Answer:
The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].
Explanation:
Given that,
Semi major axis [tex]a= 4.3\times10^{8}[/tex]
Orbital period T=1.516 days
Using Kepler's third law
[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]
[tex]M=\dfrac{4\pi^2}{GT^2}a^3[/tex]
Where, T = days
G = gravitational constant
a = semi major axis
Put the value into the formula
[tex]M=\dfrac{4\times(3.14)^2}{6.67\times10^{-11}(1.516\times24\times60\times60)^2}(4.3\times10^{8})^3[/tex]
[tex]M=2.7\times10^{27}\ kg[/tex]
Hence, The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].
Given the frequency of an electromagnetic wave, what else can you find immediately?
Answer:
wavelength
Explanation:
An electromagnetic waves is produced due to interaction of oscillating electric and magnetic field when they interacts perpendicular to each other. For example, Gamma rays, X rays , ultraviolet rays, visible radiations, infrared rays, micro waves, radio waves.
All the electromagnetic waves have velocity equal to velocity of light.
If the frequency of electromagnetic wave is unknown then we find the wavelength of wave. the formula used is
Wave speed = wavelength x frequency
Fluids at rest possess no flow energy. a)-True b)-False
I believe the answer is A: True
A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a block of mass M =11.5 kg that is at rest. After the collision, the 4.8-kg block reverses direction, and its new speed is Vf=2.5m/s. What is V, the speed of the 11.5-kg block? 5.6 m/s
6.5 m/s
3.7 m/s
4.7 m/s
4.1 m/s
Answer:
The speed of the 11.5kg block after the collision is V≅4.1 m/s
Explanation:
ma= 4.8 kg
va1= 7.3 m/s
va2= - 2.5 m/s
mb= 11.5 kg
vb1= 0 m/s
vb2= ?
vb2= ( ma*va1 - ma*va2) / mb
vb2= 4.09 m/s ≅ 4.1 m/s
A charged isolated metal sphere of diameter 15 cm has a potential of 8500 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.
Answer:
[tex]u = 0.057 J/m^3[/tex]
Explanation:
Energy density near the surface of the sphere is given by the formula
[tex]u = \frac{1}{2}\epsilon_0 E^2[/tex]
also for sphere surface we know that
[tex]E = \frac{V}{R}[/tex]
R = radius of sphere
V = potential of the surface
now we have
[tex]u = \frac{1}{2}\epsilon_0 (\frac{V^2}{R^2})[/tex]
now from the above formula we have
[tex]u = \frac{1}{2}(8.85 \times 10^{-12})(\frac{8500^2}{0.075^2})[/tex]
[tex]u = 0.057 J/m^3[/tex]
If the current density in a wire or radius R is given by J-k+5,0F wire? R, what is the current in the wire?
Answer:
I = [tex]R^{2}[/tex](K+5)
Explanation:
Given :
J = k+5
Now selecting a thin ring in the wire of radius "r" and thickness dr.
Current through the thin ring is
dI = J X 2πrdr
dI = (K+5) x 2πrdr
Now integrating we get
I = [tex]\int_{0}^{R} = (K+5).2\pi rdr[/tex]
I = (K+5) 2π[tex]\int_{0}^{R} rdr[/tex]
I = (K+5) 2π [tex]\frac{R^{2}}{2}[/tex]
I = [tex]R^{2}[/tex](K+5)
A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V at a time 3.21 s after charging begins. Find R.
Answer:
655128 ohm
Explanation:
C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F
V₀ = Voltage of the battery = 9 Volts
V = Potential difference across the battery after time "t" = 4.20 Volts
t = time interval = 3.21 sec
T = Time constant
R = resistance
Potential difference across the battery after time "t" is given as
[tex]V = V_{o} (1-e^{\frac{-t}{T}})[/tex]
[tex]4.20 = 9 (1-e^{\frac{-3.21}{T}})[/tex]
T = 5.11 sec
Time constant is given as
T = RC
5.11 = (7.8 x 10⁻⁶) R
R = 655128 ohm
A completely reversible refrigerator is driven by a 10-kW compressor and operates with thermal energy reservoirs at 250 K and 310 K. Calculate the rate of cooling provided by this refrigerator.
