Answer:
Option: e) Glaciers
Explanation:
Historically, the Inland South viewed as a backwater of the United States. The Inland South is known for its Appalachians mountains, swampland. Flatland, sandy soils, and meandering rivers are some of its features. The climate is hot and humid, which allow the area to grow as Pantanal floodplains. Glaciers will be one of the landforms that one will not expect to see in the Inland South.
what will you use to measure the length of a copper wire of 50cm long?
Answer:
BS33
Explanation:
its a cable length meter used for measuring length of all kinds of wires.
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and emerges with a speed of 910 m/s. The target is free to slide on a smooth horizontal surface. What is the targetâs speed just after the bullet emerges?
Answer:
The velocity is [tex]v_t = 0.02175 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m_b = 0.024 \ kg[/tex]
The initial speed of the bullet is [tex]u_b = 1200 \ m/s[/tex]
The mass of the target is [tex]m_t = 320 \ kg[/tex]
The initial velocity of target is [tex]u_t = 0 \ m/s[/tex]
The final velocity of the bullet is is [tex]v_b = 910 \ m/s[/tex]
Generally according to the law of momentum conservation we have that
[tex]m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t[/tex]
=> [tex]0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t[/tex]
=> [tex]v_t = 0.02175 \ m/s[/tex]
Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area shown in figure 2
Answer:
centroid: (x, y) = (81.25 mm, 137.5 mm)I = 8719.31 mm^2 for unit massExplanation:
Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.
The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...
y1 = 0.8x +125
The lower boundary is given in terms of y, but we can solve for y to find ...
100x = y^2
y2 = 10√x
Then our differential of area is ...
dA = (y1 -y2)dx
__
The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.
The area will be ...
[tex]\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx[/tex]
The y-coordinate of the centroid is ...
[tex]\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5[/tex]
Similarly, the x-coordinate is ...
[tex]\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25[/tex]
That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.
__
The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for [tex]\overline{x}[/tex], but multiplies dA by x^2 instead of x.
The attachment shows that value to be ...
I ≈ 8719.31 mm^2 (normalized by area)
The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.
Suppose you place your face in front of a concave mirror. Which one of the following statements is correct? A) If you position yourself between the center of curvature and the focal point of the mirror, you will see an enlarged image of your face. B) No matter where you place yourself, a real image will be formed. C) Your image will always be inverted. D) Your image will be diminished in size. E) None of these statements are true.
Answer:
If you position yourself between the center of curvature and the focal point of the mirror, you will see an enlarged image of your face.
Explanation:
A concave mirror is a curved mirror. The position of an object placed at any point in front of the mirror and the corresponding image formed by the concave mirror are depicted in appropriate ray diagrams.
If I place my face between the centre of curvature and the principal focus of the concave mirror, the image formed will be enlarged, real and inverted.
What is the emf of a cell consisting of a Pb2 / Pb half-cell and a Pt / H / H2 half-cell if [Pb2 ]
Answer:
0.083 V
Explanation:
The equation of the reaction is;
Pb(s) + 2H^+(aq) -------> Pb^2+(aq) + H2(g)
E°cell = E°cathode - E°anode
E°cathode = 0.00 V
E°anode = -0.13 V
E°cell = 0.00-(-0.13)
E°cell = 0.13 V
Q= [Pb^2+] pH2/[H^+]^2
Q= 0.1 × 1/[0.05]^2
Q= 0.1/0.0025
Q= 40
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
Ecell= 0.13 - 0.0592/2 log (40)
Ecell = 0.13 - 0.047
Ecell= 0.083 V
A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm from its equilibrium position, it is observed to have a speed of 0.400 m/s.Find
(a) the total energy of the object at any point in its motion,
(b) the amplitude of the motion, and (c) the maximum speed attained by the object during its motion.
Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²
[tex]A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m[/tex]
(c) the maximum speed attained by the object during its motion
[tex]E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s[/tex]
a. The total energy of the toy at any point in its motion is 0.0416 Joules.
b. The amplitude of the motion is equal to 0.0167 meter.
c. The maximum speed attained by the toy during its motion is 0.577 m/s.
Given the following data:
Mass of toy = 0.250 kgSpring constant = 300 N/mDistance = 0.0120 mSpeed = 0.400 m/sa. To find the total energy of the toy at any point in its motion:
Mathematically, the total energy of an object undergoing simple harmonic motion (SHM) is given by:
[tex]E = \frac{1}{2} MV^2 + \frac{1}{2} kx^2[/tex]
Where:
k is the spring constant.x is the distance.M is the mass of an object.V is the speed of an object.Substituting the given parameters into the formula, we have;
[tex]E = \frac{1}{2} \times 0.25 \times 0.400^2 + \frac{1}{2} \times 300 \times 0.0120^2\\\\E = 0.125 \times 0.16 + 150 \times 0.000144\\\\E=0.02+0.0216[/tex]
E = 0.0416 Joules.
b. To find the amplitude of the motion, we would use this formula:
[tex]A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2 \times 0.0416}{300} } \\\\A = \sqrt{2.77 \times 10^{-4}}[/tex]
A = 0.0167 meter.
c. To find the maximum speed attained by the object during its motion:
[tex]V_{max} = \sqrt{\frac{2E}{M} } \\\\V_{max} = \sqrt{\frac{2 \times 0.0416}{0.25} } \\\\V_{max} = \sqrt{2.77 \times 10^{-4}}[/tex]
Maximum speed = 0.577 m/s
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A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of the rock is 3200kg/m3. The mass of the wire is small enough that its effect on the tension in the wire can be ignored. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 42.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 28.0 Hz. What is the density of the liquid?
Answer:
ρ_liquid = 1,778 10³ kg/m³
Explanation:
In a wire the speed of the wave produced is
v = √T/μ
wave speed is also related to frequency and wavelength
v = λ f
as they indicate that the frequency is the fundamental, there must be a single antinode in the center
L = 2λ
λ = 2L
Using the translational equilibrium equation
T -W = 0
T = W
let's substitute
2L f = √ W /μ
μ = W / 4 L² f²
μ = 164 / (4 3² 42²)
μ = 2.5825 10⁻³ kg / m
this is the linear density of the wire
Now let's analyze when the rock is submerged in a liquid, the thrust acts on the rock
B = ρ g V
when writing the equilibrium equation we have
T + B -W = 0
T = W - B
T = W - ρ_liquid g V (1)
to find the volume of the rock we use the concept of density
ρ_body = M / V
V = M /ρ_body
W = M g
V = W / g ρ_body
V = 164 / (9.8 3200)
V = 0.00523 m³
V = 5.23 10⁻³ m³
The speed of a wave on a string is
v = √ T /μ
speed is also related to wavelength and frequency
v = λ f
indicate that the frequency formed is the fundamental one, therefore it has a single antinode in the center and two nodes at the fixed points
L = λ/ 2
λ= 2L
we substitute and look for tension
2L f = √T /μ
T = 4L² f² μ
T = 4 3² 28² 2.5825 10⁻³
T = 72.888 N
We already have the volume of the rock and the tension on the rope, we can substitute in equation 1 and find the density of the liquid
T= W – ρ_liquid g V
ρ_liquid = (W -T) / gV
ρ_liquid = ( 164 – 72,888) / ( 9,8 5,23 10⁻³)
ρ_liquid = 1,778 10³ kg/m³
Explanation:
When Ina wire the speed of the wave produced isThen v = √T/μAfter that, the wave speed is also related to the frequency and also wavelengthThen v = λ fAlso, as they indicate that the frequency is fundamental, then there must be a single antinode in the centerAfter that L = 2λThen λ = 2LNow we are Using the translational equilibrium equation is T -W = 0 T = WThen let's substitute 2L f = √ W /μ μ = W / 4 L² f² μ = 164 / (4 3² 42²) μ = 2.5825 10⁻³ kg / m this is the linear density of the wireThen Now let's analyze when the rock is submerged in a liquid, the thrust acts on the rock B = ρ g VAfter that when