Answer:
negatively charged object attract other negatively objects
Explanation:
opposites attract
Answer:
negativelycharged objects attract other negatively charged objects
Explanation:
unlike charges attract like charges repel
Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 232 and 194 newtons, forming an angle of 67
Answer:
408N at 89.89°
Explanation:
This problem requires that we resolve the force vectors into
x- and y
-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.
Finally, we will convert the resultant force into standard form and find the equilibrant.
Resolve into components:
F1x =F1cos 180°= 232(−1)=−232N
F1y=F1sin180°=0N
F2x=F2cos(−140°)=194(−0.766)=−148.6N
F1y=F1sin(−140°)=232(−0.643)=−149.17N
Note the change of the angle used to give the direction of
F2. Standard angles (rotation from thex
-axis; counterclockwise is +) should be used to avoid sign errors in the results.
Now, we add the components:
Fx=F1x+F2x=−380.6N
Fy=F1y+F1y=−148.17N
Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:
F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N
θ=tan−1(−148.17−380.6)
=89.89°
Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?
Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=
Answer:
Electric potential at position, x = -20 m, = -100 V
Electric potential at position, x = 0 m, = 0
Electric potential at position, x = 10 m, = 50 V
Electric potential at position, x = 11 m, = 55 V
Electric potential at position, x = 99 m, 495 V
Explanation:
Given;
magnitude of electric field, E = 5.0 V/m
at position x = 10 m, electric potential = 50 V
Electric potential at position, x = -20 m
V = Ex
V = 5 (-20)
V = -100 V
Electric potential at position, x = 0 m
V = Ex
V = 5(0)
V = 0
Electric potential at position, x = 10 m
V = Ex
V = 5(10)
V = 50 V
Electric potential at position, x = 11 m
V = Ex
V = 5(11)
V = 55 V
Electric potential at position, x = 99 m
V = Ex
V = 5(99)
V = 495 V
Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge, is then placed near #2. Now a copper wire is momentarily connected between #1 and #2 and then removed. Finally, sphere #3 is removed.
In this final state
a) spheres #1 and #2 are still uncharged.
b) sphere #1 carries negative charge and #2 carries positive charge.
c) spheres #1 and #2 both carry positive charge.
d) spheres #1 and #2 both carry negative charge.
e) sphere #1 carries positive charge and #2 carries negative charge
Answer:
sphere #1 carries positive charge and #2 carries negative charge
This is because from the laws of static electricity, disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charged
An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. What is the ratio of the masses X/Y of the asteroids
Explanation:
It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.
As they are in equilibrium, the force of gravity due to each other is same. So,
[tex]\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847[/tex]
So, the ratio of masses X/Y is 0.0847
A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward
Answer:
The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s
Explanation:
Given;
speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s
speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s
Determine the speed of the ball relative to the ground if the ball is thrown forward.
If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.
[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]
Therefore, the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.
have an electrical charge of +1, while
have an electrical charge of -1.
A. Neutrons, electrons
B. Protons, electrons
C. Electrons, neutrons
D. Electrons, protons
Answer:
B
Explanation:
Protons have a positive electrical charge of +1,
Electrons have a negative charge of -1,
Neutrons have a neutral charge of about 0.
A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is
Answer:
0.0081T
Explanation:
The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e
B ∝ I [tex]\frac{N}{L}[/tex]
B = μ₀ I [tex]\frac{N}{L}[/tex] ----------------(i)
Where;
μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²
Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by
L = 2π r
L = 2π (0.042)
L = 0.084π
The number of turns N = 1000
The current in the toroid = 1.7A
Substitute these values into equation (i) to get the magnetic field as follows;
B = 4π x 10⁻⁷ x 1.7 x [tex]\frac{1000}{0.084\pi }[/tex] [cancel out the πs and solve]
B = 0.0081T
The magnetic field along the central radius is 0.0081T
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low-frequency sound is originating from.
(a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 20 cm apart, and the speed of sound generated is 340 m/s. How long (in s) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear?
(b) Assume the same person was scuba diving and a low-frequency sound source was to the right of the scuba diver. How long (in ) is the interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s? S
(c) What is significant about the time interval of the two situations?
Answer:
(a) 0.59 ms
(b) 0.15 ms
(c) The significance is that the speed of sound in different media determines the time interval of perception by the ears, which are at constant distance apart.
Explanation:
(a) distance between ears = 20 cm = 0.2 m
speed of sound generated = 340 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{340}[/tex]
= 5.8824 × [tex]10^{-4}[/tex]
= 0.59 ms
The time interval of the arrival of the sound at the right ear to the left ear is 0.59 ms.
