Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
Answers
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) \(F_{net} \text {on wire }3=0\)
\(\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm\)
position of wire = 50 - 1.2
= 48.8cm
b) \(F_{net} \text {on wire }1=0\)
\(\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A\)
Direction ⇒ downward
Related Questions
why the ocean near Christchurch is a different temperature than we’d expect for its latitude
Answers
Why the ocean near Christchurch is a different temperature than we'd expect for its latitude (distance from the equator)? Water moving from the equator is warmer than would be expected based on latitude, and so is warmer than the air it passes.
Changes to prevailing winds affect ocean currents. Changes to ocean currents affect how much energy is brought to (or taken away from) a location. In El Niño years, the prevailing winds that normally drive a warm current from the Equator past New Zealand are disrupted and may stop or even reverse.
A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
Answers
. A 5cm tall object is placed perpendicular to the principal axis of a convex lens of focal
length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position
and size of the image. Also find its magnification
Answers
The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.
Given:
Object height (h) = 5 cm
Focal length of the convex lens (f) = 10 cm
Object distance (u) = 15 cm (positive since it's on the same side as the incident light)
To determine the nature, position, and size of the image, we can use the lens formula:
1/f = 1/v - 1/u
Substituting the given values:
1/10 = 1/v - 1/15
To simplify the equation, we find the common denominator:
3v - 2v = 2v/3
Simplifying further:
v = 30 cm
The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.
To find the magnification (M), we use the formula:
M = -v/u
Substituting the values:
M = -30 / 15 = -2
The magnification is -2, indicating that the image is inverted and twice the size of the object.
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A string of length 30cm has one end attached to a fixed point and the other to a mass of 100g which revolves in a horizontal circle 80 times per minute. Calculate (i) the angle of inclination of the string to the vertical, and (ii) the tension T in the string.
Answers
(i) The angle of inclination of the string to the vertical is 65⁰.
(ii) The tension in the string is determined as 2.11 N.
Angular speed of the string
The angular speed of the string is calculated as follows;
ω = 2πN/60
ω = 2π x 80/60
ω = 8.38 rad/s
Linear speed of the stringv = ωr
v = 8.38 x 0.3 m
v = 2.514 m/s
Angle of inclination of the string to the verticaltanθ = v²/rg
tan θ = (2.514²) / (0.3 x 9.8)
tan θ = 2.15
θ = arc tan(2.15)
θ = 65⁰
Tension in the stringT = mv²/r
T = (0.1 x 2.514²) / (0.3)
T = 2.11 N
Thus, the tension in the string is determined as 2.11 N.
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Show that in the presence of a time-independent electric and magnetic fields, the general equation of motion for a non-relativistic particle of mass m and charge q can be written as d^2/dt^3 v = − ω^2c v + q^2/m^2 [ (ExB) + (B.v)B) where ωc =qB/m is the cyclotron frequency.
Answers
d^3/dt^3 v = - ωc^2 v + (q^2/m^2) ((E x B) + (B.v)B) equation describes the motion of a non-relativistic particle in the presence of a time-independent electric and magnetic field.
To derive the equation of motion, we start with the Lorentz force equation:
F = q(E + v x B)
where F is the force acting on the particle, E is the electric field, B is the magnetic field, q is the charge of the particle, and v is its velocity.
Assuming the fields are time-independent, we can differentiate this equation with respect to time to get:
dF/dt = q(dE/dt + (d/dt)(v x B))
Using the vector identity d/dt(v x B) = (d/dt)v x B + v x (d/dt)B, we can simplify this expression to:
dF/dt = q(dE/dt + v x (d/dt)B + (d/dt)v x B)
Next, we use Newton's second law, F = ma, to substitute F with m(dv/dt):
mdv/dt = q(dE/dt + v x (d/dt)B + (d/dt)v x B)
Using the cross product identity a x (b x c) = (a.c)b - (a.b)c, we can simplify the right-hand side of the equation:
mdv/dt = q(dE/dt + (B.v)B - (B.B)v + (E x B))
Dividing both sides by m and taking another derivative with respect to time, we get:
d^2/dt^2 v = (q/m)(dE/dt + (B.v)B + (E x B))
Finally, taking one more derivative with respect to time, we arrive at the desired equation:
d^3/dt^3 v = - ωc^2 v + (q^2/m^2) ((E x B) + (B.v)B)
where ωc = qB/m is the cyclotron frequency.
