Using the acceleration due to gravity on the planet, we calculate its mass to be 8.97 x 10²⁴ kg. For the mass of the star, using Kepler's third law, we find it to be approximately 1.99 x 10³⁰ kg, assuming the planet's orbit is about 1 AU.
To find the mass of the planet and the star, we can use Newton's form of Kepler's third law and Newton's universal law of gravitation. Let's denote the mass of the star as M and the mass of the planet as m.
Mass of the Planet
The acceleration due to gravity (g) on the surface of the planet is given by:
g = Gm / r²
Where:
G is the gravitational constant (approximately 6.674 x 10⁻¹¹ N m2 kg-²).
m is the mass of the planet.
r is the radius of the planet, which is half of its diameter.
Rearranging the formula to solve for m:
m = g r² / G
Substituting the given values (g = 59.7 m/s² and r = 9 x 106 m), we get:
m = 59.7 x (9 x 106)² / 6.674 x 10⁻¹¹
= 8.97 x 10²⁴ kg
Mass of the Star
Using Newton's version of Kepler's third law, we get:
M = 4π² a3 / G T²
Where: a is the semi-major axis of the planet's orbit, which is the average distance from the planet to the star.
T is the orbital period of the planet around the star.
Assuming that the planet's orbit is approximately 1 Astronomical Unit (AU) since the period is close to 1 Earth year, and converting 402 Earth days to seconds, we find M.
T = 402
days x 24 hours/day x 3600 seconds/hour = 3.47 x 10⁷ seconds
a = 1 AU = 1.496 x 10¹¹ m
Then the mass of the star M is:
M = 4π² (1.496 x 1011)3 / (6.674 x 10⁻¹¹ x (3.47 x 10⁷)²) = 1.99 x 10³⁰ kg
When performing the pupillary reflexes, describe the position of the subject and environment
the subject should be comfortable and relaxed in a well lit area
the subject should be seated, relaxed in a dimly lit area with one eye closed
the subject should be seated, relaxed in a well lit area with one eye closed
the subject should be comfortable and relaxed in a dimly lit area, the subject should shield their eyes by placing their hand vertically between the eye and nose
Answer:
The pupillary light reflex (PLR) or photopupillary reflex is a reflex that controls the diameter of the pupil, in response to the intensity (luminance) of light that falls on the retinal ganglion cells of the retina in the back of the eye, thereby assisting in adaptation of vision to various levels of lightness/darkness.
A greater intensity of light causes the pupil to constrict (miosis/myosis; thereby allowing less light in), whereas a lower intensity of light causes the pupil to dilate (mydriasis, expansion; thereby allowing more light in)
Explanation:
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide box having mass m2 and they move together. Find the velocity of boxes.
Answer:Expression given below
Explanation:
Given mass of spring[tex]\left ( m_1\right )=0.5 kg[/tex]
Compression in the spring[tex]\left ( x\right )=20 cm[/tex]
Let the spring constant be K
Using Energy conservation
potential energy stored in spring =Kinetic energy of Block[tex]\left ( m_1\right )[/tex]
[tex]\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2[/tex]
[tex]v=x\sqrt{\frac{k}{m_1}}[/tex]
now conserving momentum
[tex]m_1v=\left ( m_1+m_2\right )v_0[/tex]
[tex]v_0=\frac{m_1}{m_1+m_2}v[/tex]
where [tex]v_0[/tex] is the final velocity
What is a gravational lens?
Final answer:
A gravitational lens occurs when a massive object like a galaxy causes the light from a more distant object to bend and magnify, much like how an optical lens works. Einstein predicted this based on general relativity, and its discovery has provided critical verification of this theory. The first gravitational lens was found in 1979, showing double images of a distant quasar.
Explanation:
A gravitational lens is a phenomenon where the light from a distant object, such as a quasar or galaxy, is bent and magnified by the gravity of a massive object, like a galaxy or galaxy cluster, that lies between the observer and the distant object. This effect is analogous to the way a conventional lens, like those found in glasses, bends light. Albert Einstein predicted this effect based on his theory of general relativity but thought it would be very unlikely for us to observe it. However, astronomers have since observed numerous instances of gravitational lensing, which provide significant evidence for general relativity. These observations include multiple images, arcs, or even rings of the same astronomical object, as gravity distorts the space around the massive intervening object.
The first gravitational lens was discovered in 1979 and presented dual images of the same distant object. This discovery and subsequent observations using powerful telescopes like the Hubble Space Telescope have allowed astronomers to study the universe's most distant objects and the distribution of dark matter. The bend of light due to massive objects can create spectacular effects such as Einstein rings, where a perfect alignment between the observer, massive object, and distant source results in a ring-like structure of light.
An aquarium 3 m long, 2 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg/m3, and use 9.8 m/s2 for g.)
