Answer:
Here we only need to look at the vertical problem, so first, let's look at the forces acting vertically on the wallet.
When the wallet starts to fall, the only force acting on it will be the gravitational force (where we are ignoring the effects of air friction).
Then the acceleration of the wallet will be equal to the gravitational acceleration, g = 9.8m/s^2
Then we can write:
a(t) = (-9.8m/s^2)
Where the negative sign is because this acceleration is downwards.
To get the vertical velocity equation of the wallet we need to integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the constant of integration, and in this case is the initial velocity of the wallet, which we know is equal to 6m/s, then the velocity equation is:
v(t) = (-9.8 m/s^2)*t + 6m/s
To get the position equation we need to integrate over time again, we get:
p(t) = (1/2)*(-9.8 m/s^2)*t^2 + (6m/s)*t + p0
Where p0 is the initial vertical position, in this case, is the height at which the wallet is dropped, which is also the altitude of your jet pack when the wallet falls.
Now we want to know two things:
Determine the speed of your wallet when it hits the ground at t = 8s
Here we just need to evaluate the velocity equation in t = 8s.
v(8s) = (-9.8 m/s^2)*8s + 6m/s = -72.4 m/s
We also want to determine the altitude of the jet pack (when the wallet drops).
To find this, we can use the fact that the wall hits the ground at t = 8s.
The wallet hits the ground when it's vertical position is equal to zero, then:
p(8s) = 0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0
Now we can solve this for p0.
0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0
(1/2)*(9.8 m/s^2)*(8s)^2 - (6m/s)*8s = p0
265.6m = p0
This means that the altitude of the jet pack when the wallet drops is 265.6m
A hammer with mass m is dropped from rest from a height h above the earth's surface. This height is not necessarily small compared with the radius RE of the earth. If you ignore air resistance, derive an expression for the speed v of the hammer when it reaches the surface of the earth. Your expression should involve h, RE, and mE, the mass of the earth.
Answer:
v = √{2(GmE(1/RE - 1/(RE + h)]}
Explanation:
From the law of conservation of energy, the total kinetic + potential energy at h = total kinetic + potential energy at the surface of the earth(h = 0)
So, K + U = K' + U'
where K = kinetic energy of hammer at h = 0
U = gravitational potential energy of hammer at h = -GmEm/(RE + h)
K' = kinetic energy of hammer at the earth's surface (h = 0) = 1/2mv²
U = gravitational potential energy of hammer at earth's surface (h = 0) = -GmEm/RE
So, K + U = K' + U'
0 + [-GmEm/(RE + h)] = 1/2mv² + [-GmEm/RE]
0 - GmEm/(RE + h) = 1/2mv² - GmEm/RE
collecting like terms,we have
1/2mv² = GmEm/RE - GmEm/(RE + h)
Factorizing GmE and multiplying both sides by 2, we have
v² = 2(GmE(1/RE - 1/(RE + h)]
taking square-root of both sides, we have
v = √{2(GmE(1/RE - 1/(RE + h)]}
The expression for the speed of the hammer will be:
[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]
What will be the speed of the hammer?The law of conservation says that the total energy at the height h and the total energy at the surface of the earth will remain constant.
From the law of conservation of energy
[tex]KE+PE=KE'+PE'[/tex]
where
KE = kinetic energy of hammer at h = 0
PE = gravitational potential energy of hammer at h height
[tex]PE=\dfrac{-Gm_em}{R_e+h}[/tex]
KE' = kinetic energy of hammer at the earth's surface (h = 0)
[tex]KE'=\dfrac{1}{2} mv^2[/tex]
PE' = gravitational potential energy of hammer at earth's surface (h = 0)
[tex]PE'=\dfrac{Gm_em}{R_e}[/tex]
Now by putting the values
[tex]KE+PE=KE'+PE'[/tex]
[tex]0+ \dfrac{-Gm_em}{R_e+h}[/tex] [tex]=\dfrac{1}{2} mv^2[/tex] [tex]-\dfrac{Gm_em}{R_e}[/tex]
[tex]\dfrac{1}{2} mv^2=\dfrac{Gm_em}{R_e} -\dfrac{Gm_em}{R_e+h}[/tex]
[tex]v^2=2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )[/tex]
[tex]v=\sqrt{2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )}[/tex]
Thus the expression for the speed of the hammer will be:
[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]
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We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardthat has arigid“sail” attached. The other team members will throw object at the sail to propel the skateboard and rider. They have balls and globs of clay that are the same size and have the same mass. Will they have better results throwing the clay or the balls? Explain!