The rate of cooling provided by a reversible refrigerator, working with a 10-kW compressor and thermal energy reservoirs at 250 K and 310 K, is calculated using the refrigerator's Carnot coefficient and the power of the compressor. With these specific conditions, the cooling rate is calculated to be 50 kW.
Explanation:The question pertains to the functioning of a reversible refrigerator, specifically the cooling rate provided by the appliance which uses a 10-kW compressor and operates with thermal energy reservoirs at 250 K and 310 K. To determine the cooling rate, we must first understand some basics about the refrigerator's operation.
A reversible refrigerator absorbs heat Qc from a cold reservoir and discards it into a hot reservoir, while work W is done on it. This work is represented by the power exerted by the compressor. The refrigerator functions by moving a coolant through coils, which absorbs heat from the contents of the refrigerator and releases it outside.
The coefficient of performance or Carnot coefficient (KR) of the refrigerator can be computed using the formula KR = Tc / (Th - Tc) where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this situation, KR = 250 / (310 - 250) = 5.
Since the work done W is the compressor power P, we can use the formula Qc = KR * P to find the cooling rate. Substituting the known values, Qc = 5 * 10 kW = 50 kW. Therefore, the rate of cooling provided by this refrigerator is 50 kW.
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The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration? Serway, Raymond A.; Jewett, John W.. Physics for Scientists and Engineers (MindTap Course List) (Page 121). Cengage Learning. Kindle Edition.
Answer:
Acceleration= 1,59 (meters/(second^2))
Direction= NE; 65,22° above the east direction.
Explanation:
Resulting force= ( ((180N)^2) + ((390N)^2) ) ^ (1/2) = 429,53 N
Angle obove the east direction= ((cos) ^ (-1)) (180N / 429,53 N) = 65,22°
Acceleration= Resulting force / mass = (429,53 N) / (270 kg) =
= (429,53 kg × (meters/(second^2))) / (270kg) = 1,59 (meters/(second^2))
The center of mass of a pitched baseball or radius 3.87 cm moves at 36.6 m/s. The ball spins about an axis through its center of mass with an angular speed of 112 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.
To calculate the ratio of rotational energy to translational kinetic energy for a pitched baseball, use the formulas for rotational kinetic energy and translational kinetic energy. The moment of inertia is found by using the radius of the ball, and the rotational and translational kinetic energies can be determined using the mass, radius, and velocity of the ball. Finally, calculate the ratio of the two energies to get the answer.
Explanation:To calculate the ratio of the rotational energy to the translational kinetic energy, we need to find the rotational kinetic energy and the translational kinetic energy of the pitched baseball.
The rotational kinetic energy can be calculated using the formula:
Rotational Kinetic Energy = (1/2) * I * ω²
Where I is the moment of inertia and ω is the angular speed.
The translational kinetic energy can be calculated using the formula:
Translational Kinetic Energy = (1/2) * m * v²
Where m is the mass of the ball and v is the velocity.
We know that the radius of the ball is 3.87 cm and it moves at a velocity of 36.6 m/s. The angular speed is 112 rad/s. Since the ball is treated as a uniform sphere, its moment of inertia can be calculated using the formula:
Moment of Inertia = (2/5) * m * r²
Substitute the given values into the formulas and calculate the rotational and translational kinetic energies. Then, find the ratio of the rotational energy to the translational energy.
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(a) Write the energy equation for an elastic collision. (b) For an inelastic collision.
Answer:
Explanation:
There are two types of collision.
(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.
In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.
The momentum of the system before collision = the momentum of system after collision
The kinetic energy of the system before collision = the kinetic energy after the collision
(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.
In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.