To find the work needed to pump half of the water out of the aquarium, you need to calculate the volume of the aquarium and the mass of the water. The work done is equal to the force (weight of the water) multiplied by the distance (height of the aquarium). Therefore, the work needed is 29400 Nm or J (Joules).
Explanation:To find the work needed to pump half of the water out of the aquarium, we first need to calculate the volume of the aquarium. The volume is given by length x width x height, which equals 3m x 2m x 1m = 6m³. Half of the water in the aquarium would be 6m³/2 = 3m³.
The density of water is given as 1000 kg/m³. So the mass of the water that needs to be pumped out is 3m³ x 1000 kg/m³ = 3000 kg.
The work done to pump the water out is given by the formula work = force x distance. In this case, the force required is the weight of the water, which is mass x gravity. The distance is the height of the aquarium, which is 1m.
Therefore, the work needed to pump half of the water out of the aquarium is 3000 kg x 9.8 m/s² x 1m = 29400 Nm or J (Joules).
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A solenoid has 100 turns per cm and a total resistance of 60 Ω center of the solenoid when it is connected to a 12 volt battery Determine the magnetic field in the 21 A flat c
Answer:
2.5 x 10^-3 T
Explanation:
n = 100 turns / cm = 10000 turns / m, R = 60 ohm, V = 12 V
i = V / R = 12 / 60 = 0.2 A
Magnetic field due to solenoid is
B = μ0 x n x i
B = 4 x 3.14 x 10^-7 x 10000 x 0.2
B = 2.5 x 10^-3 T
Aluminum has a magnetic susceptibility χ = 1.7 × 10-5 at T = 300 K. What will be the magnetization of a small sample of aluminum placed in a 1.5 T magnetic field at T = 150 K?
Answer:
The magnetization of a small sample of aluminum is 40.6 A/m
Explanation:
Given that,
Magnetic susceptibility [tex]\chi_{1}=1.7\times10^{-5}[/tex] at 300 K
Magnetic field = 1.5 T
We need to calculate the magnetic susceptibility at 150 K
[tex]\dfrac{\chi_{2}}{\chi_{1}}=\dfrac{T_{1}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{\chi_{2}}{1.7\times10^{-5}}=\dfrac{300}{150}[/tex]
[tex]\chi_{2}=\dfrac{300\times1.7\times10^{-5}}{150}[/tex]
[tex]\chi_{2}=3.4\times10^{-5}[/tex]
We need to calculate the magnetization
Using formula of magnetization
[tex]I=\chi_{2}\times H[/tex]
Where, H = magnetic intensity
Formula of magnetic intensity
[tex]H=\dfrac{B}{\mu_{0}}[/tex]
Where, B = magnetic field
Put the value of H into the formula of magnetization
[tex]I=\chi_{2}\times\dfrac{B}{\mu_{0}}[/tex]
[tex]I=3.4\times10^{-5}\times\dfrac{1.5}{4\pi\times10^{-7}}[/tex]
[tex]I=40.6\ A/m[/tex]
Hence, The magnetization of a small sample of aluminum is 40.6 A/m
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between the sled and surface is 0.229. What is the speed of the sled after it has traveled a distance of 3.89 m?
The final speed of the sled is calculated by first calculating the frictional force using the coefficient of kinetic friction and weight of the sled. This force is then used to calculate the deceleration due to friction. Finally, an equation of motion is employed to calculate the sled's final speed after it has traveled the given distance.
Explanation:To answer this question, we first need to calculate the force of friction, which can be found using the formula F_friction = µN, where µ is the coefficient of kinetic friction and N is the normal force. In this case, as the sled is moving on a horizontal surface, the normal force is equal to the weight of the sled, N = mg. So the frictional force F_friction = µmg.
Next, due to Newton's second law (F = ma), the deceleration (negative acceleration) is calculated by a = F_friction/m. With this acceleration, we can now use one of the equations of motion, V_f = V_i + 2a*d, to calculate the final velocity, where V_f is the final velocity, V_i is the initial velocity, a is acceleration and d is the distance.
This method allows us to calculate the final speed of the sled after traveling a certain distance considering kinetic friction.
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The final speed of the sled is calculated by finding the work done by friction and applying the work-energy theorem to find the change in kinetic energy. The kinetic energy is equated to its initial value minus the work done by friction, and solved for the final speed.
Explanation:To tackle this question, we should first calculate the total work done by the friction force (kinetic friction) which is known to act in the opposite direction of the sled's movement. The work done by friction is given by the equation W = Fd in which 'F' stands for friction force and 'd' for distance. The friction force can be calculated by the formula F = μN, where 'μ' is the coefficient of friction and 'N' is the normal force. Since the sled moves across a horizontal surface, N is equal to the weight of the sled (mg, where 'm' is the mass and 'g' is the acceleration due to gravity).