Answer:
greater speed will be obtained for the elastic collision,
Explanation:
To answer this exercise we must find the speed that the sail acquires after each impact.
Let's start by hitting a ball of clay.
The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.
initial instant. before the crash
p₀ = m v₀
where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest
final instant. After the crash
the mass of the candle is M
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m+M} \ v_o[/tex]
for when n balls have collided
v = [tex]\frac{m}{n \ m + M}[/tex] v₀
Now let's analyze the case of the bouncing ball (elastic)
initial instant
p₀ = m v₀
final moment
p_f = m v_{1f} + M v_{2f}
p₀ = p_f
m v₀ = m v_{1f} + M v_{2f}
m (v₀ - v_{1f}) = M v_{2f}
this case corresponds to an elastic collision whereby the kinetic energy is conserved
K₀ = K_f
½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²
v₁ = v_{1f} v₂ = v_{2f}
m (v₀² - v₁²) = M v₂²
let's use the identity
(a² - b²) = (a + b) (a-b)
we write our equations
m (v₀ - v₁) = M v₂ (1)
m (v₀ - v₁) (v₀ + v₁) = M v₂²
let's divide these equations
v₀ + v₁ = v₂
Let's look for the final speeds
we substitute in equation 1
m (v₀ - v₁) = M (v₀ + v₁)
v₀ (m -M) = (m + M) v₁
v₁ = [tex]\frac{m-M}{m + M}[/tex] v₀
we substitute in equation 1 to find v₂
[tex]\frac{M}{m}[/tex] v₂ = v₀ - [tex]\frac{m-M}{m+M}[/tex] v₀
v₂ = [tex]\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o[/tex]
v₂ = [tex]\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o[/tex]
v₂ = [tex]\frac{2m}{m +M} \ v_o[/tex]
Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.
In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.
Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.
The gauge pressure in your car tires is 3.00 ✕ 10^5 N/m2 at a temperature of 35.0°C when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to
−42.0°C?
Assume the tires have not gained or lost any air.
Explanation:
Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.
So, we have:
P1/T1=P2/T2
Solving for P2, we have:
P2=(P1*T2)/T1
NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!
Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:
P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]
P2=228930.7156 N/m^2
Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2
Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!
Hope this helps!
Mark as brainlist...
Please please help me please
Answer:
Could I get brainest answer
I think it is B and e
Answer:
The answer is D hope it helps
Explanation:
A 10Ω and a 15Ω resistor are connected in series across a 110V potential difference. (Can you find them) please help
A) what is the total resistance of the circuit?
B) what is the current through each resistor?
C) what is the voltage drop across each resistor
Answer:
(A) The total resistance of the circuit is 25 Ω
(B) The current through each resistor is 4.4 A
(C) For 10Ω: Potential drop = 44 V
For 15Ω: Potential drop = 66 V
Explanation:
Given;
potential difference, V = 110V
resistors in series, = 10Ω and a 15Ω
(A) The total resistance of the circuit is calculated as follows;
Rt = 10Ω + 15Ω = 25Ω
(B) The current through each resistor;
Same current will flow through the two resistors since they are in series.
I = V/Rt
I = 110 / 25
I = 4.4 A
(C) The voltage drop across each resistor;
For 10Ω: Potential drop = IR₁ = 4.4 x 10 = 44 V
For 15Ω: Potential drop = IR₂ = 4.4 x 15 = 66 V
Using Newton's 2nd Law of Motion Formula (F=MA) answer the following.
the net force on a vehicle that is accelerating at a rate of 1.5m/s^2 is 1,800 newtons. what is the mass of the vehicle to the nearest kilogram?
Explanation:
1200 is your answer for this question
someone please help!! tysm
Hi there!