The momentum of the system before collision = the momentum of system after collision
The total mechanical energy of the system before collision = total mechanical of the system after the collision
Water ia boled at 1 at pressure in a coffe aker equpped with an immension-type electric heating element. The coffee maker intially contains 1 kg of water. Once boiling started, it is observed that half of he water in the coffee maker evaporated in 10 minutes the heat loss from the cofee maker is negigible, the power rating of the heating element is: a)-1.9kw b)-16 kw c)-0.8 kW d)-2.2 kw e)-3.8KW
-- Water cannot be boled.
-- There is no such thing as a coffe aker.
-- There is no such thing as an immension-type heating element.
Be that as it may, and it very likely still is, as it were . . .
-- The latent heat of vaporization of water at 100°C is about 2250 kilo-joules per kg.
-- To evaporate half of the kg of water in the coffee maker requires 1125 kJ of heat energy.
-- To supply that amount of heat energy over a period of 10 minutes (600 seconds), it must be supplied at a rate of
(1,125,000 Joules / 600 seconds) = 1,875 joules/second.
-- That's 1,875 watts, or 1.875 kilowatts.
-- Choice-a is the choice when the solution is rounded.
At a football game, an air gun fires T-shirts into the crowd. The gun is fired at an angle of 46 degree from the horizontal with an initial speed of 27 m/s. A fan who is sitting 60 m horizontally from the gun, but high in the stands, catches a T-shirt. A) How long does it take for the T-shirt to reach the fan? B) At what height h is the fan from the ground?
Answer:
a) [tex]t=3.199 seconds[/tex]
b) [tex]h = 11.97 m[/tex]
Explanation:
Since this problem belongs to the concept of projectile motion
a) we know,
[tex]Vcos\theta=\frac{R}{t}[/tex]
Where,
V = initial speed
Θ = angle with the horizontal
R = horizontal range
t = Time taken to cover the range 'R'
Given:
V = 27m/s
R = 60m
Θ = 46°
thus,
the equation becomes
[tex]27\times cos46^o=\frac{60}{t}[/tex]
or
[tex]t=\frac{60}{27\times cos46^o}[/tex]
[tex]t=3.199 seconds[/tex]
b)The formula for height is given as:
[tex]h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\[/tex]
where,
g = acceleration due to gravity = 9.8m/s²
substituting the values in the above equation we get
[tex]h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\[/tex]
or
[tex]h = 62.124-50.14[/tex]
or
[tex]h = 11.97 m[/tex]
It takes 2.22 seconds for the T-shirt to reach the fan. The fan is located at a height of 24.57 meters from the ground.
Explanation:To find the time it takes for the T-shirt to reach the fan, we need to solve for the time in the horizontal motion. The horizontal distance from the gun to the fan is given as 60 m. Since the horizontal motion is constant velocity, we can use the equation d = vt and solve for t. Plugging in the values, we get t = 60 m / 27 m/s = 2.22 s.
To find the height of the fan from the ground, we need to solve for the vertical motion. The equation for vertical motion is y = yt + (1/2)gt^2, where y is the vertical displacement, yt is the initial vertical velocity, and g is the acceleration due to gravity. In this case, the vertical displacement is unknown, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values and solving for y, we get y = (1/2)(9.8 m/s^2)(2.22 s)^2 = 24.57 m.
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A wheel, originally rotating at 126 rad/s undergoes a constant angular deceleration of 5.00 rad/s2. What is its angular speed after it has turned through an angle of 628 radians
Answer:
97.96 [tex]\frac{rad}{s}[/tex]
Explanation:
The initial angular velocity is [tex]w_{0}[/tex] = 126 rad / s.
The constant angular deceleration is 5.00 rad / s2.
The constant angular deceleration is, by definition: dw / dt.
[tex]\frac{dw}{dt}=-5 \frac{rad}{s^{2} }[/tex]
Separating variables
[tex]dw=-5 dt[/tex]
Integration (limits for w: 0 to W0; limits for t: 0 to t)
[tex]w= w_{0}-5t[/tex]
W is by definition [tex]\frac{d\alpha }{dt}[/tex], where [tex]\alpha[/tex] is the angle.