Then, using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy, we can solve for the final speed. The kinetic energy is given by KE = 0.5mv², where 'm' is the mass and 'v' is the speed. So, the initial and the final kinetic energies of the sled are KE_initial = 0.5 * 2.12kg * (5.49 m/s)² and KE_final = KE_initial - Work_done_by_friction. We can solve the latter for the final speed of the sled, v_final = sqrt((2*KE_final)/m).
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(I) What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 3.2 mm of mica?
As per the details given, the capacitance of the pair of circular plates is approximately [tex]\(1.11 \times 10^{-11}\) Farads.[/tex]
A capacitor is a device made to store electric charge, and one of its basic electrical characteristics is capacitance.
When a voltage difference (potential difference) exists between a capacitor's two conducting plates, the capacity of the capacitor to store and hold electrical energy is measured.
Capacitance is a measurement of how much charge can be held on a capacitor's plates for a specific voltage.
The capacitance (C) of a parallel plate capacitor is given by the formula:
[tex]\[ C = \dfrac{\varepsilon_0 \cdot A}{d} \][/tex]
Where:
[tex]\( \varepsilon_0 \)[/tex] is the vacuum permittivity ([tex]\(8.85 \times 10^{-12} \, \text{F/m}\)[/tex])
A is the area of the plates (in square meters)
d is the distance between the plates (in meters)
Given:
Radius of the circular plates (r) = 5.0 cm
= 0.05 m
Distance between the plates (d) = 3.2 mm
= 0.0032 m
Calculate the area of one of the plates:
[tex]\[ A = \pi \cdot r^2 \\\\= \pi \cdot (0.05 \, \text{m})^2 \][/tex]
Calculate the capacitance (C):
[tex]\[ C = \dfrac{\varepsilon_0 \cdot A}{d} \\\\= \dfrac{8.85 \times 10^{-12} \, \text{F/m} \cdot \pi \cdot (0.05 \, \text{m})^2}{0.0032 \, \text{m}} \][/tex]
[tex]\[ C \approx 1.11 \times 10^{-11} \, \text{F} \][/tex]
Thus, the capacitance of the pair of circular plates is approximately [tex]\(1.11 \times 10^{-11}\)[/tex] Farads.
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The capacitance of a pair of circular plates with a radius of 5.0 cm separated by 3.2 mm of mica is 10.3 pF. This is calculated using the formula for the capacitance of a parallel plate capacitor and the area of a circle.
Explanation:To determine the capacitance of a pair of circular plates of a parallel plate capacitor with a given radius, separated by a specific dielectric material, we use the formula:
C = ε0 εr (A/d)
Where:
C is the capacitance,
ε0 is the vacuum permittivity (8.85 × 10-12 F/m),
εr is the relative permittivity of the dielectric (for mica, it's approximately 6),
A is the area of one of the plates, and
d is the separation between the plates.
To find the area A of a circular plate with radius r, we use the formula A = πr2. For a radius of 5.0 cm (or 0.05 m), A = π(0.05 m)2 = 7.85 × 10-3 m2.
Substituting into the capacitance formula, with d = 3.2 mm (or 0.0032 m), we get:
C = (8.85 × 10-12 F/m) × 6 × (7.85 × 10-3 m2/0.0032 m) = 1.03 × 10-11 F
This result is usually expressed in picofarads (pF), so the capacitance is 10.3 pF.
A potter's wheel of radius 11 cm starts from rest and rotates with constant angular accel- eration until at the end of 35 s it is moving with angular velocity of 19 rad/s. What is the angular acceleration? Answer in units of rad/s2.
Final answer:
The angular acceleration of the potter's wheel is approximately 0.543 rad/s^2, calculated by dividing the final angular velocity by the time taken to reach it.
Explanation:
To calculate the angular acceleration of a potter's wheel that starts from rest and reaches an angular velocity of 19 rad/s in 35 seconds, we can use the following kinematic equation for rotational motion:
\(\omega = \omega_0 + \alpha \times t\)
Where:
\(\omega\) is the final angular velocity.
\(\omega_0\) is the initial angular velocity (0 rad/s since the wheel starts from rest).
\(\alpha\) is the angular acceleration.
\(t\) is the time taken to reach the final angular velocity.
Substituting the given values:
19 rad/s = 0 rad/s + \(\alpha\) \times 35 s
To solve for \(\alpha\), we rearrange the equation:
\(\alpha = \frac{19 rad/s}{35 s}\)
\(\alpha = \frac{19}{35} rad/s^2\)
\(\alpha \approx 0.543 rad/s^2\)
Therefore, the angular acceleration of the potter's wheel is approximately 0.543 rad/s2.
