Answers
When thrown from the balloon, the sandbag has a constant horizontal/forward velocity. As the text puts it, "the sandbag continues forward as if it had not been dropped." What this means is that the sandbag has no acceleration in the horizontal plane of motion, so it moves in this direction at a constant speed. So for any time [tex]t[/tex], the horizontal velocity is given by
[tex]v_x(t) = 12\dfrac{\rm m}{\rm s}[/tex]
In the vertical plane of motion, the sandbag has no vertical velocity until it's released and allowed to fall. As it's falling, however, the sandbag is being pulled down by gravity and is accelerating with magnitude [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex], so its vertical/downward velocity at time [tex]t[/tex] is given by
[tex]v_y(t) = -gt[/tex]
So, after 3 seconds have passed, the sandbag has horizontal velocity [tex]v_x = 12\frac{\rm m}{\rm s}[/tex] and vertical velocity [tex]v_y = -g(3\,\mathrm s) = -29.4\frac{\rm m}{\rm s}[/tex]. If the question is specifically asking about speed you would just ignore the negative signs.
Related Questions
A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their velocities remain constant during the chase?
Answers
The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.
What time will the dog catch the rabbit?The time that the dog will catch up with the rabbit is given as follows:
Let the distance covered by the rabbit be x.
Distance covered by dog = x + 30
Time taken = distance/speedThe time taken will be the same T
Time taken by dog, T = (x + 30)/10Time taken by rabbit, T = x/5Equating both times.
(x + 30)/10 = x/5
x = 30 m
Solving for T in equation (ii);
T = 30/5 = 6 minutes
In conclusion, time is obtained as a ratio of distance and speed.
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Which class of hazards is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects
Answers
Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards in the form of blast pressure waves, shrapnel and fragmentation, and incendiary thermal effects.
There are different classes of Hazards
Class 1 - Explosives
Class 2 - Gases
Class 3 - Flammable liquids
Class 4 - Flammable solids
Class 5 - Oxidizers
Class 6 - Toxic materials
Class 7 - Radioactive materials
Class 8 - Corrosive materials
Class 9 - Miscellaneous dangerous goods
Any substance or item, including a gadget, that is intended to function by explosion or that, through a chemical reaction inside itself, is capable of functioning similarly even if not intended to function by explosion, falls within the category of explosive materials (class 1).
Hence, Class 1 (Explosives) is the class of hazards that is characterized by thermal and mechanical hazards.
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A footballer kicks a ball at angle teta to horizontal with some initial velocity a) what are the two types of independent motion takes place at the same time? b.how long is the ball in the air?
Answers
a) The motion along the vertical direction and the motion along the horizontal direction.
b) The object remains in the air for a time period of 2usin(θ)/g.
Any object that is thrown in the air when gravity is acting on it is called a projectile. The motion of this projectile is called projectile motion.
When the projectile is thrown in the air at some angle θ, then there are two independent motions taking place at the same time. First is the component of motion along the vertical direction along which gravity acts. Second is the component of motion along the horizontal direction along which the object moves with a constant velocity. No force acts along the horizontal direction. The horizontal motion does not affect the vertical motion and the converse is also true. So these are independent of each other.
The time of flight is the time during which a projectile remains in the air. This time of flight is calculated using the formula,
T=2usin(θ)/g
where T is the time of flight, u is the initial velocity and g is the acceleration due to gravity.
Hence, the object remains in the air for a time period of 2usin(θ)/g.
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A container of PS5s with a mass of 451kg is loaded onto a Walmart truck using a ramp. The ramp is 6.12m long and
the bed of the truck is 1.53m above the ground. A force of 2025N applied parallel to the ramp moves the precious
consoles at a constant speed up the ramp. Find the efficiency of the ramp.
a) 32%
Ob) 54.65 %
O c) 46.4%
d) 54.6 %
Answers
D. The efficiency of the ramp is 54.6 %.
Velocity ratio of the rampV.R = distance moved by effort/distance moved by load = L/h
V.R = (6.12)/(1.53) = 4
Mechanical advantage of the rampM.A = Load/Effort
M.A = (451 x 9.8)/(2025)
M.A = 2.183
Efficiency of the rampE = (M.A / V.R) x 100%
E = (2.183 / 4) x 100%
E = 54.6 %
Thus, the efficiency of the ramp is 54.6 %.