[tex]\frac{d\alpha}{dt}=w_{0} -5t[/tex]
Separating variables
[tex]d\alpha=(w_{0} -5t )dt[/tex]
Integration (limits for [tex]\alpha[/tex]: 0 to 628; limits for t: 0 to t)
[tex]\alpha =w_{0}t-(\frac{5}{2})t^{2}[/tex]
[tex]128=126t-(\frac{5}{2})t^{2}[/tex]
Put in on the typical form of a quadratic equation:
[tex]\frac{5}{2}t^{2}-126t+628=0[/tex]
Solve by using the quadratic equation formula and discard the higher result because it lacks physical sense.
t=5.608 s
Evaluate at this time the angular velocity:
[tex]w(t=5.608)=126-5*5.608[/tex]
[tex]w(5.608)=97.96 \frac{rad}{s}[/tex]
To find the final angular speed of a rotating wheel given the initial speed, angular deceleration and the angle turned, we use an equation for angular motion. We substitute the given values and solve the equation to find the final angular speed.
Explanation:
To solve this question, we will be using the equation for angular motion,
ω² = ω0² + 2αθ
, where ω is the final angular speed we want to find, ω0 = 126 rad/s is the initial angular speed, α = -5.00 rad/s² is the angular deceleration, and θ = 628 rad is the angle turned through. This equation is analogous to the equation for linear motion, v² = u² + 2as, where v is final velocity, u initial velocity, a acceleration and s distance. We substitute the given values into the equation to find the final angular speed. Keep in mind, because we are dealing with deceleration, our α value is a negative.
Solve the equation: ω² - ω0² = 2αθ, which gives: ω² = (ω0² + 2αθ), ω = sqrt((ω0² + 2αθ)), ω = sqrt((126rad/s)² + 2*(-5.00 rad/s²)*628 rad). So, when you calculate the square root of the total, you will find the answer for the final angular speed.
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3.A robot is on the surface of Mars. The angle of depression from a camera in the robot to a rock on the surface of Mars is 13.78°. The camera is 184.0 cm above the surface. How far is the camera from the rock? (Round to the nearest tenth as needed.)
Answer:
750.25 cm
Explanation:
θ = 13.78°, h = 184 cm
Let the distance between rock and camera is d.
Tan θ = h / d
tan 13.78 = 184 / d
d = 750.25 cm
By utilizing trigonometry and the tangent of an angle in a right triangle, it is determined that the camera (at a height of 184.0 cm from the Mars surface with an angle of depression of 13.78° to a rock) is approximately 761.4 cm away from the rock.
Explanation:This question involves trigonometry, specifically the tangent of an angle in a right triangle. The robot captures an image of a rock with an angle of depression of 13.78°. The camera is 184.0 cm above the Mars surface. We can create a right triangle where the angle at the camera is 13.78°, the opposite side is 184.0 cm (the height of the camera), and the adjacent side is the distance between the camera and the rock, which we will call x.
By definition, tan(angle) = opposite/adjacent. Here, the angle is 13.78°, the opposite side is 184.0 cm and the adjacent side is x (unknown). To find x, we can use the following formula: x = opposite/tan(angle).
Therefore, x = 184.0 cm / tan(13.78°). Using a calculator, x is approximately 761.4 cm (to the nearest tenth). So the camera is approximately 761.4 cm away from the rock.
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In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.74 m by the horizontal 21 N force from the broom and then has a speed of 1.2 m/s, what is the coefficient of kinetic friction between the book and floor?
Answer:
0.51
Explanation:
m = mass of the book = 3.5 kg
F = force applied by the broom on the book = 21 N
a = acceleration of the book
v₀ = initial speed of the book = 0 m/s
v = final speed of the book = 1.2 m/s
d = distance traveled = 0.74 m
Using the equation
v² = v₀² + 2 a d
1.2² = 0² + 2 a (0.74)
a = 0.973 m/s²
f = kinetic frictional force
Force equation for the motion of the book is given as
F - f = ma
21 - f = (3.5) (0.973)
f = 17.6 N
μ = Coefficient of kinetic friction
Kinetic frictional force is given as
f = μ mg
17.6 = μ (3.5 x 9.8)
μ = 0.51
700*.135An inventor claims to have developed a heat engine that receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting the waste heat to a sink at 290 K. Is this a reasonable claim?