writing the equilibrium equation we have T + B -W = 0 T = W - B T = W - ρ_liquid g V (1)then to find the volume of the rock we use the concept of density is ρ_body = M / V V = M /ρ_body W = M g V = W / g ρ_body V = 164 / (9.8 3200) V = 0.00523 m³ V = 5.23 10⁻³ m³Then The speed of a wave on a string is v = √ T /μWhen the speed is also related to wavelength and frequency v = λ fThen indicate that the frequency formed is the fundamental one, also, therefore, it has a single antinode in the center and also that two nodes at the fixed points L = λ/ 2 λ= 2LThen we substitute and also look for tension 2L f = √T /μ T = 4L² f² μ T = 4 3² 28² 2.5825 10⁻³ T = 72.888 NThen We already have the volume of the rock and also the tension on the rope, we can substitute in equation 1 and also that find the density of the liquid T= W – ρ_liquid g V ρ_liquid = (W -T) / gV ρ_liquid = ( 164 – 72,888) / ( 9,8 5,23 10⁻³) ρ_liquid = 1,778 10³ kg/m³
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A proton is released from rest in a uniform electric field. After the proton has traveled a distance of 10.0 cm, its speed is 1.4 x 10^6 m/s in the positive x direction.
Required:
a. Find the magnitude and direction of the electric field.
b. Find the speed of the proton
Answer:
Explanation:
distance travelled s = 10 cm
speed v = 1.4 x 10⁶ m /s
v² = u² + 2as
u = 0
v² = 2as
( 1.4 x 10⁶ )² = 2 x a x .10
a = 9.8 x 10¹² m /s²
force on proton = mass x acceleration
= 1.67 x 10⁻²⁷ x 9.8 x 10¹²
= 16.366 x 10⁻¹⁵ N .
If magnitude of electric field be E
force on proton
= E x charge on proton
= E x 1.6 x 10⁻¹⁹
E x 1.6 x 10⁻¹⁹ = 16.366 x 10⁻¹⁵
E = 10.22 x 10⁴ N/C
The direction of electric field will be positive x - direction .
b )
Speed of proton = 1.4 x 10⁶ m /s .
A cylindrical container of water has a height of 12 in and opens to atmosphere it squirts horizontally from a hole near the bottom. (Assume for parts a and b that the cross sectional area of this cylinder is much larger than the cross sectional area of the hole).
a) What are the values of the pressure at
1. The surface of water
2. Just outside the hole
3. At the bottom of the container, ignoring the velocity at which the water level moving. I
b) Determine the speed at which the water leaves the hole when the water level is 6 cm above the hole.of
c) Now assume that the ratios of the cross sectional areas of the cylinder to that of the hole is 10 times, calculate the velocity at which the surface of water dropping down, when the water level is 6 cm above the hole.
Answer:
a) 1 & 2The pressure at the surface of water and just outside the hole are both atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}
3. P(bottom) = 2987.04 Pa
b) v = 1.0844 m/s
c) v₁ = 0.11 m/s
Explanation:
a)
1) The pressure at the surface of water is same as the atmospheric pressure { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}
2) The pressure at just outside the hole is also same as the atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}
3) At the bottom of the container, the gauge pressure is given by:
P(bottom) = pgh
where h = 12in = 0.3084m
P(bottom) = 1000 kg/m³ × 9.8 m/s² × 0.3084m
P(bottom) = 2987.04 Pa
b)
The velocity at which the water leaves the hole can be obtained from Bernouilli's equation. Point 1 is at the surface of water, point 2 is just outside the hole;
p₁ + 1/2pv₁² + pgh₁ = p² + 1/2pv₂² + pgh₂
Pressures on both sides are the same and the velocity of the water at the surface is approximately zero (since the hole's cross sectional area is much smaller than the container's). The difference in depths h1 - h2 is just the height of the water level
so
pgh = 1/2pv₂²
v = √(2gh)
we substitute
v = √( 2 × 9.8 m/s² × 0.06 m
v = 1.0844 m/s
c)
Now the velocity of the water level can't be neglected, we can use the continuity equation:
v₂ = (A₁/A²)V₁
so Back to Bernouilli's:
1/2pv₁² + pgh₁ = 1/2 (10v₁)² + pgh₂
phg = 99/2pv₁²
v₁ = √(2/99gh)
v₁ = √(2/99 × 9.8 m/s² × 0.06 m)
v₁ = 0.11 m/s
Where did the gold of the Inca civilization come from?