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The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the average depth of the water in kilometers. What is the approximate depth of water for a tsunami traveling at 200 kilometers per hour?.
Answers
The speed of tsunami is a.0.32 km.
Steps involved :
The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?
Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32
As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.
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A duck flies in a straight line for 500m (60° N of W)
a.) What angle is this in polar coordinates?
b.) What is the North/South and East/West components of its displacement?
Please help me t-t
Answers
(A) The angle of this in polar coordinates is 30⁰.
(B) The North/South and East/West components of its displacement is 250 m and 433 m respectively.
The angle is this in polar coordinates60° N of W lies in the fourth quadrant;
θ = 90 - 60 = 30⁰
in polar coordinate = (r,θ) = (500 m, 30⁰)
Vertical component of the displacement (North/South)Dy = D sinθ
Dy = 500 m x sin(30) = 250 m
Horizontal component of the displacement (East/West)Dx = D cosθ
Dx = 500 m x cos(30) = 433 m
Thus, the angle of this in polar coordinates is 30⁰.
The North/South and East/West components of its displacement is 250 m and 433 m respectively.
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what are the wave model and particle model of light
Answers
Answer:
Newton's corpuscular theory stated that light consisted of particles that travelled in straight lines. Huygens argued that if light were made of particles, when light beams crossed, the particles would collide and cancel each other. He proposed that light was a wave.
What is the gravitational potential energy stored in a rock of 100 kg
mass, placed on top of a hill, which is 100 m high?
O 45 kJ
O 98 kJ
O 50 kJ
O 100 kJ
Answers
Consider a circuit with a main wire that branches into two other wires. If the current is 10 A in the main wire and 4 A in one of the branches, how much current is in the other branch
Answers
If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.
According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.
Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.
Consequently, the second wire's power may be expressed as;
I = I1+ I2 [ where I= total current (10A);
I1= current in one branch (4A) &
I2= current in another branch]
⇒I2 = I - I1
⇒I2 = 10A - 4A
⇒I2 = 6A
Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.
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Which hrase describes what atoms are made of? OA. A cloud of protons and neutrons surrounded by a nucleus of electrons OB. A nucleus of neutrons and electrons surrounded by a cloud of protons OC. A nucleus of protons and neutrons surrounded by a cloud of electrons OD. A nucleus of protons and electrons surrounded by a cloud of neutrons
Answers
Answer:
(OC) a nucleus of protons and neutrons surrounded by a cloud of electrons - modern description of an atom - where the nucleus may be hundreds of times smaller in diameter than the cloud of electrons
The law of combining volumes only applies under these conditions EXCEPT :_________.
i. the pressure must be the same. ii. the temperature must be the same. iii. the reactants must be liquids. iv. all reactants must be gases.
Answers
Answer:
measured at constant temperature and pressure
Hope this helps :)
A vehicle comes to a stop 10.0 s after the brakes are applied. While the brakes were applied, the vehicle travelled a distance of
75.0 m. If the vehicle's acceleration remained constant during the braking, what was the vehicle's initial speed?
Answers
Since acceleration is constant, the average and instantaneous accelerations are the same, so that
[tex]a = a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = -\dfrac{v_i}{10.0\,\rm s}[/tex]
By the same token, we have the kinematic relation
[tex]v^2 - {v_i}^2 = 2a\Delta x[/tex]
where [tex]v[/tex] is final speed, [tex]v_i[/tex] is initial speed, [tex]a[/tex] is acceleration, and [tex]\Delta x[/tex] is displacement.
Substitute everything you know and solve for [tex]v_i[/tex] :
[tex]0^2 - {v_i}^2 = 2\left(-\dfrac{v_i}{10.0\,\rm s}\right)(75.0\,\mathrm m)[/tex]
[tex]\implies {v_i}^2 - \left(15.0\dfrac{\rm m}{\rm s}\right) v_i = 0[/tex]
[tex]\implies v_i \left(v_i - 15.0\dfrac{\rm m}{\rm s}\right) = 0[/tex]
[tex]\implies v_i = \boxed{15.0\dfrac{\rm m}{\rm s}}[/tex]
If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?