Answer:
output work is not possible to have more than 294 kJ value so this is not reasonable claim
Explanation:
As we know that the efficiency of heat engine is given as
[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]
now we will have
[tex]T_2 = 290 K[/tex]
[tex]T_1 = 500 K [tex]
[tex]\eta = 1 - \frac{290}{500}[/tex]
[tex]\eta = 0.42[/tex]
now we know that efficiency is defined as
[tex]\eta = \frac{Work}{Heat}[/tex]
[tex]0.42 = \frac{W}{700}[/tex]
[tex]W = 294 kJ[/tex]
So output work is not possible to have more than 294 kJ value
To determine if the inventor's claim of developing a heat engine is reasonable, we can calculate the maximum theoretical efficiency of the engine and compare it to the claimed net work output and heat input.
Explanation:A heat engine operates between a hot reservoir and a cold reservoir, absorbing heat from the hot reservoir and converting some of it into work, while rejecting the remaining heat to the cold reservoir. The efficiency of a heat engine is determined by the temperature of the reservoirs. In this case, the inventor claims that the heat engine receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting waste heat to a sink at 290 K. To determine if this claim is reasonable, we can calculate the theoretical maximum efficiency of the heat engine using the Carnot efficiency formula.
The Carnot efficiency formula is given by:
Efficiency = 1 - (Tc / Th)
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this case, Tc = 290 K and Th = 500 K.
Plugging in these values into the formula, we have:
Efficiency = 1 - (290 K / 500 K) = 1 - 0.58 = 0.42 or 42%
Therefore, the maximum theoretical efficiency for a heat engine operating between these temperatures is 42%. Since the claimed net work output of the heat engine is 300 kJ, we can calculate the maximum heat input by dividing the net work output by the efficiency:
Maximum heat input = Net work output / Efficiency = 300 kJ / 0.42 = 714.3 kJ
Since the claimed heat input is 700 kJ, which is slightly less than the maximum calculated heat input, it is reasonable to say that the inventor's claim is possible.
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A 100 kg individual consumes 1200 kcal of food energy a day. Calculate
(a) the altitude change, in m, if the food energy content was converted entirely into lifting the indi-
vidual under normal earth gravity.
(b) the velocity, in m/s, if the food energy content was converted entirely into accelerating the indi-
vidual from rest.
(c) the final temperature, in ◦C, of a 100 kg mass of liquid water initially at the normal human body
temperature and heated with the energy content of the food. You can use a liquid water specific
heat of 4.1 kJ/kg K.
Answer:
(a) 5142.86 m
(b) 317.5 m/s
(c) 49.3 degree C
Explanation:
m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J
(a) Let the altitude be h
Q = m x g x h
504 x 10^4 = 100 x 9.8 x h
h = 5142.86 m
(b) Let v be the speed
Q = 1/2 m v^2
504 x 10^4 = 1/2 x 100 x v^2
v = 317.5 m/s
(c) The temperature of normal human body, T1 = 37 degree C
Let the final temperature is T2.
Q = m x c x (T2 - T1)
504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)
T2 = 49.3 degree C
In a model of the hydrogen atom, the electron travels in circular orbits around the proton. What is the electric potential, in volts, due to the proton on an electron in an orbit with radius 2.08 x 10^-10 m?
Answer:
6.93 volts
Explanation:
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = radius of the orbit = 2.08 x 10⁻¹⁰ m
V = Electric potential due to proton on electron
Electric potential due to proton on electron is given as
[tex]V = \frac{kq}{r}[/tex]
Inserting the values
[tex]V = \frac{(9\times 10^{9})(1.6\times 10^{-19})}{2.08\times 10^{-10}}[/tex]
V = 6.93 volts
Final answer:
The electric potential due to the proton on an electron in an orbit with a radius of 2.08 x 10^-10 m in a hydrogen atom model is approximately -13.6 volts.
