Answer:
[tex]F=208.83N[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=2.4m/s^2[/tex]
Force of Branch [tex]F=260N[/tex]
Generally the Newton's equation second law for Force is mathematically given by
[tex]ma=F-mg[/tex]
[tex]m=\frac{260}{2.4+9.8}[/tex]
[tex]m=21.31kg[/tex]
Therefore
[tex]F=mg[/tex]
[tex]F=(21.31)(9.8)[/tex]
[tex]F=208.83N[/tex]
If you buy an amateur-sized reflecting telescope, say around 10 inches (25cm) aperture, it'll have something in it that sends the gathered starlight out the side of the telescope tube. What do we call this thing
Answer: objective lens
Explanation:
Light enters a refra
Light enters a telescope through a lens at the upper end, which focuses the light near the bottom of the telescope. An eyepiece then magnifies the image so that it can be viewed by the eye, or a detector like a photographic plate can be placed at the focus. The upper end of a reflecting telescope is open, and the light passes through to the mirror located at the bottom of the telescope. The mirror then focuses the light at the top end, where it can be detected. Alternatively, a second mirror may reflect the light to a position outside the telescope structure, where an observer can have easier access to it.
Which one is the better material to use for an inexpensive compass? hard iron, soft iron, or any conductor
Answer:
Soft iron
Explanation:
A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g>cm3 (Fig. E12.19). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column
Answer:
665.25 Pa
Explanation:
Given data :
Weight of the disk, w = 45 N
Diameter, d = 30 cm
= 0.30 m
Therefore, radius of the disk,
[tex]$r=\frac{d}{2}$[/tex]
[tex]$r=\frac{0.30}{2}$[/tex]
= 0.15 m
Now, area of the cylindrical disk,
[tex]$A=\pi r^2$[/tex]
[tex]$A=3.14 \times (0.15)^2$[/tex]
[tex]$=0.07065 \ m^2$[/tex]
∴ The gauge pressure at the top of the oil column is :
[tex]$p=\frac{w}{A}$[/tex]
[tex]$p=\frac{47}{0.07065}$[/tex]
= 665.25 Pa
Therefore, the gauge pressure is 665.25 Pa.
The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa
The pressure is defined by the relationship between perpendicular force and area.
[tex]P = \frac{F}{A}[/tex]
where P is pressure, F is force, and A is area.
They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.
The area is:
A = π r² = [tex]\pi \frac{d^2}{4}[/tex]
In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.
B-W = 0
B = W
The force applied to the liquid is the weights of the cylinder. Let's replace.
[tex]P= \frac{W}{A} \\P = W \frac{4}{\pi d^2 }[/tex]
Let's calculate.
[tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²
P = 636.6 Pa
In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa.
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Pure water is an example of alan
A. Insulator
B. Metalloid
C. Conductor
D. Nonmetal
Answer: I think its A or C I'm not sure though sorry.
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?
Explain the following observations:
a) A balloon filled with hydrogen gas floats in air;
B) A ship made of steel floats on water.
Answer and Explanation:
a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.
b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.
If the length of the rod is 2.65 m, and the mass of the bob and the rod are both 1.4 kg, what is the period of this pendulum
Answer:
T = 5.66 s
Explanation:
The system formed by the bar plus ball forms a physical pendulum
w = [tex]\sqrt{mgd/I}[/tex]
the moment of inertia of a rod held at one end is
I = [tex]\frac{1}{3}[/tex] m L²
we substitute
w = [tex]\sqrt{\frac{d \ d}{ 3 L^2 } }[/tex]
in this case the turning distance and the length of the rod are equal
d = L
w = [tex]\sqrt{\frac{g}{3L} }[/tex]
angular velocity and period are related
w = 2π / T
2π / T = [tex]\sqrt{\frac{g}{3L} }[/tex]
T = 2π [tex]\sqrt{3L/g}[/tex]
let's calculate
T = 2π [tex]\sqrt{3 \ 2.65 / 9.8}[/tex]
T = 5.66 s
Find the amount og work done
Answer:
100j
Explanation:
In the early 1900's ____
began leading the automobile exploration in the US automotive industry.
-Karl Benz
-Henry Ford
-Gottlieb Daimler
-None of the above
Answer:
Henry Ford
Explanation:
he built the first ford
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
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A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the instant you make measurements on the glider, it is moving at 0.835 m/sm/s and is 4.00 cmcm from its equilibrium point.
Required:
a. Use energy conservation to find the amplitude of the motion.
b. Use energy conservation to find the maximum speed of the glider.
c. What is the angular frequency of the oscillations?