[tex]\large\boxed{5N}[/tex]
From the given diagram, 20 N is going left, and 15 N is going right.
To find the net force, we must subtract the two since the two forces go in opposite directions:
20 N - 15 N = 5N.
How many protons, electrons, and neutrons are there in the following atoms and ions?
Answer:
19 protons, 20 neutrons and 18 electrons.
Explanation:
The atomic number gives the number of protons 19
p
=
19
The atomic mass is the sum of the protons and neutrons
p
+
n
=
39
p
=
19
put p into the equation and solve for n the neutrons.
19
+
n
=
39
Subtract 19 from both sides
19
−
19
+
n
=
39
−
19
n
=
20
The number of electrons equals the number of protons in a neutral atom. The positive charge equals the negative charge. The negative charge is the number of electrons. This ion has a charge of +1. So solve for the negative charge.
−
19
+
1
=
−
18
The negative charge is -18 so
e
=
18
If the distance is 16m and the time is 4 seconds what is the speed?
Answer:
4m/s
Explanation:
Speed=distance/time
So the speed will be 16m/4s which gives 4m/s
The half-life of argon-44 is 12 minutes. Suppose you start with 20 atoms of
argon-44 and wait 12 minutes. How many atoms of argon-44 will be left?
A. 20 atoms
B. 40 atoms
C. 10 atoms
D. 5 atoms
SUBMIT
Answer:
C. 10
Explanation:
The half-life of argon-44 is 12 minutes, 10 atoms of argon-44 will be left. The correct option is C.
A radioactive substance's half-life is the amount of time it takes for half of the atoms in a sample to decay.
The half-life of argon-44 in this situation is 12 minutes. Starting with 20 argon-44 atoms, half of them will have disintegrated within 12 minutes, leaving 10 atoms.
This happens because radioactive decay is an exponential process with a constant rate of decay. Every half-life cuts the number of atoms in half.
So, if we waited another 12 minutes, half of the remaining 10 atoms would decay, leaving us with 5 atoms. This process is repeated, with half of the remaining atoms dying with each half-life.
Thus, the correct option is C.
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what is inside a black hole
[tex]\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}[/tex]
A black hole is a tremendous amount of matter crammed into a very small — in fact, zero — amount of space. The result is a powerful gravitational pull, from which not even light can escape — and, therefore, we have no information or insight as to what life is like inside. A black hole is not empty, It's actually a lot of matter condensed into a single point. This point is known as a singularity.
What is the average speed of the bicyclist's ride?
A.45m/s
B.7.5m/s
C45mi/hr
D.7.5mi/hr
sorry i was gone hope you dident mess me hers a question
Identify the organ that brings oxygen into the body.
brain
chest
heart
lungs
Answer:
lungs
Explanation:
lungs bring oxygen
Answer:
lungs
Explanation:
i took the quiz
A vertical spring scale can measure weights up to 225 N . The scale extends by an amount of 12.5 cm from its equilibrium position at 0 N to the 225 N mark. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.65 Hz. Ignoring the mass of the spring, what is the mass m of the fish?
Answer:
The mass of the fish is 6.493 kilograms.
Explanation:
The spring-mass system experiments a Simple Harmonic Motion, then the angular frequency ([tex]\omega[/tex]), in radians per second, is determined by the following equation:
[tex]\omega = 2\pi\cdot f[/tex] (1)
Where [tex]f[/tex] is the frequency, in hertz.
In addition, the mass of the fish ([tex]m[/tex]), in kilograms, is:
[tex]m = \frac{k}{\omega^{2}}[/tex] (2)
Where [tex]k[/tex] is the spring constant, in newtons per meter.
And the spring constant is determined by Hooke's Law:
[tex]k = \frac{F}{\Delta x}[/tex] (3)
Where:
[tex]F[/tex] - Elastic force, in newtons.
[tex]\Delta x[/tex] - Spring elongation, in meters.