Explanation:
In the context of the model of a hydrogen atom, the electric potential due to the proton on an electron in an orbit with radius 2.08 x 10-10 m can be calculated using the formula for the electric potential V due to a point charge, which is V = k * Q / r. Here, k is the Coulomb's constant (approximately 8.99 x 109 N m2/C2), Q is the charge of the proton (1.602 x 10-19 C), and r is the distance from the proton to the electron, which is the radius of the orbit.
To find the electric potential, simply plug these values into the formula:
V = (8.99 x 109 N m2/C2) * (1.602 x 10-19 C) / (2.08 x 10-10 m)
After calculating, the electric potential is found to be approximately -13.6 volts, with the negative sign indicating that the potential energy associated with the electron is negative, which is common for bound states such as electrons in an atom.
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial Kinetic energy. What is the mangnitude of the change in momentum of the stone?
Answer:
change in momentum of the stone is 1.635 kg.m/s
Explanation:
let m = 0.500kg ball and M be the mass of the stone, v be the velocity of the ball and V be the velocity of the stone
the initial kinetic energy of the ball is 1/2(0.500)(20^2) = 100 J
the kinetic energy of the ball after rebounding is 70/100(100) = 70 J
Kb = 1/2mv^2
v = \sqrt{2k/m} = \sqrt{2(70)/0.500} = 16.73 m/s
from the conservation of linear momentum, we know that:
mvi + MVi = mvf + MVf
MVf - MVi = mvi - mvf
MVf - MVi = (0.500)(20) - (0.500)(16.73)
= 1.635 kg.m/s
therefore, the change is momentum of the stone is 1.635 kg.m/s
ly charged particles are held 24 x 103m apart and then released from rest. The initial acceleration of the first particle is observed to be 7.0 m/s2 and that of the second to be 9.0 m/s2. The mass of the first particle is 6.3 x 107 kg. (a) What is the mass of the second particle? kg (b) What is the magnitude of the charge of each particle?
Answer:
Part a)
[tex]m_2 = 4.9 \times 10^7 kg[/tex]
Part b)
[tex]q_1 = q_2 = 5312.6 C[/tex]
Explanation:
Part a)
As we know that both charge particles will exert equal and opposite force on each other
so here the force on both the charges will be equal in magnitude
so we will have
[tex]F = m_1a_1 = m_2a_2[/tex]
here we have
[tex]6.3 \times 10^7(7) = m_2(9)[/tex]
now we have
[tex]m_2 = 4.9 \times 10^7 kg[/tex]
Part b)
Now for the force between two charges we can say
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we have
[tex](6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}[/tex]
now we have
[tex]q_1 = q_2 = 5312.6 C[/tex]
A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 30.0 m/s2, what is the radius of the planet
Answer:
5.44×10⁶ m
Explanation:
For a satellite with period t and orbital radius r, the velocity is:
v = 2πr/t
So the centripetal acceleration is:
a = v² / r
a = (2πr/t)² / r
a = (2π/t)² r
This is equal to the acceleration due to gravity at that elevation:
g = MG / r²
(2π/t)² r = MG / r²
M = (2π/t)² r³ / G
At the surface of the planet, the acceleration due to gravity is:
g = MG / R²
Substituting our expression for the mass of the planet M:
g = [(2π/t)² r³ / G] G / R²
g = (2π/t)² r³ / R²
R² = (2π/t)² r³ / g
R = (2π/t) √(r³ / g)
Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:
R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)
R = 5.44×10⁶ m
Notice we didn't need to know the mass of the satellite.
What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic field of 3.0 T/s causes a 6.0 A current to flow?The resistance of the circuit that contains the solenoid is 17 ȍ. The only emf source forthe circuit is the induced emf.