(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is
W = - (1/2 kx ² - 1/2 k (0.0400 m)²)
(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)
By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so
W = ∆K = 0 - 1/2 m (0.835 m/s)²
Solve for x :
- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²
==> x ≈ 0.0493 m ≈ 4.93 cm
(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is
W = - 1/2 k (0.0400 m)²
If v is the glider's maximum speed, then by the work-energy theorem,
W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²
Solve for v :
- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²
==> v ≈ 1.43 m/s
(c) The angular frequency of the glider's oscillation is
√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz
The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s
From the given information;
the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 mUsing energy conservation E, the amplitude of the motion can be calculated by using the formula:
[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]
[tex]\mathbf{E =0.194 \ J}[/tex]
Similarly, we know that:
[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]
Making amplitude A the subject, we have:
[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]
[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]
[tex]\mathbf{A =0.049 \ cm}[/tex]
Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:
[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]
[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]
The angular frequency of the oscillation can be computed by using the expression:
[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]
[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]
ω = 29.02 rad/s
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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 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
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.
An electron in a hydrogen atom is in a p state. Which of the following statements is true?
a.
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
b.
The electron has an energy of -13.6 eV.
c.
The electron has a total angular momentum of ħ.
d.
The electron has a z-component of angular momentum equal to sqrt(2)* ħ.
Answer:
The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).
Explanation:
We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.
P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.
Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node
A closely wound, circular coil with radius 2.70 cm has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 T
Answer:
Approximately 4.029 A
Explanation:
We can use the formula that the B field of a few loops all with current in same direction is permeability of free space (mu)* current * Number or loops divided by 2*radius. You are given B field, radius(convert into meters), number of loops and mu is 4pi * 10^-7. Solve for current and you get 4.029 Amperes.
A wire 54.6 cm long carries a 0.480 A current in the positive direction of an x axis through a magnetic field with an x component of zero, a y component of 0.000420 T, and a z component of 0.0130 T. Find the (a) x, (b) y, and (c) z components of the magnetic force on the wire.
Answer:
wire 66.0 cm long carries a 0.750 A current in the positive direction of an x axis through a magnetic field $$\vec { B } = ( 3.00 m T ) \hat { j } ...
Top answer · 1 vote
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Answer:
Explanation:
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot
Answer:
The distance of a mass 25g from the pivot is 18cm
Explanation:
Given
[tex]m_1 = 10g[/tex]
[tex]d_1 = 45cm[/tex]
[tex]m_2 = 25g[/tex]
Required
Distance of m2 from the pivot
To do this, we make use of:
[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses
So, we have:
[tex]10 * 45= 25* d_2[/tex]
[tex]450= 25* d_2[/tex]
Divide both sides by 25
[tex]18= d_2[/tex]
Hence:
[tex]d_2 = 18[/tex]
An airplane increases its speed at the average rate of 15 m/s2. How much time does it take to increase its speed from 100 m/s to 160 m/s
Answer:
4 s
Explanation:
Acceleration (a) = 15 m/s²Initial velocity (u) = 100 m/sFinal velocity (v) = 160 m/sWe are asked to calculate time taken (t).
By using the first equation of motion,
[tex]\longrightarrow[/tex] v = u + at
[tex]\longrightarrow[/tex] 160 = 100 + 15t
[tex]\longrightarrow[/tex] 160 - 100 = 15t
[tex]\longrightarrow[/tex] 60 = 15t
[tex]\longrightarrow[/tex] 60 ÷ 15 = t
[tex]\longrightarrow[/tex] 4 s = t
You are holding one end of a horizontal stretched string. Flicking your wrist will send a pulse down the string. Which actions will make the pulse travel faster
Answer:
Use a lighter string of the same length, under the same tension.
Stretch the string tighter to increase the tension
Explanation:
The wave speed depends on propertices of the medium, not on how you generate the wave. For a string
Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.
A pulse will be sent down a horizontal extended thread if a person flicks the wrist. To raise the tension, pull the string tighter.
Define pulse.Pulse is the same thing as monitoring heartbeat. The pulse rate could also be measured via auscultation, which includes hearing to the heart rhythm using a stethoscope then counting it for one minute.
A pulse will be sent down a horizontal extended thread if a person holds one end of the rope and flicks their wrist.
To raise the tension, pull the string tighter.
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A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J
Answer:
a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy is -18279.78 J
c) heat lost by the gas is zero or 0
d) the work done on the gas is -18279.78 J
Explanation:
Given the data in the question;
P[tex]_i[/tex] = 1 atm = 101325 pascal
P[tex]_f[/tex] = ?
