = 0.500 (98.0 N) = 49.0 N
==> - (49.0 N) = (10.0 kg) a
==> a = - 4.90 m/s²
The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that
v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)
==> v₂² = 490 m²/s²
and thus the block has total/kinetic energy
E (C) = 1/2 (10.0 kg) v₂² = 2450 J
(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so
2450 J = (10.0 kg) (9.80 m/s²) h
==> h = 25.0 m
(d) At half the maximum height, the block has speed v₃ such that
2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²
==> v₃ ≈ 15.7 m/s
The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by
v = v₁ + at = 22.6 m/s - (4.90 m/s²) t
The block comes to a rest when v = 0 :
0 = 22.6 m/s - (4.90 m/s²) t
==> t ≈ 4.61 s
It covers a distance x after time t of
x = v₁t + 1/2 at ²
so when it comes to a complete stop, it will have moved a distance of
x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m
(e) The block crosses the rough region
(52.0 m) / (2.00 m) = 26 times
Which of the following categories of motion is mutually exclusive with each of the others? A. Translational motion B. Rectilinear motion C. Rotational motion D. Curvilinear motion
Answer:
C. Rotational motion
Explanation:
The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).
Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.
What is Rotational motion?"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."
Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.
What is Curvilinear motion?Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.
The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.
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A new car manufacturer advertises that their car can go from zero to sixty mph in 8 [s]. This is a description of
Answer:
Acceleration
Explanation:
The fact that new can go from zero to 60mph in 8 secs is a description of its pick-up or in physics, it's called acceleration.
Here initial velocity u= 0
final velocity v = 60 mph = 1m/minute.
or v =1609.344/60 = 26.82m/s
and time taken to do so is 8 sec
Acceleration a = (v-u)/t
a = (26.82-0)/8 = 3.35 m/s^2
Therefore, acceleration of the car a = 3.35 m/s^2.
93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2
and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?
Answer:
dont be lose because the person who lose will win the match
A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this
value express?
Answer:
the SI unit (meter per second square) indicates a linear type of motion.
Explanation:
Given;
acceleration of the car, a = 5.2 m/s² North
the SI unit of the car, = m/s²
The SI unit of the given value (acceleration), indicates a linear type of motion.
Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.
Therefore, the SI unit (meter per second square) indicates a linear type of motion.
Answer:
displacement
Explanation:
Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2.
(a) When a constant force is applied to object 1, it accelerates through a distance d. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (Select all that apply.)
K1 < K2 p1 = p2 p1 < p2 p1 > p2 K1 > K2 K1 = K2
(b) When a force is applied to object 1, it accelerates for a time interval ?t. The force is removed from object 1 and is applied to object 2. Which statements are true after object 2 has accelerated for the same time interval ?t? (Select all that apply.)
K1 > K2 K1 = K2 p1 = p2 p1 > p2 K1 < K2 p1 < p2
Answer:
Look at explanation
Explanation:
a) Kinetic energy= ΔW. W=Fd, and since in both scenarios the same force and same distance is travelled. K1=K2. I am assuming that the objects are at non zero height so by P=mgh, P1>P2
b. Again I am assuming that the objects are at non zero height so by P=mgh, P1>P2. A heavier mass, a constant force means a smaller acceleration. So a1<a2. We can then use x-x0=v0t+1/2at² and since v0=0, x-x0(d)=1/2at². Solve for t²=2d/a. Since t is the same for both but a1<a2, d1<d2. And since Kinetic Energy=ΔW, W=Fd and F is constant while d1<d2, K1<K2.
According to the question,
Potential energy be "P".Kinetic energy be "K".(a)
Word done towards both the block will be similar.
So,
→ [tex]P1 = P2[/tex]
→ [tex]K1= K2[/tex]
(b)
We know,
→ [tex]a = \frac{F}{M}[/tex]
or,
→ [tex]V = a\times t[/tex]
Now,
→ [tex]K = \frac{1}{2} MV^2[/tex]
[tex]= 0.5\times M\times V^2[/tex]
[tex]=0.5\times M\times (\frac{F^2}{M^2} )\times t^2[/tex]
[tex]= 0.5\times F^2\times \frac{t^2}{M}[/tex]
The force and t will be same. So K of the smaller mass will be greater than the larger mass.
hence,
→ [tex]K1<K2[/tex]
Thus the above responses are correct.
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A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s
Answer:
a
Explanation:
you take 32,000kg ÷2.0m
A box-shaped metal can has dimensions 8 in. by 4 in. by 10 in. high. All of the air inside the can is removed with a vacuum pump. Assuming normal atmospheric pressure outside the can, find the total force on one of the 8-by-10-in. sides
Answer:
The force on the side is 5252 N.
Explanation:
Area, A = 8 in x 10 in = 80 in^2 = 0.052 m^2
height, h = 10 in
The force on the area is
F = P x A
where, P is the atmospheric pressure and A is the area.
P = 1.01 x 10^5 Pa
Force = 1.01 x10^5 x 0.052 = 5252 N
A ball drops from a height, bounces three times, and then rolls to a stop when it reaches the ground the fourth time.
At what point is its potential energy greatest?
At what points does it have zero kinetic energy?
At what point did it have maximum kinetic energy?
Answer:
Greatest potential: moment before being dropped
Zero Kinetic: when it comes to rest
Greatest Kinetic: moment before first bounce
Explanation:
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