This equation describes the motion of a non-relativistic particle in the presence of a time-independent electric and magnetic field. The first term on the right-hand side represents the centripetal acceleration due to the magnetic field, while the second term represents the force due to the electric field. The equation shows that the acceleration of the particle is proportional to the cross product of the electric and magnetic fields, as well as the dot product of the velocity and the magnetic field.
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Sort the tiles into the appropriate categories.
releases a lot of energy
Nuclear Fission
Both
Nuclear Fusion
produces a heavier atom
combines two atoms
produces two lighter atoms
splits the nucleus
Done
Answers
1. (NF)- splits the nucleus
2 lighter atoms
(BOTH)- releases lots of energy
2.(NF)- produces a heavy atom
combines 2 atoms
i’ve been up for 20 hours my dritydrops
What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring
Answers
Answer:
The orbital velocity \(v = 16.4 \ km/s\)
The period is \(T = 14.8 \ hours\)
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the planet , this is mathematically represented as
\(\frac{GM_s * m }{r^2 } = m w^2 r\)
=> \(w = \sqrt{ \frac{GM}{r^3} }\)
Here G is the gravitational constant with value \(G = 6.67*10^{-11}\)
\(M_s\) is the mass of with value \(M_s =5.683*10^{26} \ kg\)
r is the is distance from the center of the to the outer edge of the A ring
i.e r = R + D
Here R is the radius of the planet with value \(R = 60300 \ km\)
D is the distance from the equator to the outer edge of the A ring with value \(D = 80000 \ kg\)
So
\(r =80000 + 60300\)
=> \(r =140300 \ km = 1.4*10^{8} \ m\)
So
=> \(w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }\)
=> \(w = 1.175*10^{-4} \ rad/s\)
Generally the orbital velocity is mathematically represented as
\(v = w * r\)
=> \(v = 1.175*10^{-4} * 1.4*10^{8}\)
=> \(v = 1.64*10^{4} \ m /s = 16.4 \ km/s\)
Generally the period is mathematically represented as
\(T = \frac{2 \pi }{w }\)
=> \(T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }\)
=> \(T = 53473 \ second = 14.8 \ hours\)
Answer:
The orbital velocity \(v = 16.4 \ km/s\)
The period is \(T = 14.8 \ hours\)
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as
\(\frac{GM_s * m }{r^2 } = m w^2 r\)
Part 3: Energy Conversions 7. Record your data in the chart and include at least 5 potential-kinetic energy conversions shown in your device's construction. Example Item Description of potential-kinetic energy conversion Example Book The book had gravitational potential energy when it was on the table. Then as the book fell off the table, it was in motion and had kinetic energy. 1 2 3 4 5
Answers
Here are five potential-kinetic energy conversions that could be shown in the construction of a device: Pendulum, Roller Coaster, Wind-up Toy, Elastic Slingshot, Windmill.
Pendulum: A pendulum consists of a weight attached to a string or rod, suspended from a fixed point. When the weight is lifted to a certain height, it possesses gravitational potential energy.
As the weight is released, it swings back and forth, converting the potential energy into kinetic energy. At the highest point of each swing, the weight briefly comes to a stop and has maximum potential energy, which is then converted back to kinetic energy as it swings downward.
Roller Coaster: In a roller coaster, potential-kinetic energy conversions occur throughout the ride. When the coaster is pulled up to the top of the first hill, it gains gravitational potential energy.
As the coaster descends, the potential energy is converted into kinetic energy, resulting in a thrilling and high-speed ride. Subsequent hills and loops continue to convert potential energy into kinetic energy and vice versa as the coaster moves along the track.
Wind-up Toy: Wind-up toys typically have a spring mechanism inside. When the toy is wound up, potential energy is stored in the wound-up spring. As the spring unwinds, it transfers its potential energy into kinetic energy, causing the toy to move or perform actions. The kinetic energy gradually decreases as the spring fully unwinds.
Elastic Slingshot: With an elastic slingshot, potential-kinetic energy conversions are evident when the slingshot is stretched. As the user pulls back on the elastic band, potential energy is stored.
Windmill: Windmills harness the kinetic energy of the wind and convert it into other forms of energy. As the wind blows, it imparts kinetic energy to the blades of the windmill. The rotating blades then transfer this kinetic energy into mechanical energy, which can be used for various purposes such as grinding grains or generating electricity.
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a plastic rod is rubbed with cotton and it is observed that the rod acquires a negative charge. the same cotton is brought near the cap of a positively charged electroscope. state the observation and explain the observation on the leaf of the electroscope
Answers
This observation demonstrates the principle of electrostatics, which is the study of electric charges at rest. It highlights the fact that charged objects can interact with each other, and that opposite charges attract while like charges repel.
When a plastic rod is rubbed with cotton, electrons from the rod are transferred to the cotton, leaving the rod with an excess of positive charges, and the cotton with an excess of negative charges. This results in the rod acquiring a negative charge.
When the same cotton is brought near the cap of a positively charged electroscope, the leaves of the electroscope will diverge. This happens because the positively charged cap attracts the negative charges on the cotton, causing the electrons to move towards the cap. As a result, the leaves of the electroscope acquire a negative charge, and they repel each other due to their like charges.
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What is revolution?
A. The motion of an object along a curved path
B. The spinning of an object on its axis
C. The orbit of a satellite around a central body
D. The motion of two objects around each other
Answers
Answer:
A. the motion of an object along a curved path
a. When throwing a ball vertically upward, my hand moves through a distance of about 1.0 m before the ball leaves my hand. The 0.80 kg ball reaches a maximum height of about 20 m above my hand. while the ball is in my hand after the ball leaves my hand
Answers
The required, it experiences a downward force due to gravity and a force due to air resistance.
What is the projectile motion?Projectile motion is the movement of an entity projected into space. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory.
Here,
When throwing a ball vertically upward, there is a displacement of about 1.0 m from the initial position of the hand to the position where the ball leaves the hand. The mass of the ball is 0.80 kg and it reaches a maximum height of about 20 m above the initial position of the hand. While the ball is in the hand after it leaves, it experiences a downward force due to gravity and a force due to air resistance.
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Which graph represents a function with direct variation? A coordinate plane with a line passing through (negative 4, 0) and (0, negative 2). A coordinate plane with a line passing through (negative 5, 4) and (0, 3).
Answers
Answer:
Answer:
A line passing through the points (-1,-4),(0,0) and (1,4)
Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form y/x=ky/x=k or y=kxy=kx
In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin
Verify each case
Part 1) A line passing through the points (-4,0) and (0,-2)
This line not represent a direct variation, because the line not passes through the origin.
Part 2) A line passing through the points (-5,4) and (0,3)
This line not represent a direct variation, because the line not passes through the origin.
Part 3) A line passing through the points (-4,-6) and (0,3)
This line not represent a direct variation, because the line not passes through the origin.
Part 4) A line passing through the points (-1,-4),(0,0) and (1,4)
The line passes through the origin
Find out the value of k
k=y/xk=y/x
For the point (-1,-4)
substitute
k=-4/-1=4k=−4/−1=4
For the point (1,4)
substitute
k=4/1=4k=4/1=4
The linear equation is y=4xy=4x
This line represent a direct variation
2. A 0.85kg block on a spring is oscillating with an amplitude of
A = 0.42m and a frequency of f = 3.5Hz.
a. What is the spring constant of the spring?
Answers
The spring constant of the spring is 1197.48 N/m.
What is a spring constant?
The spring constant of a spring can be determined from the relationship between the force required to compress or stretch the spring and the displacement of the spring. The force required to compress or stretch a spring is proportional to the displacement and is given by Hooke's law:
f = -kx
where f is the force, x is the displacement, and k is the spring constant.
In the case of a mass on a spring undergoing simple harmonic motion, the force required to compress or stretch the spring is equal to the weight of the mass. Therefore, the spring constant can be determined using the following equation:
\(k = m * (2 * π * f)^2\)
where m is the mass of the block (0.85 kg), f is the frequency of oscillation (3.5 Hz), and π is pi (3.14).
Plugging in the values, we get:
\(k = 0.85 kg * (2 * 3.14 * 3.5 Hz)^2\\k = 0.85 kg * (37.67)^2\\k = 0.85 kg * 1412.57\\k = 1197.48 N/m\)
So, the spring constant of the spring is 1197.48 N/m.
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What is also know When you know both the speed and direction of an objects motion.
Answers
Answer:
When you know both the speed and direction of an object's motion, you know the velocity of the object.
If a quarterback gets hit by a defensive lineman with a mass of 100 kg and accelerating at a rate of 1m/s2 at what force is the quarterback getting hit?
Answers
The quarterback is getting hit with a force of 100 Newtons.
How to calculate the force with which the quarterback is getting hit
We can use Newton's second law of motion:
Force = Mass * Acceleration
Given that the mass of the defensive lineman is 100 kg and the acceleration is 1 m/s², we can substitute these values into the equation:
Force = 100 kg * 1 m/s²
Force = 100 N
Therefore, the quarterback is getting hit with a force of 100 Newtons.
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Object 1 with mass 1=3.25 kg
is held in place on an inclined plane that makes an angle
of 40.0∘
with the horizontal. The coefficient of kinetic friction between the plane and the object is 0.535.
Object 2 with mass 2=4.75 kg
is connected to object 1 with a massless string over a massless, frictionless pulley. The objects are then released.
Calculate the magnitude
of the initial acceleration.
Calculate the magnitude
of the tension in the string once the objects are released.
Answers
The magnitude of the initial acceleration of the object is 4.2 m/s².
The tension in the string once the object starts moving is 13.65 N.
What is the magnitude of the initial acceleration?The magnitude of the initial acceleration of the object is calculated by applying Newton's second law of motion as follows;
F(net) = ma
m₂g - μm₁g cosθ = a(m₁ + m₂)
where;
m₁ and m₂ are the masses of the blocksg is acceleration due to gravityμ is coefficient of frictionθ is the angle of inclinationa is the acceleration(4.75 x 9.8) - (0.535 x 3.25 x 9.8 x cos40) = a(3.25 + 4.75)
33.5 = 8a
a = 33.5/8
a = 4.2 m/s²
The tension in the string once the object starts moving is calculated as;
T = m₁a
T = 3.25 x 4.2
T = 13.65 N
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What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m
Answers
What does the lower scale read? Answer in units of N
Answers
We will have the following:
First, we are given:
*Mass of the breaker: 1.1kg
*Mass of water: 3.3 kg
*Mass of metallic alloy: 4.2kg
*Density of the alloy: 5300kg/m^3
*Density of water: 1000kg/m^3
Now, we find the volume of water displaced by the alloy:
\(V_{\text{w}}=4.2\operatorname{kg}\cdot\frac{m^3}{5300\operatorname{kg}}\Rightarrow V_w=\frac{21}{26500}m^3\Rightarrow V_w\approx7.92\cdot10^{-4}m^3\)Then, from the reading in the hanging scale we will have the force experienced by the alloy due to the upthrust when placed in water, that is:
\(R=mg-\rho Vg\)So:
\(R=(4.2\operatorname{kg})(9.8m/s^2)-(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)\)\(\Rightarrow R=33.39396226\ldots N\Rightarrow R\approx33.4N\)The reading on the lower scale is due to the weight of the water in the breaker and upthrust on the scale:
\(R=g(m_1+m_2)+\rho Vg\)Finally:
\(R=(9.8m/s^2)(1.1\operatorname{kg}+3.3\operatorname{kg})+(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)\)\(\Rightarrow R=50.886003774\ldots N\Rightarrow R\approx50.9N\)So, the readin on the lower scale is approximately 50.9N.
Given the functions f(x)=(1/x-3)+1 and g(x) = (1/1+4)+3
Which statement describes the transformation of the graph of function f onto the graph of function g?
O The graph shifts 2 units right and 7 units down.
O The graph shifts 7 units left and 2 units up.
O
e graph shifts 7 units right and 2 units down.
O The graph shifts 2 units left and 7 units up.
Answers
The statement that describes the transformation of the graph of function f onto the graph of function g is: The graph shifts 2 units right and 7 units down.
To determine the transformation of the graph of function f onto the graph of function g, we compare the two functions f(x) and g(x) and observe the changes in the equations.
The function f(x) = (1/x - 3) + 1 represents a reciprocal function that is shifted vertically 1 unit up and horizontally 3 units to the right. The reciprocal function is reflected about the line y = x.
The function g(x) = (1/(1 + 4)) + 3 simplifies to g(x) = 4 + 3 = 7, which is a constant function representing a horizontal line at y = 7.
By comparing the equations, we can see that the transformation from f(x) to g(x) involves the following changes:
The term 1/x in f(x) is replaced by the constant 1/(1 + 4) in g(x), resulting in a vertical shift of 7 units up.
The term -3 in f(x) is replaced by 3 in g(x), resulting in a vertical shift of 3 units up.
The +1 in f(x) is replaced by +3 in g(x), resulting in an additional vertical shift of 2 units up.
Therefore, the overall transformation is a shift of 2 units to the right and 7 units down.
Hence, the correct statement is: The graph shifts 2 units right and 7 units down.
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why is heat called a form of energy?
Answers
Answer:
Because heat can be converted from one form to another.
Heat is also does work, for example it sets water molecules into motion when boiling ( convection currency ).
since energy is the ability to do work, heat does work.
A rotating heavy wheel is used to store energy as kinetic energy. If it is designed to store 1.00 x 106 J of kinetic energy when rotating at 64 revolutions per second, find the moment of inertia (rotational inertia) of the wheel. (Hint: Start with the expression for rotational kinetic energy.)
Answers
We know, \(1\ rpm = \dfrac{2\pi}{60} \ rad/s\) .
\(64\ rpm\ is = \dfrac{2\pi}{60}\times 64\ rad/s\\\\= \dfrac{32\pi}{15}\ rad/s\)
We know, kinetic energy is given by :
\(K.E = \dfrac{I\omega^2}{2}\\\\I = \dfrac{2(K.E)}{\omega^2}\\\\I = \dfrac{2\times 10^6}{\dfrac{32}{15}\times \pi}\\\\I = 298415.52 \ kg \ m^2\)
Hence, this is the required solution.
In the diagram, q1 = +3.25*10^-9 C.
What is the potential difference when
you go from point B to point A?
q1
A
B
0.150 m
0.250 m
Answers
Answer:
ΔV = 78 V
Explanation:
The potential at a point due to a point charge is given as:
\(V = \frac{kq}{r}\)
where,
V = Potential at the point
k = Colomb's Constant = 9 x 10⁹ Nm²/C²
q = charge = 3.25 x 10⁻⁹ C
r = distance
AT POINT A:
\(V_A = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.25\ x\ 10^{-9}\ C)}{0.15\ m} \\\\V_A = 195\ V\)
AT POINT B:
\(V_B = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.25\ x\ 10^{-9}\ C)}{0.25\ m} \\\\V_B = 117\ V\)
Now, the potential difference will be:
\(\Delta V = V_A-V_B = 195\ V - 117\ V\)
ΔV = 78 V
Answer:
121.7
Explanation:
credit to the comment above^
The first P-wave of an earthquake travels 5600 kilometers from the epicenter and arrives at a seismic station at 10:05 a.m. At what time did this earthquake occur?
Ahhhhhh I have a Regent's test in 2 hours and I don't know how to solve this type of question! Any help would be appreciated.
Anyone know what the steps to do this are? I dont even need an answer, just how to get to it. Thank you!
Answers
The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.
The p-waves travel with a constant velocity of 7 km/s
The time can be calculated by using the formula
t = d / v
where
T1 = 10:05 a.m
d is the distance they take to travel from the epicenter
v is the speed of the p-waves
On average, the speed of p-waves is
v = 7 km/s
d = 5600 km (given)
Substituting the values in the formula;
t = d / v
t = 5600 ÷ 7
t = 800 seconds
Converting into minutes,
t = 800 ÷ 60
t = 13.3
≈ 13 mins
T1 - 13 mins = T2
10:05 - 13 mins = 9.52 am
It means the earthquake occurred prior 13 minutes, that is at 9.52 am.
Therefore, the earthquake occurred at 9.52 am.
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20 examples of scalar quantity
Answers
Answer:
Length
Time
Mass
Temperature
Energy
Direct Current (DC)
Frequency
Volume
Speed
Amount of substance
Luminous Intensity
Density
Concentration
Refractive Index
Work
Pressure
Power
Charge
Electric Potential
Entropy
Edwin Hubble discovered that:
Galaxies were moving away from us
Some stars got brighter and dimmer in patterns.
Triangulation was not a reliable method for measuring distances
All of the above.
Answers
Answer:
galaxies were moving away from us
Explanation:
i got 100% on my quiz
You find an old book and decide to estimate the thickness of one of the sheets or leafs. Neglecting the front and back cover, if all the sheets of book are 3 inches thick, and has 1024 marked pages (numbered 1 to 1024), what is the approximate thickness of the sheet in units of mm.
Answers
Explanation:
There are 1024/2 = 512 SHEETS (each is numbered on two sides)
3 inches * 25.4 mm/inch / 512 sheet = .1488 mm per sheet
1. Which distance is the greatest?
O 7000 meters
O 99.99 meters
O 4.5 x 10³ meters
O 9.0 x 10² meters
Answers
Answer:
7000 meters
Explanation:
obviously 99.99 is less than 7000
10 cubed is 10 × 10 × 10, 1000, × 4.5 is 4500, which is less than 7000
10 squared is 100, which × 9 is only 900, which is also less than 7000
have a good day
A car is heading North at 18 m/s.
The driver flicks the wheel, and within 3 seconds he has the car headed South at 12
m/s.
What is the acceleration?
Answers
Answer:
10 m/s south
Explanation:
From 18 m/s North to 12 m/s South is a total change of 30 m/s towards the South.
So, Change in velocity = 30 m/s South
Time = 3 s
30/3=10 so 10 m/s South
1. A charge of 6.4 C passes through a cross-sectional area or conductor in 2s. How much charge will pass through a cross sectional area of the conductor in 1 min?
Answers
The amount of charge that will pass through the cross-sectional area of the conductor in 1 min is 192 C.
What is the amount of charge?
We can use the formula Q = I * t,
where;
Q is the amount of charge, I is the current, and t is the time.Given that a charge of 6.4 C passes through a cross-sectional area of the conductor in 2 s, we can find the current using the formula:
I = Q / t = 6.4 C / 2 s = 3.2 A
So, the current through the conductor is 3.2 A.
To find the amount of charge that will pass through the cross-sectional area of the conductor in 1 min (60 s), we can use the same formula:
Q = I * t = 3.2 A * 60 s = 192 C
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In the Roman soldier model for refraction, what angle do the soldiers now appearing to be walking at (remember they entered at 45 )?
A. 45
B. Less than 45
C. Greater than 45
D. Not enough info
Answers
Snell's law states:
n₁ * sin(θ₁) = n₂ * sin(θ₂),
where n₁ and n₂ are the refractive indices of the first and second medium, respectively, and θ₁ and θ₂ are the angles of incidence and refraction, respectively.
Without knowing the refractive indices of the two media involved, we cannot determine the angle of refraction (θ₂) and cannot answer the question.In the Roman soldier model for refraction, the soldiers appear to be walking at an angle that is different from their actual direction of motion due to the bending of light at the interface between two media with different refractive indices.
Assuming that the soldiers entered the interface at an angle of 45 degrees, the angle at which they appear to be walking will depend on the refractive indices of the two media and the angle of incidence.
Since we are not given any information about the refractive indices of the media, we cannot determine the angle at which the soldiers appear to be walking. Therefore, the answer is D. Not enough info.