Answer:
ocean-continental convergence
If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will
Answer:
Therefore the answer is the precision in the speed DECREASES
Explanation:
In quantum mechanics, we have the uncertainty principle that establishes that when the accuracy of the position increases the accuracy the speed decreases, being related by the expression
Δx Δv ≥ h'/ 2
h' = h/2π
Therefore the answer is the precision in the speed DECREASES
A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.80 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is 340 m????s?
Answer:
[tex]5.997m/s[/tex]
Explanation:
We were told to calculate the speed of the ball,
Given speed of sound as 340 m
And we know that the sound of the ball hitting the pins is at 2.80 s after the ball is released from his hands.
Speed of ball = distance traveled/(time of hearing - time the sound travels).
Speed= S/t
Where S= distance traveled
t= time of hearing - time the sound travels
time=time for ball to roll+timefor sound to come back.
time of sound=16.5/340
=0.048529secs
solving for speedof ball
Then,Speed of ball = distance traveled/(time of hearing - time the sound travels).
=16.5/(2.80-0.048529) m/s = 5.997m/s
Therefore, the speed of the ball is
5.997m/s
Question : Is it possible for heat to transfer from T3 to T1 and why?
Answer:
no its a negative
Explanation:
because they both are positive
A hollow, thick-walled, conducting cylinder carries a current of 11.2 A and has an inner radius ri = r and outer radius ro = 3r/2, where r = 4.90 mm. Determine the magnitude of the magnetic field at the following distances from the center of the cylinder.
(a) ra = r/2 T
(b) rb = 5r/4 T
(c) rc = 2r T
Answer:
a. 0 T b. 1.03 × 10⁻⁴ T c. 2.29 × 10⁻⁴ T
Explanation:
a. Using ampere's law ∫B.ds = μ₀i
for ra = r/2, i = 0 (since no current is enclosed) and
∫B.ds = B∫ds = B(2πr/2) = Bπr
So, Bπr = 0
B = 0 T
b. rb = 5r/4
WE find the current enclosed between r = r and r = 5r/4. The total current density in the holow thick-walled conducting cylinder is J = i/[π(3r/2)² - πr²] = i/[9πr²/4 - πr²] = i/8πr²/4 = i/2πr².