If we know that [tex]F = 225\,N[/tex], [tex]\Delta x = 0.125\,m[/tex] and [tex]f = 2.65\,hz[/tex], then the mass of the fish is:
[tex]\omega = 2\pi\cdot f[/tex]
[tex]\omega \approx 16.650\,\frac{rad}{s}[/tex]
[tex]k = \frac{F}{\Delta x}[/tex]
[tex]k = 1800\,N[/tex]
[tex]m = \frac{k}{\omega^{2}}[/tex]
[tex]m = 6.493\,kg[/tex]
The mass of the fish is 6.493 kilograms.
This question involves the concepts of Hooke's Law, and the frequency of spring-mass system.
The mass of the fish is "6.5 kg".
First, we will calculate the angular speed of the system:
[tex]\omega=2\pi f[/tex]
where,
ω = angular speed = ?
f = frequency = 2.65 Hz
Therefore,
[tex]\omega = 2\pi(2.65\ Hz)\\\omega = 16.65\ rad/s\\[/tex]
Now, we will use the Hooke's Law to find out the spring constant of the spring:
[tex]K =\frac{F}{\Delta x}[/tex]
where,
K = spring constant = ?
F = force applied = maximum weight = 225 N
Δx = extension = 12.5 cm = 0.125 m
Therefore,
[tex]k=\frac{225\ N}{0.125\ m}\\\\k = 1800\ N/m\\[/tex]
Now, we will use the formula for the angular speed of a spring-mass system to find out the mass of the fish:
[tex]\omega = \sqrt{\frac{k}{m}}\\\\m = \frac{k}{\omega^2}\\\\m=\frac{1800\ N/m}{(16.65\ rad/s)^2}[/tex]
m = 6.5 kg
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The attached picture illustrates Hooke's Law.
Although the use of absorbances at 450 nm provided you with maximum sensitivity, the absorbances at, say, 400 nm or 500 nm are not zero and could have been used throughout this experiment. Would the same value of K be obtained at one of these wavelengths
Answer:
Yes, the value will be the same.
Explanation:
Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.
In the diagrams, the solid green line represents Earth's rotational axis and the dashed red line represents the magnetic field axis. Which diagram accurately shows Earth's magnetic field?
Answer: Picture 2 or B!!!!!
Answer:
B
Explanation:
the answer as said is picture 2
An astronaut brings a cube from the Earth to the Moon.
What is true about the inertial mass and weight of the cube?
Note that the gravitational acceleration on the Moon is about 1.6m/s^2
A) The inertial mass increases but the weight decreases.
B) Both the inertial mass and weight decrease.
C) The inertial mass decreases but the weight increases.
D) The inertial mass remains constant but the weight decreases.
Answer:
Answer D. Inertial mass constant, weight decreases.
Explanation:
Khan Academy
Electricity involves transfer of charge. The charge(s) involved in electrical forces are composed of which?
- Protons
- Neutrons
- Electrons
- Quarks
- Protons & Electrons
What turns the drive shaft of the generator?
Help
Answer:
Produces 60-cycle AC electricity; it is usually an off-the-shelf induction generator. High-speed shaft: Drives the generator. Low-speed shaft: Turns the low-speed shaft at about 30-60 rpm. Nacelle: Sits atop the tower and contains the gear box, low- and high-speed shafts, generator, controller, and brake.
Explanation:
What does the balloon of the air capacitor represent in an electrical capacitor?
Answer:
The balloon prohibits the flow of air through the air capacitor.
Explanation:
Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.
The balloon analogy is frequently used in electrical capacitors to assist visualise the notion of the capacitor's behaviour.
The balloon illustrates the capacitor's physical structure, specifically the two conducting plates and the dielectric material between them.
The capacitor's stored energy is equivalent to the air pressure within the balloon. When the voltage is withdrawn, the stored charge is released, and the capacitor returns to its uncharged condition, exactly as the balloon deflates when the air exits.
The balloon analogy aids in understanding how a capacitor stores electrical energy in its electric field and how that energy may be released when linked to a circuit.
Thus, it depicts the relationship between a capacitor's voltage, charge, and capacitance.
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a 200 kg car is travelling at 33m/s. what is the kinectic energy of the car
-2 m
2m-2m
Determine the reactions on the beam as shown.
270 kN
67.5 KN.m
0.3 m
760
B
-2.1 m
3 m-
-1.2 m
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How many radians did the disk turn while stopping ? C. how many revolutions?