The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is μs = 0.41, determine the greatest slope θ the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.
The greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.
The given parameters;
mass of the car, m = 1.6 Mgcoefficient of static friction, [tex]\mu_ s[/tex] = 0.41The greatest slope of the shoulder is determined by taking net force on the car as shown below;
[tex]\Sigma F= ma \\\\Wsin\theta - \mu_s Wcos \theta = ma[/tex]
at constant velocity, acceleration, a, = 0
[tex]Wsin\theta - \mu_s W cos \theta = 0\\\\W sin\theta = \mu_s Wcos \theta \\\\\mu_s = \frac{Wsin\theta }{W cos \theta } \\\\\mu_s = tan \theta \\\\\theta = tan^{-1}(\mu_s)\\\\\theta = tan^{-1}(0.41)\\\\\theta =22.3\ ^0[/tex]
Thus, the greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.
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The greatest slope the car can be without slipping or tipping over is approximately 22.4 degrees where weight due to gravity is balanced by static friction.
Explanation:This problem involves the concept of static friction, which prevents an object from sliding down a slope. Here, the car is held in place on a slope due to static friction between the tires and the road. First, we need to calculate the force of gravity acting on the car, which is the weight of the car mg. The mass of the car is given as 1.6 Mg = 1.6 x 10^6 grams. Therefore, the weight of the car is 1.6 x 10^6 g * 9.8 m/s² = 1.568 x 10^7 N.
Next, the force of static friction is given by the equation μsN, where N is the normal force and μs is the coefficient of static friction which is 0.41. Since the car is not moving, the net force on it must be zero. This implies that the force of static friction must balance out the force of gravity. So, we equate the forces and solve for the slope angle θ, getting a relationship: tan θ = μs.
So, θ = tan^-1 (0.41) ≈ 22.4°. Therefore, the greatest slope the shoulder can have without causing the car to slip or tip over is approximately 22.4 degrees.
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A marble rolls off a tabletop 1.0 m high and hits the floor at a point 1.2 m away from the table’s edge in the horizontal direction. What is the speed of the marble when it leaves the table’s edge?
A. 1.2 m/s
B. 9.8 m/s
C. 2.7 m/s
D. 3.7 m/s
Answer:
C. 2.7 m/s
Explanation:
Consider the motion of the marble along the vertical direction
y = vertical displacement = height of tabletop = 1.0 m
a = acceleration = 9.8 m/s²
t = time taken to hit the floor = ?
v₀ = initial speed of the marble along vertical direction = 0 m/s
Using the equation
y = v₀ t + (0.5) a t²
1.0 = (0) t + (0.5) (9.8) t²
t = 0.45 s
Consider the motion of the marble along the horizontal direction :
v = speed with which the marble leaves the table
x = horizontal displacement = 1.2 m
t = 0.45 s
horizontal displacement is given as
x = v t
1.2 = v (0.45)
v = 2.7 m/s
A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil has 100 turns how much current must pass through the coil?
Answer:
The current pass through the coil is 6.25 A
Explanation:
Given that,
Diameter = 25 cm
Magnetic field = 1.0 mT
Number of turns = 100
We need to calculate the current
Using the formula of magnetic field
[tex]B =\dfrac{\mu_{0}NI}{2\pi r}[/tex]
[tex]I=\dfrac{B\times2\pi r}{\mu N}[/tex]
Where, N = number of turns
r = radius
I = current
Put the value into the formula
[tex]I=\dfrac{1.0\times10^{-3}\times2\pi\times12.5\times10^{-2}}{4\pi\times10^{-7}100}[/tex]
[tex]I=6.25\ A[/tex]
Hence, The current passes through the coil is 6.25 A
Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter. The resistivity of copper is 1.68×10−8Ω⋅m and the resistivity of aluminum is 2.65×10−8Ω⋅m.
Answer: 0.258
Explanation:
The resistance [tex]R[/tex] of a wire is calculated by the following formula:
[tex]R=\rho\frac{l}{s}[/tex] (1)
Where:
[tex]\rho[/tex] is the resistivity of the material the wire is made of. For aluminium is [tex]\rho_{Al}=2.65(10)^{-8}m\Omega[/tex] and for copper is [tex]\rho_{Cu}=1.68(10)^{-8}m\Omega[/tex]
[tex]l[/tex] is the length of the wire, which in the case of aluminium is [tex]l_{Al}=12m[/tex], and in the case of copper is [tex]l_{Cu}=30m[/tex]
[tex]s[/tex] is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:
[tex]s=\pi{(\frac{d}{2})}^{2}[/tex] (2) Where [tex]d[/tex] is the diameter of the circumference.