Since the current density is constant, we find the current, i' enclosed between r = r and r = 5r/4. J = i'//[π(5r/4)² - πr²] = i'/[25πr²/16 - πr²] = i'/9πr²/16 = 16i'/9πr²
So, i/2πr² = 16i'/9πr²
i' = 9i/32
Using ampere's law ∫B.ds = μ₀i'
∫B.ds = = B∫ds = B × 2π(5r/4) = 5Bπr/2
5Bπr/2 = μ₀(9i/32)
B = 9μ₀i/32πr × 2/5
B = 9μ₀i/80πr
Substituting r = 4.90 mm = 4.90 × 10⁻³ m and i = 11.2 A, we have
B = 9μ₀i/80πr
= 9 × 4π × 10⁻⁷ H/m × 11.2 A/80π(4.90 × 10⁻³ m)
= 403.2/392 × 10⁻⁴ T
= 1.029 × 10⁻⁴ T
≅ 1.03 × 10⁻⁴ T
c. rc = 2r
Using ampere's law ∫B.ds = μ₀i
∫B.ds = = B∫ds = B × 2π(2r) = 4Bπr
4Bπr = μ₀i (since i = current enclosed = 11.2 A)
B = μ₀i/4πr
= 4π × 10⁻⁷ H/m × 11.2 A/4π(4.90 × 10⁻³ m)
= 2.286 × 10⁻⁴ T
≅ 2.29 × 10⁻⁴ T
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side of firecracker A. You see two flashes of light, from the two explosions, at exactly the same instant of time. Define event 1 to be firecracker A explodes and event 2 to be firecracker B explodes. According to your lab partner, based on measurements he or she makes, does event 1 occur before, after, or at the same time as event 2? Explain
Answer:
Explanation:
A and B are 600 m apart . firecrackers explodes and make two waves , sound waves and light waves . they spread out on all sides . The person sitting in the middle , observes two lights simultaneously and after some time hears two sound simultaneously . It is so because waves coming from A and B covers equal distance to reach the middle point . Same will be the case for sound waves .
But for observer sitting on the other side of A , both light and sound wave will not reach simultaneously . Light wave coming from A will reach earlier because distance covered by light will be less . Similarly sound coming from A will reach earlier . So he / she will observe cracker at A to have exploded earlier than that at B .
A thermal conductivity detector is used in all of the GCs in our lab to measure the amount of each component (analyte) in a sample. Which statement best describes how a thermal conductivity detector works when an analyte flows over the wire filaments in the detector
Answer:
The thermal conductivity detector (TCD) is a widespread detector dependent on the estimation of the thermal conductivity of a gas. It measures the distinction in heat conductivity between unadulterated carrier gas and carrier gas containing test parts.
Explanation:
When you have two parallel tubes both containing gas and coils being heated, thermal conductivity occurs. The gases are inspected by comparing the rate of heat loss in the heated coils in the gas. One of the cylinders has a reference gas and the other cylinder contains the sample to be tested. A thermal conductivity detector will sense the change in the thermal conductivity and compares it to a reference stream of carrier gas. An analyte washed from the section causes the reduction of the thermal conductivity of the effluent and produces a detectable signal. The thermal conductivity of most compounds is less than the thermal conductivity of hydrogen and helium.
A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright red fringes spaced 6.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 6.19 mm apart. What is the wavelength of light produced by the pointer?
Answer:
The wavelength is [tex]\lambda_R = 649 *10^{-9}\ m[/tex]
Explanation:
From the question we are told that
The wavelength of the red laser is [tex]\lambda_r = 632.8 \ nm = 632.8 *10^{-9}\ m[/tex]
The spacing between the fringe is [tex]y_r = 6.00\ mm = 6.00*10^{-3} \ m[/tex]
The spacing between the fringe for smaller laser point is [tex]y_R = 6.19 \ mm = 6.19 *10^{-3} \ m[/tex]
Generally the spacing between the fringe is mathematically represented as
[tex]y = \frac{D * \lambda }{d}[/tex]
Here [tex]D[/tex] is the distance to the screen
and d is the distance of the slit separation
Now for both laser red light light and small laser point D and d are same for this experiment
So
[tex]\frac{y_r}{\lambda_r} = \frac{D}{d}[/tex]
=> [tex]\frac{y_r}{\lambda_r} = \frac{y_R}{\lambda_R}[/tex]
Where [tex]\lambda_R[/tex] is the wavelength produced by the small laser pointer
So
[tex]\frac{6.0 *10^{-3}}{ 632.8*10^{-9}} = \frac{ 6.15 *10^{-9}}{\lambda_R}[/tex]
=> [tex]\lambda_R = 649 *10^{-9}\ m[/tex]
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to zero.
If the wire is 2.25 mm in diameter, how much charge moves past a point in the coil during this operation? The resistivity of copper is 1.68Ã10â8Ωâm.