(b) distance between ears = 20 cm = 0.2 m
speed of sound in water = 1530 m/s
time = ?
speed = [tex]\frac{distance covered}{time taken}[/tex]
⇒ time taken, t = [tex]\frac{distance covered}{speed}[/tex]
= [tex]\frac{0.2}{1530}[/tex]
= 1.4815 × [tex]10^{-4}[/tex]
= 0.15 ms
The sound heard by the right ear of the diver would arrive at the left 0.15 ms latter.
(c) The significance is that the speed of sound in different media, determines the time interval of perception by the ears, which are at constant distance apart.
A) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear is; t = 0.588 × 10⁻³ seconds
B) The time interval between when the sound arrives at the right ear and the sound arrives at the left ear if the speed of sound in water is 1,530 m/s is; t = 0.131 × 10⁻⁵ seconds
C) The significance about the time interval of the two situations is that;
Transmission of sound varies with different mediums.
A) We are given;
Distance between the two ears; d = 20 cm = 0.2 m
Speed of sound; v = 340 m/s
Since the sound source is placed at the right ear, the time interval for it to get to the left ear is;
t = d/v
t = 0.2/340
t = 0.588 × 10⁻³ seconds
B) We are now told that the speed of sound in water is 1530 m/s. Thus;
t = 0.2/1530
t = 0.131 × 10⁻⁵ seconds
C) We can see that in answer A and B, the time interval is different even when the distance remained the same. This means that, the time interval of hearing a sound changes with respect to the medium of transmission.
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A bowling ball of mass 5 kg rolls off the edge of a building 20 meters tall. What is the work done by gravity during the fall, in Joules
Answer:
1000j
Explanation:
work done = force x distance
w = 5 x 10 x 20 = 1000joules
A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.
Answer:
The time rate of change of the electric field between the plates is [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 2.8 \ cm = 0.028 \ m[/tex]
The distance of separation is [tex]d = 1.1 \ mm = 0.0011 \ m[/tex]
The current is [tex]I = 5 \ A[/tex]
Generally the electric field generated is mathematically represented as
[tex]E = \frac{q }{ \pi * r^2 \epsilon_o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value
[tex]\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So the time rate of change of the electric field between the plates is mathematically represented as
[tex]\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }[/tex]
But [tex]\frac{q}{t } = I[/tex]
So
[tex]\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }[/tex]
substituting values
[tex]\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }[/tex]
[tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
As more energy from fossil fuels and other fuels is released on Earth, the overall temperature of Earth tends to rise. Discuss how temperature equilibrium explains why Earth’s temperature cannot rise indefinitely.
Answer:
processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Explanation:
The temperature of planet Earth is due to two main types of process, internal and external.
Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect
External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.
These are the two terms that heat the Earth
In addition there are several processes so the planet loses energy,
* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission
* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system
* liquid (water) system that absorbs large amounts of heat to change state and temperature.
These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value
1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tank. Viscous effects are negligible.
Answer:
3.26m
Explanation:
See attached file
Waves from two slits are in phase at the slits and travel to a distant screen to produce the second minimum of the interference pattern. The difference in the distance traveled by the wave is:
Answer:
Three halves of a wavelength I.e 7lambda/2
Explanation:
See attached file pls
Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, and thus the magnetic field it produces. The second coil is connected only to an ammeter. The ammeter will indicate that a current is flowing in the second coil:______.
a. only if the second coil is connected to the power supply by rewiring it to be in series with the first coil.
b. only when the current in the first coil changes.
c. whenever a current flows in the first coil.
d. only when a steady current flows in the first coil.
Answer:
B. only when the current in the first coil changes
Explanation:
This is because for current to be induced in the second coil they must be an change in current inyhe first coil in line with Faraday's first law of electromagnetic induction. Which state that whenever their is a change in magnetic lines of flux an emf is induced
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The moment about the origin O of the force F is Mo
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. The amount of charge that passes per unit time is called
The amount of charge that passes per unit time is called electric current .
Current has dimensions of [Charge] / [Time] .
It's measured and described in units of ' Ampere ' .
1 Ampere means 1 Coulomb of charge passing a point every second.
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit is:
Answer:
Explanation:
The whole wave front may be divided into two halves , the upper half and the lower half . Waves coming from top of the slit or top of upper half and top of lower half or from the mid point of slit can form minima at given point only when there is phase difference of π radian or path difference of λ or one wavelength. Every other point in upper half and corresponding point in lower half will interfere destructively at that point and will form dark spot at the given point . In this way minima will be formed at that point
Hence the phase difference between the Huygens wavelet from the top of the slit and the wavelet from the midpoint of the slit at first minima is π radian .