For aluminium wire the diameter is [tex]d_{Al}=2.5mm=0.0025m[/tex] and for copper is [tex]d_{Cu}=1.6mm=0.0016m[/tex]
So, in this problem we have two transversal areas:
For aluminium:
[tex]s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}[/tex]
[tex]s_{Al}=0.000004908m^{2}[/tex] (3)
For copper:
[tex]s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}[/tex]
[tex]s_{Cu}=0.00000201m^{2}[/tex] (4)
Now we have to calculate the resistance for each wire:
Aluminium wire:
[tex]R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}[/tex] (5)
[tex]R_{Al}=0.0647\Omega[/tex] (6) Resistance of aluminium wire
Copper wire:
[tex]R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}[/tex] (6)
[tex]R_{Cu}=0.250\Omega[/tex] (7) Resistance of copper wire
At this point we are able to calculate the ratio of the resistance of both wires:
[tex]Ratio=\frac{R_{Al}}{R_{Cu}}[/tex] (8)
[tex]\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}[/tex] (9)
Finally:
[tex]\frac{R_{Al}}{R_{Cu}}=0.258[/tex] This is the ratio
One piece of copper jewelry at 111°C has exactly twice the mass of another piece, which is at 28°C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter (c of copper = 0.387 J/g·K)?
Answer:
83.33 C
Explanation:
T1 = 111 C, m1 = 2m
T2 = 28 C, m2 = m
c = 0.387 J/gK
Let the final temperature inside the calorimeter of T.
Use the principle of calorimetery
heat lost by hot body = heat gained by cold body
m1 x c x (T1 - T) = m2 x c x (T - T2)
2m x c X (111 - T) = m x c x (T - 28)
2 (111 - T) = (T - 28)
222 - 2T = T - 28
3T = 250
T = 83.33 C
Thus, the final temperature inside calorimeter is 83.33 C.
_________is defined as the how many cycles are completed in one second by an oscillating object. Frequency Period Linear Velocity Momentum
Answer:
Frequency
Explanation:
The number of cycles completed in one second by an oscillating object is called its frequency. The term frequency shows number of times an oscillation is occurring. It is denoted by [tex]\nu[/tex]. The reciprocal of frequency is called the time period of the wave i.e.
[tex]\nu=\dfrac{1}{T}[/tex]
T = time period
So, the correct option is (a) "frequency".
Final answer:
Frequency is the number of oscillations per second, measured in hertz (Hz), and is inversely related to the period (T), which is the time it takes to complete one oscillation.
Explanation:
Frequency is defined as the number of cycles completed in one second by an oscillating object. In physics, frequency (f) is the number of oscillations that take place per unit of time. The SI unit for frequency is the hertz (Hz), which is equivalent to one oscillation per second. The relationship between frequency and period (T) is given by f = 1/T, meaning that as the period increases, the frequency decreases, and vice versa.
Periodic motion, such as that of a mass suspended by a wire undergoing simple harmonic motion, has a constant time to complete one cycle, known as the period (T). Its units are usually seconds. Period refers to the time it takes to complete one oscillation, while frequency is the number of these oscillations within a given timeframe.
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point located 6.00 cm from the axis of rotation of the centrifuge with a speed of 99 m/s when the centrifuge is at full speed moves (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up? (b) How many revolutions does the centrifuge make as it goes from rest to its final angular speed?
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
An ordinary glass is filled to the brim with water at 100oc. How much water could be added to the glass if the temperature is lowered to 20oC? Assume that the coefficient of volume expansion for glass is 2.7 x 10-5 K-1 and for water it is 2.1 x 10-4 K-1
Answer:
14.64% of the volume of the glass, more water can be added
Explanation:
given:
Initial temperature, = 100°C = 373K
Final temperature = 20°C = 293K
Coefficient of volume expansion for glass, = 2.7 x 10⁻⁵ K⁻¹
Coefficient of volume expansion for Water, = 2.1 x 10⁻⁴ K⁻¹
The apparent Coefficient of volume expansion for Water, γ = 2.1 x 10⁻⁴ K⁻¹ - 2.7 x 10⁻⁵ K⁻¹ = 1.83 × 10⁻⁴ K⁻¹
Change in temperature, ΔΘ = Final Temperature - Initial Temperature = 293K - 373K = -80K
Now, the Coefficient of volume expansion is given as:
[tex]\Delta V = -{\gamma}{V \Delta \theta}\\[/tex]
where,
V = initial Volume
ΔV = change in the volume
Thus,
[tex]\Delta V = -{1.83\times 10^{-4} K^{-1} }\times {V \times -80K}\\[/tex]
or
[tex]\Delta V = 146.4\times 10^{-4}\times V[/tex]
or
[tex]\frac{\Delta V}{V} = 146.4\times 10^{-4}[/tex]
Multiplying both sides by 100
we get
[tex]\frac{\Delta V}{V}\times 100 = 146.4\times 10^{-4}\times 100[/tex]
or
change in volume with respect to the initial volume = 1.464%
thus, 1.464 % of the original volume water can be added.