Answer:
Radius of cross section, r = 0.24 m
Explanation:
It is given that,
Number of turns, N = 180
Change in magnetic field, [tex]\dfrac{dB}{dt}=3\ T/s[/tex]
Current, I = 6 A
Resistance of the solenoid, R = 17 ohms
We need to find the radius of the solenoid (r). We know that emf is given by :
[tex]E=N\dfrac{d\phi}{dt}[/tex]
[tex]E=N\dfrac{d(BA)}{dt}[/tex]
Since, E = IR
[tex]IR=NA\dfrac{dB}{dt}[/tex]
[tex]A=\dfrac{IR}{N.\dfrac{dB}{dt}}[/tex]
[tex]A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}[/tex]
[tex]A=0.188\ m^2[/tex]
or
[tex]A=0.19\ m^2[/tex]
Area of circular cross section is, [tex]A=\pi r^2[/tex]
[tex]r=\sqrt{\dfrac{A}{\pi}}[/tex]
[tex]r=\sqrt{\dfrac{0.19}{\pi}}[/tex]
r = 0.24 m
So, the radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.
A second's pendulum is taken to the moon. What will be the time period of the pendulum at the moon? (Acceleration due to gravity at the surface of the moon is 1/6 on the surface of earth).
Answer:
4.89 seconds
Explanation:
The time period of a pendulum is given by
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
For a second pendulum on earth, T = 2 second
[tex]2 = 2\pi \sqrt{\frac{l}{g}}[/tex] ...... (1)
Now the time period is T when the pendulum is taken to moon and gravity at moon is 1/6 of gravity of earth
[tex]T = 2\pi \sqrt{\frac{l}{\frac{g}{6}}}[/tex] ...... (2)
Divide equation (2) by equation (1)
[tex]\frac{T}{2} = \sqrt{6}[/tex]
T = 4.89 seconds
Blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up in a row on a frictionless table. All three are pushed forward by a 13 N force applied to the 3.0 kg block. How much force does the 4.0 kg block exert on the 5.0 kg block?
Answer:
Fc = 5.41 N
Explanation:
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Newton's second law for the set of the three blocks
F = 13 N
m=3.0 kg+ 4.0 kg+5.0 kg = 12 kg
F = m*a
13 = 12*a
a = 13 / 12
a = 1.083 m/s² : acceleration of the set of the three blocks
Newton's second law for the 5.0 kg block
m= 5.0 kg
a = 1.083 m/s²
Fc: Contact force of the 4 kg block on the 5 kg block
Fc = 5.0 kg * 1.083 m/s²
Fc = 5.41 N
The 4.0 kg block exerts a force of 5.4 N on the 5.0 kg block when a 13 N force is applied to the entire system of blocks and accelerates the system at 1.08 m/s² on a frictionless surface.
Explanation:We are dealing with a problem related to Newton's Third Law of Motion where a 13 N force is applied to a system of three blocks in contact with each other on a frictionless surface.
To find the force that the 4.0 kg block exerts on the 5.0 kg block, we need to calculate the acceleration of the system first and then use Newton's second law for just the last two blocks.
The total mass of the system is the sum of the masses of the three blocks: 3.0 kg + 4.0 kg + 5.0 kg = 12.0 kg. The total force provided by the 13 N force gives us an acceleration of the system:
a = F_{total} / m_{total} = 13 N / 12.0 kg = 1.08 m/s² (rounded to two decimal places)
Now, using Newton's second law, F = ma, for the 5.0 kg block, which is being accelerated by the force from the 4.0 kg block (F_{45}), we find:
F_{45} = m_{5.0kg} × a = 5.0 kg × 1.08 m/s²= 5.4 N
Therefore, the 4.0 kg block exerts a force of 5.4 N on the 5.0 kg block.
Learn more about Force on a Block here:https://brainly.com/question/29668172
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A constant force of 120 N pushes a 55 kg wagon across an 8 m level surface. If the wagon was initially at rest, what is the final kinetic energy of the wagon?