Express your answer to three significant figures and include the appropriate units.
Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
A 90 kg man lying on a surface of negligible friction shoves a 63 g stone away from himself, giving it a speed of 4.4 m/s. What speed does the man acquire as a result
Answer:
the speed acquired by the man is: [tex]0.00308\,\,\frac{m}{s}[/tex]
Explanation:
Use conservation of linear momentum to solve this problem, and make sure you start by converting the 63 g stone mass into kilograms = 0.063 kg.
Now from conservation of linear momentum, the momentum imparted to the stone (which is the product of the stone's mass time its speed) must equal in magnitude that of the 90 kg man. Then we need to find the unknown speed of the man:
[tex]m_1\.v_1=m_2\,v_2\\0.063\,(4.4)\,\frac{kg\,m}{s} =v_2\,(90\,\,kg)\\0.2772\,\frac{kg\,m}{s}=v_2\,(90\,\,kg)\\v_2=\frac{0.2772}{90} \,\frac{m}{s} \\v_2=0.00308\,\,\frac{m}{s}[/tex]
therefore the speed acquired by the man is: [tex]0.00308\,\,\frac{m}{s}[/tex]
blending three primary colors of light
Answer: white light
Explanation:
if you mix red and blue light together, you get magenta and when you mix blue and green light together, you get cyan and when you mix red and green light together, you get yellow but when you mix all three primary colors of light together, you end up with white light or as we see it.
English units of distance include the
Answer:mile
Explanation: heres a hint think aboyt the distance between your house to school
The options are stationary, running, and walking
Answer:
A is walking
B is stationary
C is running
D is walking
E is stationary
A horizontal wire carries a current (conventional) straight toward you. From your point of view, the magnetic field caused by the current in the wire is doing what?
Answer:
The magnetic field circles the wire in a counter-clockwise direction.
Explanation:
This is by using Fleming right hand rule which says If you point your pointer finger in the direction the positive charge is moving, and then your middle finger in the direction of the magnetic field, your thumb points in the direction of the magnetic force pushing on the moving charge.
A force of 200 N is applied on small piston of a pascal press. What would be the
force applied on the big piston, if the diameter of the small piston is 4.37 cm and the area of the big piston is 98 cm2?
Answer:
The force applied on the big piston is 1306.67 N
Explanation:
Given;
force applied on small piston, F₁ = 200 N
diameter of the small piston, d₁ = 4.37 cm
radius of the small piston, r₁ = d₁/2 = 2.185 cm
Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²
Area of the big piston, A₂ = 98 cm²
The pressure of the piston is given by;
[tex]P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}[/tex]
Where;
F₂ is the force on big piston
[tex]F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N[/tex]
Therefore, the force applied on the big piston is 1306.67 N
A uniform disk a uniform hoop and a uniform sphere are released at the same time at the top of an inclined ramp. They all roll without slipping in what order do they reach the bottom of the ramp?
a. disk hoop, sphere
b. sphere, hoop, disk
c. hoop, sphere, disk
d. sphere, disk, hoop
e. hoop, disk, sphere
Answer:
D. The sphere the disk and the hoop
Explanation:
This is because the sphere has inertial of
2/5mR²
Disk 1/2mR²
Hope mR²
So these are moment of inertial which is resistance or opposition to rotation so since the sphere has a smaller moment to inertial it will move faster and reach the ground first then the disk then the hoop in that order
The magnetic flux through a certain coil is given by φm = (1/50π) cos 100πt
where the units are SI. The coil has 100 turns. The magnitude of the induced EMF when t = 1/200 s is:_______.