Upon cooling from 100 °C to 20 °C, the thermal contraction of the water in the glass will be greater than that of the glass itself due to higher coefficient of volume expansion for water, thereby creating some extra space in the glass. This extra space represents the amount of additional water that can be added.
Explanation:In order to find out how much more water can be added to the glass as it cools from 100 °C to 20 °C, we first need to know the volume of water and glass that will change with this temperature drop. The formula we use to calculate this is ΔV = βV₀ΔT, where β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.
First, we calculate this for the glass and then for the water, using their respective coefficients of volume expansion. The difference in the volume changes for water and glass will give us the amount of water that can be added to the glass as it cools. Since the coefficient of volume expansion for water is greater than glass, its volume will contract more due to the decrease in temperature, leaving some extra space in the glass.
Note that we make an assumption here that the glass and water were initially at thermal equilibrium at 100 °C and hence have the same initial volume. This problem illustrates the principle of thermal expansion which is a property of matter to change its volume with a change in temperature, and is dependent on the material's coefficient of volume expansion.
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A rocket takes off from Earth and reaches a speed of 105 m/s in 18.0 s. If the exhaust speed is 1,200 m/sand the mass of fuel burned is 110 kg, what was the initial mass (in kg, including the initial fuel) of the rocket?
Answer:
The initial mass of the rocket is 526.2 kg.
Explanation:
Given that,
Final speed = 105 m/s
Time = 18.0 s
Exhaust initial speed = 1200 m/s
Mass of burned fuel = 110 kg
We need to calculate the initial mass
The velocity change of rocket under gravity is defined as,
[tex]v=u\ ln(\dfrac{m_{i}}{m})-gt[/tex]....(I)
We know that,
[tex]m=m_{i}-m_{bf}[/tex]
Put the value of m in equation (I)
[tex]v=u\ ln(\dfrac{m_{i}}{m_{i}-m_{bf}})-gt[/tex]
[tex]m_{i}=\dfrac{m_{bf}}{1-e^-{\dfrac{v+gt}{u}}}[/tex]
[tex]m_{i}=\dfrac{110}{1- e^-{\frac{105+9.8\times18}{1200}}}[/tex]
[tex]m_{i}=526.2 kg[/tex]
Hence, The initial mass of the rocket is 526.2 kg.
The initial mass of the rocket including the initial fuel is 1257.14 kg.
Conservation of linear momentum
The initial mass of the rocket is determined by applying the principle of conservation of linear momentum as follows;
m₁v₁ = m₂v₂
105m = 1200 x 110
105 m = 132,000
m = 132,000/105
m = 1257.14 kg
Thus, the initial mass of the rocket including the initial fuel is 1257.14 kg.
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In a performance test, each of two cars takes 9.0 s to accelerate from rest to 28 m/s. Car A has a mass of 1379 kg, and car B has a mass of 1902 kg. Find the net average force that acts on (a) car A and (b) car B during the test.
Explanation:
It is given that,
Mass of car A, [tex]m_A=1379\ kg[/tex]
Mass of car B, [tex]m_B=1902\ kg[/tex]
In a performance test, each of two cars takes 9.0 s to accelerate from rest to 28 m/s. Their acceleration is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{28\ m/s}{9}[/tex]
a = 3.12 m/s²
(a) The net force of car A, [tex]F_A=m_Aa[/tex]
[tex]F_A=1379\ kg\times 3.12\ m/s^2[/tex]
[tex]F_A=4302.48\ N[/tex]
(b) The net force of car B, [tex]F_B=m_Ba[/tex]
[tex]F_B=1902\ kg\times 3.12\ m/s^2[/tex]
[tex]F_A=5934.24\ N[/tex]
Hence, this is the required solution.
The charge on a capacitor increases by 20 μC when the voltage across it increases from 84 V to 121 V. What is the capacitance of the capacitor?
The capacitance of a capacitor when the charge increases by 20 µC and the voltage increases from 84 V to 121 V is approximately 0.541 µF.
The question asks to determine the capacitance of a capacitor. To find the capacitance when the charge (ΔQ) increases by 20 µC as the voltage increases from 84 V to 121 V, we can use the capacitor charge formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. The change in voltage (ΔV) can be calculated as 121 V - 84 V = 37 V. The capacitance (C) can then be found using the change in charge and the change in voltage:
C = ΔQ / ΔV
Plugging in the values, we get:
C = 20 µC / 37 V ≈ 0.541 µF
Therefore, the capacitance of the capacitor is approximately 0.541 µF.