Answer:
The kinetic energy of the wagon is 967.0 J
Explanation:
Given that,
Force = 120 N
Mass = 55 kg
Height = 8 m
We need to calculate the kinetic energy of the wagon
Using newtons law
[tex]F = ma[/tex]
[tex]\dfrac{120}{55}=a[/tex]
[tex]a =2.2\ m/s^2[/tex]
Using equation of motion
[tex]v^2 =u^2+2as[/tex]
Where,
v = final velocity
u = initial velocity
s = height
Put the value in the equation
[tex]v^2=0+2\times2.2\times8[/tex]
[tex]v=5.93\ m/s[/tex]
Now, The kinetic energy is
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]K.E=\dfrac{1}{2}\times55\times(5.93)^2[/tex]
[tex]K.E=967.0\ J[/tex]
Hence, The kinetic energy of the wagon is 967.0 J
How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
Answer:
[tex]Q = 4.63 \times 10^6 J[/tex]
Explanation:
As we know that melting point of silver is
T = 961.8 degree C
Latent heat of fusion of silver is given as
L = 111 kJ/kg
specific heat capacity of silver is given as
[tex]s = 240 J/kg C[/tex]
now we will have
[tex]Q = ms\Delta T + mL[/tex]
[tex]\Delta T = 961.8 - 20 [/tex]
[tex]\Delta T = 941.8 degree C[/tex]
now from above equation
[tex]Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)[/tex]
[tex]Q = 3.1 \times 10^6 + 1.53 \times 10^6[/tex]
[tex]Q = 4.63 \times 10^6 J[/tex]
Which of the following can penetrate the deepest (Please explain)
A) 3MeV electron
B) 10MeV alpha
C) 0.1 MeV auger
D) 400keV proton
Answer: 3MeV electron
Explanation:
m_e={9.1\times 10^{-31} m_α=4\times m_e m_a={9.1\times 10^{-31}
m_p=1.67\times 10^{-27}
(a) K.E. Energy of electron =[tex]\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}[/tex]=3MeV
[tex]v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }[/tex]=1.05\times10^{18}
[tex]v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}[/tex]
(b) K.E. Energy of alpha particle =[tex]\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}[/tex]=10MeV
[tex]v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6} }{4\times9.1\times 10^{-31} }[/tex]=0.88\times10^{18}
[tex]v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}[/tex]
(c) K.E. Energy of auger particle =[tex]\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}[/tex]=0.1MeV
[tex]v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }[/tex]=0.035\times10^{18}
[tex]v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}[/tex]
(d) K.E. Energy of proton particle =[tex]\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}[/tex]=400keV
[tex]v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3} }{1.67\times 10^{-27} }[/tex]=0.766\times10^{14}
[tex]v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}[/tex]
from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.
A coil of conducting wire carries a current i. In a time interval of Δt = 0.520 s, the current goes from i1 = 3.20 A to i2 = 1.90 A. The average emf induced in the coil is e m f = 14.0 mV. Assuming the current does not change direction, calculate the coil's inductance (in mH).
Answer:
5.6 mH
Explanation:
i1 = 3.20 A, i2 = 1.90 A, e = 14 mV = 0.014 V,
Let L be the coil's inductance.
[tex]e = -L\times \frac{\Delta i}{\Delta t}[/tex]
[tex]0.014 = -L\times \frac{1.90 - 3.20}{0.52}[/tex]
L = 0.0056 H
L = 5.6 mH
A 4-kg hammer is lifted to a height of 10 m and dropped from rest. What was the velocity (in m/s) of the hammer when it was at a height of 4 m from the earth? O12 O5 O11 O109.5
Answer:
v = 10.84 m/s
Explanation:
using the equation of motion:
v^2 = (v0)^2 + 2×a(r - r0)
due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.
v^2 = 2×g(r - r0)
v = \sqrt{2×(-9.8)×(4 - 10)}
= 10.84 m/s
therefore, the velocity at r = 4 meters is 10.84 m/s
One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 24 m/s. The masses of the two objects are 3.8 and 8.4 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially.
Answer:
Part a)
v = 16.52 m/s
Part b)
v = 7.47 m/s
Explanation:
Part a)
(a) when the large-mass object is the one moving initially
So here we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]
since this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]
[tex]v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}[/tex]
[tex]v = 16.52 m/s[/tex]
Part b)
(b) when the small-mass object is the one moving initially
here also we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]
Again this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]
[tex]v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}[/tex]
[tex]v = 7.47 m/s[/tex]