If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is
Explanation:
F = kx
mg = kx
(20 kg) (10 m/s²) = (1800 N/m) x
x = 0.11 m
A 4.7 kg solid sphere made of metal, whose density is 4000 kg /m3, is suspended by a cord. The density of water is 1000 kg/m3.When the sphere is immersed in water, the tension in the cord is closest to:
Answer:
34.58 N
Explanation:
From the question,
Tension in the cord = Weight of the solid metal- Upthrust
T = W-U................... Equation 1
But,
W = mg
Where m = mass of solid, g = acceleration due to gravity.
Given: m = 4.7 kg, g = 9.8 m/s²
W = 4.7(9.8) = 46.1 N.
U = (Density of water×Volume of water displaced.)gravity
U = D×V×g.
But,
V = Mass of solid/density of solid
V = 4.7/4000
V = 1.175×10⁻³ m³.
Therefore,
U = 1.175×10⁻³(1000)(9.8)
U = 11.52 N
Therefore,
T = 46.1-11.52
T = 34.58 N
A drum rotates around its central axis at an angular velocity of 19.4 rad/s. If the drum then slows at a constant rate of 8.57 rad/s2, (a) how much time does it take and (b) through what angle does it rotate in coming to rest
Answer:
Explanation:
Using equations of motion:
(a)
v=u+at
∴0=19.4−8.57t
∴t=19.4/8.57
=2.3s
B. Using s= ut + 1/2 at²
19.4(2.3)-1/28.57(2.3)²
= 21.92rad
An inductor is connected to a 18 kHz oscillator. The peak current is 70 mA when the rms voltage is 5.4 V What is the value of the inductance L
Answer:
The value of the inductance is 0.955 mH
Explanation:
Given;
frequency of the oscillator, f = 18 kHz = 18,000 Hz
the peak current, I₀ = 70 mA = 0.07 A
the root mean square voltage, [tex]V_{rms}[/tex] = 5.4 V
The root mean square current is given as;
[tex]I_{rms}= \frac{I_o}{\sqrt{2} }[/tex]
[tex]I_{rms} = \frac{0.07}{\sqrt{2} } \\\\I_{rms} = 0.05 \ A[/tex]
Inductive reactance is given by;
[tex]X_L =\frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.05} \\\\X_L = 108 \ ohms[/tex]
Inductance is given by;
[tex]L = \frac{X_L}{2\pi f} \\\\L = \frac{108}{2\pi *18,000} \\\\L = 9.55 *10^{-4} \ H[/tex]
L = 0.955 mH
Therefore, the value of the inductance is 0.955 mH
The value of the inductance (L) for this oscillating circuit is equal to [tex]9.55 \times 10^{-4}[/tex] Henry.
Given the following data:
Oscillator frequency = 18 kHzPeak current = 70 mARms Voltage = 5.4 VTo determine the value of the inductance (L):
First of all, we would find the root mean square (rms) current by using the formula:
[tex]I_{rms} = \frac{I_o}{\sqrt{2} }\\\\I_{rms} = \frac{70 \times 10^{-3}}{1.4142} \\\\I_{rms} = 0.050 \;A[/tex]
Next, we would calculate the inductive reactance of the oscillator by using the formula:
[tex]X_L = \frac{V_{rms}}{I_{rms}} \\\\X_L = \frac{5.4}{0.050} \\\\X_L = 108 \; Ohms[/tex]
Now, we can solve for the value of the inductance (L):
[tex]L = \frac{X_L}{2\pi f}[/tex]
Where:
L is the inductance.f is the frequency.[tex]X_L[/tex] is the inductive reactance.Substituting the parameters into the formula, we have;
[tex]L = \frac{108}{2 \times 3.142 \times 18 \times 10^3} \\\\L = \frac{108}{113112}[/tex]
L = [tex]9.55 \times 10^{-4}[/tex] Henry.
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2. Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the
volume flow rate of the fluid in m^3/s *
Answer:
9.42*10^-4 m^3/s
Explanation:
d=3/100 m
=0.03 m
A=3.14*0.03^2/6
=4.71*10^-4 m^2
Volume flow rate V = A * s
= 4.71*10^-4 * 2
= 9.42*10^-4
So, the volume flw rate of fluid is 9.42*10^-4 m^3/s
Which of the following is true of children with chronic illness? a.) They are all eligible to recievie special education services. b.) Very few such children are enrolled in public schools c.) Their eligibility for soeical education services depends on whether their conditions adversely affect their educational functioning. d.) They represent a large proportion of the children eligible for speical education services.
Answer:
c.) Their eligibility for social education services depends on whether their conditions adversely affect their educational functioning.
Explanation:
Chronic Illness is a human health condition in which a particular (or number of) illness is persistent in the body and the effects on the body are long-lasting and are often resistant to treatment. The word chronic is usually used when the disease/illness/sickness and its effects stay in the body for more than three months.
The likeliest answer from the options given is option C because before social education services are given, it has to be decided if their health condition adversely affects their education.