An EM wave has a wavelength A 500 nm and a peak electric field Eo 100 i. The wave propagates along the z axis. a) What is the magnitude and direction of the magnetic field? b) What is the frequency f? c) What is the angular frequency w? d) What is the wavenmber k?
Answer:
(a) 3.33 x 10^-7 T along Y axis
(b) 6 x 10^14 Hz
(c) 3.768 x 10^15 rad/s
(d) 2 x 10^6 m^-1
Explanation:
Wavelength = 500 nm = 500 x 10^-9 m
Electric Field, E = 100 i (along + X axis)
Wave is in +Z axis
So, the magnetic field is in + Y axis.
(a) c = E / B
B = E / c = 100 / (3 x 10^8) = 3.33 x 10^-7 T along Y axis
(b) Frequency = wave speed / wavelength = ( 3 x 10^8) / (500 x 10^-9)
frequency = 6 x 10^14 Hz
(c) w = 2 x 3.14 x f = 2 x 3.14 x 6 x 10^14 = 3.768 x 10^15 rad/s
(d) k = reciprocal of wavelength = 1 / (500 x 10^-9) = 2 x 10^6 m^-1
It is noted that a sample of 134La with a half-life of 6.5 minutes has an activity of 2.6 Ci. What was its activity 72 minutes ago?
Answer:
36//
Explanation:
(½)^t
(½)^72
72/2=36//
A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N due west, while the wind exerts a force of 800 N due east and the water exerts a resistive force of 1200 N due east. What are the magnitude and direction of the boat’s acceleration?
Answer:
The boat's acceleration are: a= 0.3 m/s² in west direction.
Explanation:
Fm= 4100 N
Fwi= 800 N
Fwa= 1200 N
Fm - Fwi - Fwa= m *a
clearing a:
a= 0.3 m/s²
A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted through a power line of total resistance 1.0 Ω. Find the percentage power lost in the transmission line if the voltage is not stepped up.
Without stepping up the voltage, 50% of the power is lost in the transmission due to the high I^2R losses. Stepping up the voltage using a transformer significantly reduces these losses and makes power transmission more effective.
Explanation:The subject question pertains to the percentage power loss in a given transmission line when the voltage is not stepped up. Intuitively, power loss will be considerably greater when voltage is not stepped up due to a relatively high current flowing through the lines that enormously increases the I2R (current squared times the resistance) losses.
Firstly, calculate the initial power, PInitial using the formula P = IV, where I is current and V is voltage. In this case, PInitial = 60 A * 120 V = 7200 W. If the voltage is not stepped up, this power is transmitted at 120 V.
The power dissipated due to resistance (Ploss) can be calculated using the formula P = I2R, where I is current and R is resistance. So: Ploss = (60 A)2 * 1 Ω = 3600 W. The percentage of power loss is thus (Ploss / PInitial) * 100 = (3600 W / 7200 W) * 100 = 50%.
In contrast, when voltage is stepped up to 4500 V using a transformer, the current decreases to maintain the same power, greatly reducing power losses. This is eminently beneficial over long distances, making power transmission more efficient.
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Final answer:
the percentage power lost in the transmission line if the voltage is not stepped up is 50%
Explanation:
To calculate the percentage power lost in the transmission line if the voltage is not stepped up, we first calculate the power loss in the transmission line using the formula P = I²R, where I is the current and R is the resistance of the transmission line.
We are given that the current I is 60 A and the total resistance R is 1.0 Ω. The power loss in the transmission line would be:
P = (60 A)² × 1.0 Ω = 3600 W or 3.6 kW
The total power transmitted without stepping up the voltage is calculated as P = IV, where V is the voltage. Here, V is 120 V, so the total power transmitted is:
P = 60 A × 120 V = 7200 W or 7.2 kW
To find the percentage power lost, we divide the power lost by the total power transmitted and multiply by 100:
Percentage power lost = (3600 W / 7200 W) × 100% = 50%
Without the step-up transformer, 50% of the power would be lost in the transmission line, which is a significant loss compared to transmission at a higher voltage. This example illustrates the importance of high-voltage transmission in reducing power losses.
What is the wavelength of an ultrasound signal having a velocity of 1,500 m/sec. at 2.5 MHz?
Answer:
The wavelenght is λ= 6 * 10⁻⁴m.
Explanation:
v= 1500 m/s
f= 2.5 MHz = 2.5 *10⁶ Hz
λ=v/f
λ= 6 * 10⁻⁴ meters
A 160-kg object and a 460-kg object are separated by 3.40 m.(a) Find the magnitude of the net gravitational force exerted by these objects on a 33.0-kg object placed midway between them.