Answers
The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
Given values:
Length of non-conducting rod, l = 1.20 m
Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C
Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C
Distance from point P of each rod, x = 60 cm = 0.60 m
Calculation of Net electric force exerted on point P:
Consider an electron released at point P, then the net electric force exerted will be given as:
F = e. E_net - ( 1 )
Step 1:
The net electric field value is given as:
E_net = E₁ cos Φ + E₂ cos Φ
= 2E₁ cos Φ -( 2 )
where, E₁ & E₂ are electric fields due to positive and negative rod
respectively.
Φ is phase angle
Step 2:
The electric field due to positive rod is given as:
E₁ = k (λ/r) - ( 3 )
where, k is Coulomb's force constant
λ is linear charge density
r is distance between point P and half of the rod.
Now, the linear charge density is given as:
λ = Charge/length = Q/x
The value of r is given as:
r = √x²-a²
where, x is length of rod
a is half length of rod
Applying values in above equation, we get:
r = √x²-(x/2)²
r = √(1.20 m)²-(1.20/2)²
= √1.08
= 1.04 m
Substituting all the determined values in equation 3 we get:
E₁ = k (λ/r)
= k [(Q/x)/r]
= k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 3:
Similarly, the electric field due to negative rod is given as:
E₂ = k [ Q/xr ]
= (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]
= 1.803×10⁴ N/C
Step 4:
Consider equation 2:
E_net = 2E₁ cos Φ
From the figure we get the phase angle as:
Φ = tan⁻¹ (0.60 m/0.60 m)
= tan⁻¹ ( 1 )
= π/4
Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:
E_net = 2(1.803×10⁴ N/C) cos π/4
= 2(1.803×10⁴ N/C) (0.5)
= 18030 N/C
Step 5:
Consider equation 1 :
F = e. E_net
where, e is charge on an electron
Applying values in above equation we get:
F = (1.6 × 10⁻¹⁹ C)(18030 N/C)
= 2.885 × 10⁻¹⁵ N
Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.
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Calculate the speed of sound in air when the temperature is 20°C. 
Answers
You push a heavy crate down a ramp at a constant velocity. Only four forces act on the crate. Which force does the greatest magnitude (it does not matter positive or negative) of work on the crate
Answers
The friction force does the greatest magnitude of work on the crate
Consider all four forces. The normal force does no work at all, since there is no motion in the direction of that force, perpendicular to the ramp. The force of gravity is smaller than the force of friction, since you still need to push the crate to get constant velocity. The force of you pushing is also smaller than the force of friction, since you are moving down a ramp, and are therefore assisted by gravity. Therefore the force doing greatest magnitude of work is the force of friction. Note that, even though the frictional work is negative, it still has the greatest magnitude
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A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod we can conclude:
Answers
Answer: A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.
Explanation: To find the answer, we need to know more about the Electric charges and its properties.
What is meant by electric charge?It can be defined as the basic fundamental property of matter causes to experience a force, when it is placed in an external electric field.The charge of the particle can be positive, negative, and zero.What are the fundamental properties of electric charge?Additivity of chargesConservation of chargesQuantization of charges.Like charges repels and unlike charges attract.It's a scalar quantity.Thus, from the above, we can conclude that, when a positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away from the rod, we can conclude that the object is positively charged.
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Answer:
positively charged
Explanation:
when a positively charged object is being repelled, that means the charge should be the same for both.
ust before it exploded, the star that became supernova 1987A was a(n) Group of answer choices B3 supergiant. white dwarf. M2 supergiant. pulsar.
Answers
Option (a) B3 supergiant is correct.
Just before it exploded, the star that became supernova 1987A was a B3 supergiant.
Any star with extremely high intrinsic luminosity and relatively vast size is referred to as a supergiant star.
B3 supergiant is a blue supergiant.
An OB supergiant, often known as a blue supergiant (BSG), is a hot, brilliant star like Rigel.