A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.
Required:
What is the maximum emf induced in the loop?
Answer:
The maximum emf induced in the loop is 0.132 Volts
Explanation:
Given;
radius of the circular loop, r = 9.5 cm
intensity of the wave, I = 0.0295 W/m²
wavelength, λ = 6.40 m
The intensity of the wave is given as;
[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]
where;
B₀ is the amplitude of the field
c is the speed of light = 3 x 10⁸ m/s
μ₀ is permeability of free space = 4π x 10⁻⁵ m/A
[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]
Area of the circular loop;
A = πr²
A = π(0.095)²
A = 0.0284 m²
Frequency of the wave;
f = c / λ
f = (3 x 10⁸) / (6.4)
f = 46875000 Hz
Angular velocity of the wave;
ω = 2πf
ω = 2π(46875000)
ω = 294562500 rad/s
The maximum induced emf is calculated as;
emf = B₀Aω
= (1.572 x 10⁻⁸)(0.0284)(294562500)
= 0.132 Volts
Therefore, the maximum emf induced in the loop is 0.132 Volts
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Answer:
a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.
Explanation:
a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:
[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]
Where:
[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.
[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.
The impact experimented by the ball due to collision is:
[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]
By using the definition of momentum, the expression is therefore expanded:
[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]
Where:
[tex]m[/tex] - Mass of the ball, measured in kilograms.
[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.
If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:
[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]
The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.
b) The magnitudes of initial and final momentums of the ball are, respectively:
[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is:
[tex]\Delta p = p_{f}-p_{o}[/tex]
[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.
c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.
In a nutshell, the right choice is option D.
If d is 10% smaller than D, how do the pressure drops experienced by the fluid across region I and region II compare?
Answer:
ΔP = - 0.262 ρ v₁²
Explanation:
This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
suppose the pipe is horizontal, therefore
y₁ = y₂
P₁ -P₂ = ½ ρ (v₁² -v₂²) 1
Now let's use the continuity equation
v₁ A₁ = v₂ A₂ 2
The area of a circle is
A = π r²
also the radius is half the diameter r = d /2
A₁ = π d₁² / 4
let's substitute in equation 2
v1 π d₁² / 4 = v2 π d₂² / 4
v₁ d₁² = v₂ d₂²
v₂ = v₁ d₁² / d₂²
let's substitute in equation 1
P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )
Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1
d₂ = D -0.1D = 0.9 D
we substitute in the previous equation
P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)
P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )
P1 - P2 = ½ ρ v12 (- 0.524)
ΔP = - 0.262 ρ v₁²
in this solution we assume that the data in zone I is known
A stone is thrown vertically upward with a speed of 21.0 m/s . How fast is it moving when it is at a height of 11.0 m ?
Answer:
15.01 m/s
Explanation:
Applying newtons equation of motion,
v² = u²+2gs.................. Equation 1
Where v = final velocity, u = initial velocity, s = distance, g = acceleration due to gravity.
Given: u = 21 m/s, s = 11 m
Constant: g = -9.8 m/s² ( g is negative because the stone against the force of gravity).
Substitute these values into equation 1
v² = 21²-2(11)(9.8)
v² = 441-215.6
v² = 225.4
v = √225.4
v = 15.01 m/s
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
Express your answer numerically in kilowatts to two significant figures.
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Express the total force numerically, to two significant figures, in units of newtons.
Answer:
1) 14 kW
2) 4.67 x 10^-5 N
Explanation:
Area of solar panel = 10 m^2
Intensity of sun's radiation incident on earth = 1.4 kW/m^2
Solar power absorbed = ?
We know that the intensity of radiation on a given area is
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where I is the intensity of the radiation
P is the power absorbed due to this intensity on a given area
A is the area on which this radiation is incident
From the equation, we have
P = IA
P = 1.4 kW/m^2 x 10 m^2 = 14 kW
b) For a perfect absorbing surface, the radiation pressure is given as
p = I/c
where p is the radiation pressure
I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa
we know that Force = pressure x area
therefore force on the solar panels is
F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
352 m/s
Explanation:
The velocity of sound in air is approximated as:
v ≈ 331 + 0.6 T
where v is the velocity in m/s and T is the temperature in Celsius.
At T = 35:
v = 331 + 0.6 (35)
v = 352 m/s
Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity:_____________.
(a) stays the same.
(b) Increases.
(c) Decreases.
(d) The answer would depend on the path of motion
Answer:
(b) Increases
Explanation:
The potential energy between two point charges is given as;
[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]
Where;
k is the coulomb's constant
q₁ ans q₂ are the two point charges
r is the distance between the two point charges
Since the two charges have opposite sign;
let q₁ be negative and q₂ be positive
Substitute in these charges we will have
[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]
The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.
Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.