V[tex]_i[/tex] = 0.1550 m³
V[tex]_f[/tex] = 0.742 m³
we know that for an adiabatic process γ = 1.4
P[tex]_i[/tex]V[tex]_i^Y[/tex] = P[tex]_f[/tex]V[tex]_f^Y[/tex]
P[tex]_f[/tex] = P[tex]_i[/tex][tex]([/tex] V[tex]_i[/tex] / V[tex]_f[/tex] [tex])^Y[/tex]
we substitute
P[tex]_f[/tex] = 1 × [tex]([/tex] 0.1550 / 0.742 [tex])^{1.4[/tex]
= [tex]([/tex] 0.2088948787 [tex])^{1.4[/tex]
= 0.11166 atm
a) the initial and final temperatures
Initial temperature
T[tex]_i[/tex] = P[tex]_i[/tex]V[tex]_i[/tex] / nR
given that n = 3.10 mol
= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )
= 15705.375 / 25.7734
T[tex]_i[/tex] = 609.4 K
Final temperature
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )
= 8394.95 / 25.7734
= 325.7 K
Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy
ΔE[tex]_{int[/tex] = nC[tex]_v[/tex]ΔT
here, C[tex]_v[/tex] = ( 5/2 )R
ΔE[tex]_{int[/tex] = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J
c) the heat lost by the gas
Since its an adiabatic process,
Q = 0
Therefore, heat lost by the gas is zero or 0
d) the work done on the gas
W = ΔE[tex]_{int[/tex] - Q
= -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J
a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.
b) The change in internal energy is -18279.78 J.
c) The heat lost by the gas is zero.
d) The work done on the gas is -18279.78 J.
Given data:
The moles of sample is, n = 3.10 mol.
The initial volume of sample is, [tex]V_{1}=0.1550 \;\rm m^{3}[/tex].
The final volume of sample is, [tex]V_{2}=0.742 \;\rm m^{3}[/tex].
The initial pressure of the sample is, [tex]P_{1}=1.00 \;\rm atm[/tex].
(a)
We know that the relation between the pressure and volume for an adiabatic process is as follows,
[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]
Here, [tex]\gamma[/tex] is a adiabatic index. And for air, its value is 1.41.
Solving as,
[tex]P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm[/tex]
Now, calculate the final temperature using the ideal gas equation as,
[tex]P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K[/tex]
Similarly, calculate the initial temperature as,
[tex]P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K[/tex]
Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.
(b)
The change in internal energy is given as,
ΔE = nCΔT
here, C = ( 5/2 )R
ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J.
c)
The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.
Q = 0
Therefore, heat lost by the gas is zero or 0
d)
The work done on the gas
W = ΔE - Q
W = -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J.
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An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 19 m/s and measures a time of 24.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Answer:
1.56 m/s²
Explanation:
Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.
The total time (time of flight) of an object is given by:
T = 2usinθ / g
where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity
Since the astronaut throws a rock straight up, hence θ = 90°, u = 19 m/s, T = 24.4 s.
T = 2usinθ / g
Substituting:
24.4 = 2(19)(sin90)/g
g = 2(19)(sin90) / 24.4
g = 1.56 m/s²
the Period T of oscillation of a Single Pendulum depends on the length l, and acceleration g. Determine the exact form of the dependence.
Answer:
[tex]{ \tt{check \: in \: the \: pic}}[/tex]
A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.
Required:
What is the orbital period of this GPS satellite?
Answer:
[tex]T=66262.4s[/tex]
Explanation:
From the question we are told that:
Altitude [tex]A=2.90 *10^7[/tex]
Mass [tex]m=5.97 * 10^{24} kg[/tex]
Radius [tex]r=6.38 *10^6 m.[/tex]
Generally the equation for Satellite Speed is mathematically given by
[tex]V=(\frac{GM}{d} )^{0.5}[/tex]
[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]
[tex]V=3354.83m/s[/tex]
Therefore
Period T is Given as
[tex]T=\frac{2 \pi *a}{V}[/tex]
[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]
[tex]T=66262.4s[/tex]
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
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.
Read more on quantity of energy here: https://brainly.com/question/13439286
Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitude and directions of their resultant
Answer:
19N to the south
Explanation:
F =10N + 5N + 4N
1
An astronaut weighs 202 lb. What is his weight in newtons?
Answer:
978.6084 Newton
Explanation:
Given the following data;
Weight = 220 lbTo find the weight in Newtown;
Conversion:
1 lb = 4.448220 N
220 lb = 220 * 4.448220 = 978.6084 Newton
220 lb = 978.6084 Newton
Therefore, the weight of the astronaut in Newton is 978.6084.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, the weight of an object is given by the formula;
Weight = mg
Where;
m is the mass of the object.g is the acceleration due to gravity.Note:
lb is the symbol for pounds.N is the symbol for Newton.