The hottest stars in the universe are blue supergiants, with temperatures ranging from 10,000 K to 50,000 K or higher.
A type II supernova known as SN 1987A occurred in the Milky Way's dwarf satellite galaxy, the Large Magellanic Cloud.
Red supergiants are the most frequent supernova progenitors, and it was once thought that only red supergiants could explode as supernovae. However, because SN 1987A's parent star, was a B3 blue supergiant, this notion had to be reexamined.
Hence, just before it exploded, the star that became supernova 1987A was a B3 supergiant.
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What are the sign and magnitude in coulombs of a point charge that produces a potential of -3.50 V at a distance of 2.00 mm
Answers
The magnitude and sign of the charge are 0.8 MC and negative respectively.
To find the answer, we need to know about the electric potential of a point charge.
What's the mathematical expression of the electric potential of a point charge?Mathematically, the electric potential at a distance 'r' from a point charge 'q' is given as (Kq)/r. K is the electrical constant with value 9×10^(-9) in vaccum. What will be the magnitude and sign of a charge, if potential is -3.50V at 2mm?From the expression of electric potential, charge is
q= (potential ×r)/K
= (-3.5×0.002)/ (9×10^(-9))
= -0.8 mega coulomb.
Thus, we can conclude that the magnitude and sign of the charge are 0.8 MC and negative respectively.
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If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gravity at the surface would become.
Answers
When the mass of Earth doubles, acceleration due to gravity doubles as well.
Acceleration due to gravity
Apply Newton's second law of motion;
F = mg --- (1)
where;
m is mass of the objectg is acceleration due to gravityF = GmM/R² --- (2)
where;
M is mass of EarthR is radius of EarthSolve (1) and (2)
mg = GmM/R²
g = GM/R²
when the mass of Earth doubles, acceleration due to gravity becomes;
g' = G(2M)/R²
g' = 2(GM/R²)
g' = 2g
Thus, when the mass of Earth doubles, acceleration due to gravity doubles as well.
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1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.
Answers
Answer:
final velocity=u+at 15+2×5=25
Dinesh takes 20 minutes to cover a distance of 5 km on a bicycle. Calculate his average speed in.
A) 1m/s
B) 2km/h
Please explain in detail.
Answers
Answer:
Soln:Here
Given,
Distance(s)=5m
Time taken(t)=20min=2400seconds
Speed(s)=?
We know that,
Speed(s)=Distance/timetaken
=5m/2400sec
580m/s
Thus the average speed of his bicycle is 580 m/s..
Thank you...
Why is the potential difference higher between any two points on a single E-field line than between two points the same distance apart on different E-field lines
Answers
While moving over a separate electric field line, however, has a non-zero value. If the test charge is transferred in a direction perpendicular to E, no work is done and the electric potential does not change. A smooth equipotential surface characterizes every such path.
Why is the potential difference higher between any two points on a single E-field line than between two points the same distance apart on different E-field lines?Generally, The equation of the potential difference is mathematically given as
[tex]V = Ed cos \theta.[/tex]
Where
[tex]\theta=Angle[/tex]
E=electric field
d= the varing distance
When the charge to be tested is transferred via an electric field line. Since =0, the maximum possible velocity (Vmax) is calculated.
In conclusion, Since no work is done when the test charge is shifted in a direction perpendicular to E, the electric potential does not shift in this case. Any such route is a smooth equipotential surface.
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The work that is done when twice the load is lifted twice the distance is _______. View Available Hint(s)for Part A The work that is done when twice the load is lifted twice the distance is _______. the same three times as much twice as much four times as much
Answers
The work that is done when twice the load is lifted twice the distance is
four times as much
The net work performed by forces acting on an object equals the change in kinetic energy, according to the work-energy theorem.
when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.
Hence The work that is done when twice the load is lifted twice the distance is four times as much
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What type of circuit have you made?
Answers
The type of circuit which have been made is the series circuit in this scenario.
What is a Series circuit?
This is a complete path which involves the whole electric current flowing through the various parts such as resistor etc..
There is only one path of current in which it does not undergo any form of split during motion.