A. 1/50π V
B. 2/π V
C. 100 V
D. zero
E. 200 V
Answer:
A. 1/50π V
Explanation:
Given;
magnetic flux through the coil, φm = (1/50π) cos 100πt
t = 1/200 s
The magnitude of the induced EMF is given by;
[tex]EMF = \frac{d \phi_m}{dt} \\\\EMF =\frac{d}{dt} (1/50 \pi)cos \ 100\pi t)\\\\EMF = (1/50 \pi)cos \ 100\pi \\\\EMF =(1/50 \pi)cos \ 100 *180\\\\ EMF =(1/50 \pi)cos \ 18000\\\\ EMF =(1/50 \pi) (1)\\\\EMF =(1/50 \pi) \ V[/tex]
Therefore, the magnitude of the induced EMF is 1/50π V
The correct option is "A. 1/50π V"
Select the correct answer.
A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?
OA. 100 kilometers/hour south
OB.
200 kilometers/hour
OC.
200 kilometers/hour north
OD 100 kilometers/hour
A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?
Choose: 100 kilometers/hour south
Answer:
100 kilometers/hour its d
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube length is 25 cm. What is the magnitude of the overall magnification of the microscope?
Answer:
The value is [tex]m \approx 310[/tex]
Explanation:
From the question we are told that
The focal length of the objective is [tex]f_o = 1.0 \ cm[/tex]
The focal length of the eyepiece is [tex]f_e = 2.0 \ cm[/tex]
The tube length is [tex]L = 25 \ cm[/tex]
Generally the magnitude of the overall magnification is mathematically represented as
[tex]m = m_o * m_e[/tex]
Where [tex]m_o[/tex] is the objective magnification which is mathematically represented as
[tex]m_o = \frac{L}{f_o }[/tex]
=> [tex]m_o = \frac{25}{1 }[/tex]
=> [tex]m_o = 25[/tex]
[tex]m_e[/tex] is the eyepiece magnification which is mathematically evaluated as
[tex]m_e = \frac{L }{f_e }[/tex]
[tex]m_e = \frac{25 }{ 2}[/tex]
[tex]m_e = 12.5 \ cm[/tex]
So
[tex]m = 25 * 12.5[/tex]
[tex]m \approx 310[/tex]
Find the angle between two polarizers that will result in one-eighth of the light incident on the second polarizer passing through.
Answer:
The angle is [tex]\theta = 6.93 *10^{1}[/tex]
Explanation:
From the question we are told that
The intensity of light emerging from the second polarizer is [tex]I_2 = \frac{1}{8} * I_1[/tex]
Now the light emerging from the first polarizer is [tex]I_1 = \frac{I_o}{2}[/tex]
Where [tex]I_o[/tex] is the intensity of the unpolarized light
Now according to Malus law the intensity of light emerging from the second polarizer is mathematically represented as
[tex]I_2 = \frac{I_1}{8} = I_1 cos^2(\theta )[/tex]
=> [tex]I_2 = \frac{I_1}{8} = I_1 cos^2(\theta )[/tex]
=> [tex]\theta = cos^{-1}[0.3536][/tex]
=> [tex]\theta = 69.3^o[/tex]
=> [tex]\theta = 6.93 *10^{1}[/tex]
Both the x and y coordinates of a point execute simple harmonic motion. The frequencies are the same but the amplitudes are different. The resulting orbit might be:
Answer:
the orbit resulting an ELIPSE
Explanation:
Harmonic motion is described by the expression
x = A cos (wt +Ф₁)
In this exercise it is established that in the y axis there is also a harmonic movement with the same frequency, so its equation is
y = B cos (wt + Ф₂)
the combined motion of the two bodies can be found using the Pythagorean theorem
R² = x² + y²
R² = [A² cos² (wt + Ф₁) + B² cos² (wt + Ф₂)]
to simplify we can assume that the phase in the two movements are equal
R = √(A² + B²) cos (wt + Ф)
If the two amplitudes are equal we have a circular motion, if the two amplitudes are different in elliptical motion, the amplitudes of the two motions are
circular R² = (A² + A²)
elliptical R² = (A² + B²)
We see from the last expression that the broadly the two axes is different, so the amplitude is an ellipse.
By which the orbit resulting an ELIPSE