Answer:
The net gravitational force exerted by these objects is [tex]5.712\times10^{-8}\ N[/tex]
Explanation:
Given that,
Mass of another object m = 33.0 kg
Mass M₁ = 160 kg
Mass M₂ = 460 kg
Distance = 3.40 m
We need to calculate the net gravitational force
Using formula,
[tex]F=\dfrac{GM_{2}m}{r^2}-\dfrac{GM_{1}m}{r^2}[/tex]
[tex]F=Gm(\dfrac{M_{2}-M_{1}}{r^2})[/tex]
Where, G = gravitational constant
m = mass of another object
[tex]M_{1}[/tex]= mass of first object
[tex]M_{2}[/tex]= mass of second object
r = distance
Put the value into the formula
[tex]F=6.67\times10^{-11}\times33.0\times\dfrac{460-160}{(3.40)^2}[/tex]
[tex]F=5.712\times10^{-8}\ N[/tex]
Hence, The net gravitational force exerted by these objects is [tex]5.712\times10^{-8}\ N[/tex]
Araised flower bed requires 13 cubic feet of garden soil. You can get a good deal by importing the soil, but it is sold from the dealer in cubic meters. To the nearest hundredth, how many cubic meters of soil is needed? There are 3.28 feet in 1 meter.
Answer:
0.368 m^3
Explanation:
3.28 feet = 1 m
Cube on both the sides
35.29 feet^3 = 1 m^3
So, 13 cubic feet = 13 / 35.29 m^3
= 0.368 m^3
The heat of combustion of benzene, C6H6, is –41.74 kJ/g. Combustion of 4.50 g of benzene causes a temperature rise of 4.50°C in a certain bomb calorimeter. What is the heat capacity of this bomb calorimeter?
Answer:
Heat capacity this bomb calorimeter = [tex] 41.74 KJ/^oC [/tex]
Explanation:
Given data we have in the question:
The heat of combustion for the benzene, [tex]Q_{C_6H_6} = - 41.74 KJ/g[/tex] (Here the negative sign depicts the release of heat and only magnitude is considered fro the heat capacity)
The mass of benzene, [tex]M_{C_6H_6} = 4.50g[/tex]
The change in temperature due to the combustion, [tex]\Delta T = 4.50^oC[/tex]
Now, the formula for the heat capacity is given as:
Heat capacity (C) = [tex]\frac{Q_{C_6H_6}\times M_{C_6H_6}}{\Delta T }[/tex]
or
C = [tex]\frac{ 41.74 KJ/g\times 4.50g}{4.50^oC }[/tex]
or
C = [tex] 41.74 KJ/^oC [/tex]
The heat capacity of the bomb calorimeter, using the given values and the formula: (heat = heat capacity * change in temperature), is 41.74 kJ/°C.
Explanation:In bomb calorimetry, the heat capacity of the bomb calorimeter, or the calorimeter constant (C), can be calculated by using the formula: q = C * change in temperature. The heat (q) released by burning the benzene is the product of the mass burned and the heat of combustion, which is -41.74 kJ/g * 4.5 g, giving us -187.83 kJ. Since a negative q indicates that heat is released, we're actually dealing with 187.83 kJ of heat. The heat capacity (C) of the calorimeter can now be calculated by rearranging the formula to C = q / delta T = 187.83 kJ / 4.5°C = 41.74 kJ/°C. This means the heat capacity of the bomb calorimeter is 41.74 kJ/°C.
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A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the horizontal. What is the horizontal distance traveled when the stone returns to its original height? Ignore air resistance. Check your calculator deg/rad mode.
Answer:
The horizontal distance traveled when the stone returns to its original height = 220.81 m
Explanation:
Considering vertical motion of catapult:-
At maximum height,
Initial velocity, u = 50 sin30 = 25 m/s
Acceleration , a = -9.81 m/s²
Final velocity, v = 0 m/s
We have equation of motion v = u + at
Substituting
v = u + at
0 = 25 - 9.81 x t
t = 2.55 s
Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s
Considering horizontal motion of catapult:-
Initial velocity, u = 50 cos30 = 43.30 m/s
Acceleration , a = 0 m/s²
Time, t = 5.10 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
s = 43.30 x 5.10 + 0.5 x 0 x 5.10²
s = 220.81 m
The horizontal distance traveled when the stone returns to its original height = 220.81 m
A charge alters the space around it. What is this alteration of space called? Electric ether Electric Force Electric field Charged space
Explanation:
A charge alters the space around it. This alteration of space is called the electric field. It is also defined as the electric force acting on a charged particle per unit test charge. It is given by :
[tex]E=\dfrac{F}{q}[/tex]
Where
F is the electric force, [tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
The direction of electric field is in the direction of electric force. For a positive charge, the direction of electric field lines are outwards and for a negative charge, the direction of field lines are inwards.
Hence, the correct option is (c) "electric field".