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The positron created in pair production travels a certain distance and loses all of its kinetic energy. It finally annihilates with an electron and releases two photons. What is the angular separation between the two photons
Answers
Answer:
In pair production, after the loss of Kinetic energy, the angular separation between the two photons is 180°.
Explanation:
Pair production is the process of formation of two electrons, one negative and the other positive (positron), from a pulse of electromagnetic energy traveling through matter.It is a process of direct conversion of radiant energy to matter.The sum of the Kinetic energies of the formed particles amounts to a value of 4 MeV.When the kinetic energy is lost, emission of two photons, each with an energy of approximately 1 MeV in the form of gamma rays takes place ( in opposite direction).
Therefore, the angle of separation between the two photons is 180°.
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A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V. At what voltage will the resistor dissipate 10.0 W
Answers
50 V to dissipate a power of 10 W
Given
A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V.
Resistor and Voltage The difference in charge between two places is known as voltage. The passage of charge is measured as current. A material's propensity to oppose the flow of charge is known as resistance (current).Ohm's Law describes the relationship between current, voltage, and resistance. This asserts that, given the temperature is constant, the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance.Explanation[tex]$\mathrm{V}_{r m s}=\mathrm{P}$[/tex]
10 v provides 2 W of power whereas V provides 10 W of power.
10 V = 2 W
x V=10 W
By using cross multiplication
2 x = 100
x = 50 V
So, 50 V to dissipate a power of 10 W
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PLEASE HELP! An airplane is moving at a velocity of 90.1 m/s and requires 1435 m to reach that velocity. Determine the acceleration of the plane and the time required to reach this velocity.
Answers
The time required to reach the velocity is 15.9s.
The acceleration of the plane is 5.67m/s²
Explanation:Acceleration is solved by the given formula.
Acceleration = velocity ÷ Time
Finding Time
Time = distance ÷ speed
Time = 1435m ÷ 90.1m/s = 15.9s
Finding Acceleration
Acceleration = velocity ÷ time
Acceleration = 90.1m/s ÷ 15.9s = 5.67m/s²
Suppose you put very large-diameter tires on your car. Then, your speedometer will show a speed that is
Answers
Answer:
The speed registered by the speedometer depends on the rotations of the axle to which it is connected -
Larger diameter tires will contribute to a larger distance traveled in one rotation of the axle and hence a larger speed and the speed shown on the speedometer will be less than the actual speed.
The figure below shows a ball on a curved surface. The ball is released at point A. At which point does the ball have maximum gravitational potential energy?
Answers
Answer:
Point A
Explanation:
Because it reaches maximum height
In a science museum, a 110 kg brass pendulum bob swings at the end of a 13.9 m -long wire. the pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.2 m to the side and releasing it. because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s.
at exactly 12:00 noon, how many oscillations will the pendulum have completed?
what is its amplitude at noon?
Answers
The number of oscillations completed by the pendulum is 2736.
The amplitude of the pendulum is 3.47 m.
The given motion is an underdamped motion. So its frequency will be similar to that of a simple harmonic motion.
The frequency of oscillation is defined as the number of oscillations completed in unit time. It is calculated using the formula.
f=(1/2π)*√(l/g)
where f is the frequency, l is the length of the pendulum, and g is the acceleration due to gravity.
Given the length of the wire l=13.9 m and acceleration due to gravity g=9.8 m/s^2. The frequency of oscillation is:
f=(1/(2*3.14)) * √(13.9/9.8)
f=0.19 Hz (approximately)
Since the pendulum started oscillating at 8:00 am, 4 hours has been passed when it shows 12:00 pm. So time t=4 hours or t=4*3600. Hence t=14400 s. The total number of oscillations is then given by the formula,
n=ft
where n is the number of oscillations.
n=0.19*14400=2736.
In damping motion, the amplitude of the pendulum decreases with time. The amplitude of the pendulum is given by the formula,
A' = A exp (-b*t)
where A' is the amplitude after time t, A is the initial amplitude, b is the damping constant, and t is the time.
Here A=1.2 m, b=0.010 kg/s and t=14400 s.
A' = 1.2 exp (-0.010*14400)
A'=3.47 m (approximately)
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