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
[tex]\omega_f = \omega_i + \alpha t[/tex]
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
[tex]0\ rad/s = 3.14\ rad/s + \alpha(3\ s)[/tex]
α = - 1.047 rad/s²
B.
We can use the second equation of motion to find the angular distance:
[tex]\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2[/tex]
θ = 14.1 rad
C.
θ = (14.1 rad)(1 rev/2π rad)
θ = 2.24 rev
what is Shortening melting
physical property
chemical change
chemical property
physical change
A heat pump with a COP of 3.0 is used to heat air contained in a 1205.4 m3 of well-insulated, rigid tank. Initially the pressure and the temperature inside the gas tank are 100 kPa and 7 oC, respectively. When running, the heat pump consumes 5 kW of electric power. How long does it take for the heat pump to raise the temperature of air in the tank to 22 oC
Answer:
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Explanation:
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The solar glare of sunlight bouncing off water or snow can be a real problem for drivers. The reflecting sunlight is horizontally polarized, meaning that the light waves oscillate at an angle of 90 degrees with respect to a vertical line. At what angle relative to this vertical line should transmission axis of polarized sunglasses be oriented, if they are to be effective against solar glare
Answer:
Explanation:
The light waves in the reflected sunlight are horizontally polarized, which illustrates that they oscillate at a [tex]90^o[/tex] angle related to a vertical line.
Depending on the condition of the height of the light, the glare can be almost entirely horizontally polarized. Furthermore, all reflections from over-water surfaces are partially polarized. The water becomes more translucent when using polarized sunglasses.
If polarized sunglasses are to be efficient against solar glare, the transmission axis should be positioned at an angle of [tex]\theta = 45^{o}[/tex]
. A circular wire loop 40 cm in diameter has 100 Ohm resistance and lies in a horizontal plane. A uniform magnetic field points vertically downward, and in 25 ms it increases from 5 mT to 55 mT. Find the magnetic flux through the loop at (a) the beginning and (b) the end of 25 ms period. (c) What is the loop current during this time
Answer:
(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A
Explanation:
Given that,
The diameter of the loop, d = 40 cm
Radius, r = 20 cm
Initial magnetic field, B = 5 mT
Final magnetic field, B' = 55 mT
Initial magnetic flux,
[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]
Final magnetic flux,
[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]
Due to change in magnetic field an emf will be generated in the loop. It is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]
Let I be the current in the loop. We can find it using Ohm's law such that,
[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]
Hence, this is the required solution.
What is the electric field strength of a point charge of 10 uC at a distance of 5 cm away?
Answer:
E = 3.6 x 10⁷ N/C
Explanation:
The electric field strength due to a point charge is given by the following formula:
[tex]E = \frac{kq}{r^2}\\\\[/tex]
where,
E = Electrical Field Strenght = ?
k = Colomb's Constant = 9 x 10⁹ Nm²/C²
q = magnitude of charge = 10 μC = 1 x 10⁻⁵ C
r = distance = 5 cm = 0.05 m
Therefore,
[tex]E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(1\ x\ 10^{-5}\ C)}{(0.05\ m)^2}[/tex]
E = 3.6 x 10⁷ N/C
what is Muscular system
How much energy (in kWh) is produced in one day by a solar panel of surface area A =15
m? in a region where the average solar power density if 4.33 kWh/m²/day. Assume the
efficiency of the panel to be 18 %. Round off your result to 2 decimal places, and do not
write the unit
Question 2
20 pts
The average electricity consumption of a house in Gainesville is known to be 907 kWh in a
month (One month - 30 days). They would like to install solar panels of 12 % efficiency to
generate this electricity. Given that the average solar power density in Gainesville is 5,47
kWh/m2/day, how much surface area must the panels occupy? Calculate the result in m²
but do not write the unit. Round off you answer to a whole number (zero decimal place.)
Answer:
I am calculating the total area of a solar panel for a particular load demand by ... designing according to energy demand for a day then how will it affect total solar area? ... Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency ... would need is a 1 m2 solar panel to produce 1000 Watts of electrical energy.
Explanation:
