We will have the following:
The first law of Thermodynamics:
Energy cannot be created or destroyed, it can only be transformed.
An example is when a car suddenly breaks in order to avoid hitting a person, here the velocity of the car, all the energy that is acumulated is transformed into thermal energy and it disipated by the friction of the breaks and the tires. The total energy of the system will remain the same, but it will change the way the energy is present. This as per our understanding of the universe so far.
Alnico is _____.an alloy of metals with strong magnetic propertiesa brittle mixture of substances containing ferromagnetic elementsany material containing ironan element found in nature that behaves like a magnet
Alnico is an alloy made of iron combined with other metals, aluminum, nickel, and cobalt.
The alnico is a permanent magnet
A parallel plate capacitor is charged by connecting it to a battery. After reaching steady state, the electric energy stored in the capacitor is UE . Select the correct claim about the amount of work needed to charge the capacitor.1) The amount of work needed to charge the capacitor is UE, because the work done is equal to the final potential energy of the system.2) No work is needed to charge the capacitor, because the charges will naturally move to their equilibrium position.3) The amount of work needed to charge the capacitor is 2UE , because it takes UE of work to remove the charges from the battery and and additional UE to place the charges on the capacitor.4) The amount of work needed to charge the capacitor is UE , because integrating the equation W=∫q dV yields the equation for the energy stored on a capacitor, Ue= 1/2qV
4. IS correct, that is the for mula used to calculate the potential energy stored
1. Is more or lees correct, but is less explained than the 4
2. Is false, is needed some energy
3. Is missunderstand, cause you are transfering energy from the battery to the capacitor, you dont have to count it twice
the spaceship is flying through space far from planets and stars with the engines firing.
The astronaut shuts the engines off.
The spaceship will….
a. stop moving immediately
b. slow down gradually and stop
c. continue with whatever speed it had when the engines were cut off
d. speed up for just a little while, then slow down
Which of these is a property of an electromagnetic wave? A)magnetic and electric fields oscillate perpendicular to each other but not to the velocity of the wave B)transports energy C)has a magnetic wave but no electric wave
Electromagnetic waves :
- are transverse waves
- Can travel through a vacuum
- Tranport energy from one place to another
- can be reflected
-can be refracted
Correct options:
B)transports energy
Be the action of a force of 51N, a spring measures 39cm. Its length becomes 40.8 cm when subjected to another force of 61N. 1)Determine the empty length of the spring 2)Determine an elongation which will correspond to a force of 32N.3) So what is its length
Answer:
1) 29.82 cm
2) 5.76 cm
3) 35.58 cm
Explanation:
Part 1)
The force of a spring is equal to:
F = kΔx
Where k is the constant of the spring and Δx is the elongation. Δx = xf - xi, where xf is the length of the spring when the force is applied and xi is the empty length. Then
F = k(xf - xi)
Now, by the action of a force of 51N, a spring measures 39 cm, so
51 = k(39 - xi)
And by the action of a force of 61N, the spring length is 40.8 cm, so
61 = k(40.8 - xi)
To find the empty length, we need to solve the system of equations
51 = k(39 - xi)
61 = k(40.8 - xi)
First, solve the first equation for k
[tex]k=\frac{51}{39-x_i}[/tex]Then, replace this on the second equation and solve for xi
[tex]\begin{gathered} 61=k(40.8-x_i) \\ 61=\frac{51}{(39-x_i)}(40.8-x_i) \\ 61(39-x_i)=51(40.8-x_i) \\ 61(39)-61(x_i)=51(40.8)-51(x_i) \\ 2379-61x_i=2080.8-51x_i \\ 2379-2080.8=61x_i-51x_i \\ 298.2=10x_i \\ \frac{298.2}{10}=x_i \\ 29.82=x_i \end{gathered}[/tex]Therefore, the empty length of the spring is 29.82 cm
Part 2)
Now, we need to calculate the value of k, so replacing xi = 29.82, we get:
[tex]k=\frac{51}{39-29.82}=5.556[/tex]Therefore, the equation for the force is
F = 5.556Δx
Solving for Δx, we get:
Δx = F/5.556
Replacing the force by 32N, we can calculate the elongation as
Δx = 32/5.556 = 5.76 cm
Part 3)
Then, the length can be calculated by solving the following equation for xf
Δx = xf - xi
xf = Δx + xi
Replacing Δx = 5.76 cm and xi = 29.82 cm, we get:
xf = 5.76 cm + 29.82 cm
xf = 35.58 cm
So, its length is 35.58 cm
Therefore, the answers are
1) 29.82 cm
2) 5.76 cm
3) 35.58 cm
It is not possible to derive an equation of motion for uniform acceleration without a time variable. Is this true or false?
The third equation of motion is given as,
[tex]v^2=u^2+2as[/tex]Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is displacement.
The expression for the acceleration is given as,
[tex]a=\frac{v^2-u^2}{2s}[/tex]In the above expression, the acceleration of the body is not a function of the time variable. Therefore, it is possible to derive an equation of motion for uniform acceleration without a time variable. Hence, the given statement is false.
A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0cm. If a timer is started when it's displacement is a maximum (hence x=4cm when t=0), what is the speed of the mass when t=3s ?
Since the frequency is 4 Hz, it completes one cycle in 1/4 seconds (its period, T, is 1/4 sec).
f = 1/T
Let's draw a graph of the displacement of the mass along time.
This pattern repeats, with each crest (high point) at every 1/4 second. This means that at t = 3, which is a multiple of 1/4, the mass is at its high point.
Notice that when the spring reaches its high or low point, it changes direction.
At this instant when it changes direction, its speed momentarily becomes 0.
That means that each multiple of 1/8 seconds, the speed is momentarily 0.
The speed of the mass is 0 at t = 3.
What is the volume of a piece of iron ( = 7.9 g/cm3) that has a mass of 0.75 kg? (Enter your answer in cm3.) answer in: cm3
We will have the following:
[tex]\begin{gathered} \frac{7.9g}{cm^3}\ast\frac{1kg}{1000g}=\frac{0.75kg}{V}\Rightarrow0.0079kg/cm^3=\frac{0.75kg}{V} \\ \\ \Rightarrow V=\frac{0.75kg}{0.0079kg/cm^3}\Rightarrow V=\frac{7500}{79}cm^3 \\ \\ \Rightarrow V\approx94.9cm^3 \end{gathered}[/tex]So, the volume is 7500/79 cm^3, that is approximately 94.9 cm^3.
A pottery wheel with rotational inertia 40 kgm^2 rotates at 10 rev/s. 4 kg of clay is dropped onto the wheel 1.2 m from the axis. What angular speed will the wheel have after this?1. 55 rad/s2. 8.7 rad/s3. 70 rad/s4. 0 rad/s
Given:
• Rotational inertia = 40 kg.m²
,• Initial angula speed = 10 rev/s
,• Mass, m = 4 kg
,• Diameter, d = 1.2 m
Let's find the angular speed of the wheel.
To find the angular speed, apply the formula:
[tex]L_i=(I+md^2)*w_f[/tex]Where wf is the final angular speed
I is the rotational inertia
m is the mass
d = 1.2
Li is the angular momentum.
To find the angular momentum, we have:
[tex]\begin{gathered} L_i=40*10*2\pi \\ L_i=2513.27\text{ kg.m}^2\text{ rad/s} \end{gathered}[/tex]Now, to find the final angular speed, wf, plug in values in the first equation and solve for wf:
[tex]\begin{gathered} Li=(I+md^2)w_f \\ \\ 2513.27=(40+4*1.2^2)w_f \\ \\ 2513.27=45.76w_f \\ \\ w_f=\frac{2513.27}{45.76} \\ \\ w_f=54.9\approx55\text{ rad/s} \end{gathered}[/tex]Therefore, the final angular speed is 55 rad/s.
ANSWER:
1.) 55 rad/s
When an object falls through air, there is a drag force that depends on the product
of the surface area of the object, and the square of its velocity, i.e. F=CAv2
, where C
is a constant, determine the dimension of C and its SI unit
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
What is dimension and units ?
Despite the fact that the solution to any engineering problem must incorporate units, dimensions and units are frequently misunderstood. While units are arbitrary names that are correlated to certain dimensions to make it relative, dimensions are tangible quantities that can be measured (e.g., a dimension is length, whereas a metre is a relative unit that describes length). Through a conversion factor, all units for the same dimension are related to one another.
given in question drag force depends upon the product of the cross sectional area of the object and the square of its velocity
and drag force can be given by
F = CAv^2 ----------- (1)
It is given that
Dimension of mass = [M]
Dimension of length = [L]
Dimension of time = [T]
So, by using above dimension we can write
the dimension of force,
F = [ MLT^-2]
dimension of cross-section area,
A = [L^2]
and dimension of velocity
v = [LT^-1]
now, by putting these values in equation (1), we will get
F = CAv^2
[ MLT^-2] = C[L^2][LT^-1]^2
C = [ML^-3T^0]
C = [ML^-3]
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
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Hence, the dimension of constant C will be,
What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N
We will have the following:
[tex]\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}[/tex]So, the force is approximately 1.85*10^-6 N.
As a torque activity, your Physics TA sets up the arrangement decribed below. A uniform rod of mass mr = 158 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the r1 = 10.0 cm and r2 = 90.0 cm mark, passed over pulleys, and masses of m1 = 281 g and m2 = 177 g are attached. Your TA asks you to determine the following: (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). r3 = Fp = F = (b) Let's now remove the mass m3 and determine the new mass m4 you would need to suspend from the rod at the position r4 = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 = Fp = F = (c) Let's now remove the mass m4 and determine the mass m5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 = r5 =
a) Recall, the net torque on the rod must be zero. Thus,
Σt = 0
where
t represents torque
Thus,
t1 + t2 - tr - t3 = 0
t = rF
where
F = force
r = distance
r1F1 + r2F2 - rrFr - r3F3 = 0
r3F3 = r1F1 + r2F2 - rrFr
r3 = (r1F1 + r2F2 - rrFr)/F3
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
Thus,
r3 = (r1m1g + r2m2g - rrmrg)/m3g
g cancels out
r3 = (r1m1 + r2m2 - rrmr)/m3
From the information given,
r1 = 10 cm = 10/100 = 0.1 m
r2 = 90 cm = 90/100 = 0.9 m
rr = 100/2 = 50 cm = 50/100 = 0.5 m
m1 = 281 g = 281/1000 = 0.281 kg
m2 = 177g = 0.177 kg
mr = 158g = 0.158 kg
m3 = 200g = 0.2kg
By substituting these values into the equation,
r3 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
r3 = 0.542 m
The force exerted by the pin, Fp = mg
g = 9.8
Fp = (m3 - mr - m1 - m2)g
Fp = (0.2 + 0.158 - 0.281 - 0.177)9.8
Fp = - 0.981
Taking the absolute value,
IFpI = 0.981 N
F = - 90 degrees
b) r1F1 + r2F2 - rrFr - r4F4 = 0
r4F4 = r1F1 + r2F2 - rrFr = 0
F4 = (r1F1 + r2F2 - rrFr)/r4
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
F4 = T4 = m4g
Thus,
m4g = (r1m1g + r2m2g - rrmrg)/r4
m4g = (r1m1 + r2m2 - rrmr)/r4
r4 = 0.2
By substituting these values into the equation,
m4 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
m4 = 0.542 kg
The force exerted by pin is
Fp = (m4 + mr - m1 - m2(g
Fp = (0.542 + 0.158 - 0.281 - 0.177)9.8
Fp = 2.37 N
Fp = 2.37 N
F = 90 degrees
c) When the pin does not exert a force,
Fp = 0
F1 + F2 - Fr = F5
m1 + m2 - mr = m5
m5 = 0.281 + 0.177 - 0.158
m5 = 0.3 kg
Since the net torque on the rod is zero,
t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
r5 = r1F1 + r2F2 - ffFr)/F5
r5 = (r1m1 + r2m2 - rrmr)/m5
r5 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.3
r5 = 0.36
A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?
Given:
The mass of the first ball is,
[tex]m_1=4\text{ kg}[/tex]The initial velocity of the first ball towards West is,
[tex]u_1=25\text{ m/s}[/tex]The mass of thr second ball is,
[tex]m_2=15\text{ kg}[/tex]the second object is initially at rest.
The final velocity of the first ball is,
[tex]v_1=-8.0\text{ m/s}[/tex]we are taking West as positive.
Applying momentum conservation principle we can write,
[tex]m_1u_1+m_2\times0=m_1v_1+m_2v_2[/tex]Substituting the values we get,
[tex]\begin{gathered} 4\times25+0=4\times(-8.0)+15\times v_2 \\ v_2=\frac{100+32}{15} \\ v_2=8.8\text{ m/s} \end{gathered}[/tex]THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.
The impulse of the Second ball is,
[tex]\begin{gathered} I=m_2v_2-m_2\times0 \\ =15\times8.8 \\ =132\text{ kg.m/s} \end{gathered}[/tex]Fill in the blank: 66 in. = yd
Answer: 1.833
Explanation:
Multiply the value in inches by 0.027777777777727 (the conversion factor)
So 66 inches*.027777777777727=1.833 yards
Answer:
66inches = 1.833 yards
Explanation:
Calculate the height of an image in a converging lens with a height of 6.1 cm, an image location of 6.2 cm, and a distance of 3.6 cm for the object's placement from the lens. Final answer will have 2 decimal places and might not follow sig fig rules.
the formula we will use is
1/f = 1/p + 1/q
where f is = focal length
p is the distance away (distance from the lens to the object)
and q is the image distance
What is the electric field amplitude of an electromagnetic wave whose magnetic field amplitude is 7.9 mT ?
Given:
The amplitude of the magnetic field of the electromagnetic wave is,
[tex]B_0=7.9\text{ mT}[/tex]To find:
The amplitude of the electric field
Explanation:
Let, the amplitude of the electric field is
[tex]E_0[/tex]As we know,
[tex]\begin{gathered} \frac{E_0}{B_0}=c \\ c=3\times10^8\text{ m/s} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} \frac{E_0}{7.9\times10^{-3}}=3\times10^8 \\ E_0=3\times10^8\times7.9\times10^{-3} \\ E_0=2.37\times10^6\text{ N/C} \end{gathered}[/tex]Hence, the amplitude of the electric field is,
[tex]2.37\times10^6\text{ N/C}[/tex]Need help with this question Short straight forward answers please :)
We will have the following:
a. The gravitational potential energy will be:
[tex]P_C=(15kg)(9.8m/s^2)(6m)\Rightarrow P_C=882J[/tex]So, the gravitational potential energy of C is 882 J.
b. The velocity of C right before it hits the ground will be:
[tex]\begin{gathered} 882J=\frac{1}{2}(15kg)v^2\Rightarrow\frac{1764J}{15kg}=v^2 \\ \\ \Rightarrow v=\frac{14\sqrt{15}}{5}m/s\Rightarrow v\approx10.84m/s \end{gathered}[/tex]So, the velocity will be approximately 10.84 m/s.
c.
1. We will have that Eg at the initial position will be: B < C
2. Vfinal upon impact with ground: B = C
3. Ek right before hitting he ground: B < C
f.
1. Eg: A > B
2. V final: A > B
3. Ek: A > B
4. V at 2 meters above the ground: A > B
5. Total energy at 2 m above the ground: A > B.
Three people are driving their cars in different directions, in an open field. At one point, while they are all driving, they each measure the other drivers’ velocities. When they compare measurements afterward, they notice that they all got different measurements from each other. Why do their measurements not match?
Answer:
because of their change in momentum
For an object spinning around a central point, what will happen if its distance from the center is decreased
Answer:
Its a acceleration will increase
Explanation:
The force required to keep an object in a circular motion is given by
[tex]F=\frac{mv^2}{R}[/tex]where v is the radial velocity and R is the radius of the object with mass m.
Now our question is what happens to the above equation as we decrease R?
We can see that as R decreases the quantity mv^2 /R increases (since R is getting smaller ).
Hence, we conclude that F increases. But what if F? it is the centripetal force.
Since centripetal force has increased, so has the quantity v^2 /R (called the acceleration ).
Meaning an increase in centripetal force implies an increase n acceleration.
Since in the answer choices we are not given the option to increase our centripetal force, the next best choice is to choose 'acceleration will increase. '
PLEASE HELP
Which is not an accurate statement about Earth's gravitational pull?
A) Earth's gravitational pull helps keep it in orbit.
B) Earth's gravitational pull is the same as Jupiter's gravitational pull.
C) Earth's gravitational pull is 9.8 m/s2.
D) Earth's gravity helps keep people from floating outside of the planet.
Shown here are astronomical objects located at different distances from earth. rank the objects based on their distances from earth, from farthest to nearest.
- star on far side of Andromeda Galaxy
- star on near side of Andromeda Galaxy
- star on far side of Milky Way Galaxy
- star near center of Milky Way Galaxy
- Orion Nebula
- Alpha Centauri
- Pluto- The Sun
The distance of astronomical objects is measure very carefully. These are having a different unit. This is the astronomical unit. The distances are very huge.
The distance between objects in space is vast and very difficult to calculate. These are learned under solar system mathematics. The values for these distances are cumbersome for astronomers and scientists to manipulate. Therefore, scientists use a unit of measurement called an astronomical unit.
Let us understand the distances first.
To know the distance of stars in Andromeda Galaxy, we should first know the distance of Andromeda Galaxy. The distance of Andromeda Galaxy from Earth is 2.5 million light years away. The astronomical unit used is the light years.
Thus, from this we can conclude that,
The star near to Andromeda Galaxy must be at a distance of 2.5 million light years away.The star far side from Andromeda Galaxy will be more than 2.5 million light years away.Now to know about the stars in Milky Way Galaxy, the distance of milky way galaxy from Earth is approximately 9 light years away.
So,
The star far from Milky Way Galaxy should be more than 9 light years away from Earth.The star near to the Milky Way Galaxy should be close to 9 light years away from Earth.Orion Nebula is 1,344 light years away.Alpha Centauri is approximately 4.3 light years away from the Earth.The Pluto is approximately 5 billion km away from the Earth.The Sun is approximately 148 million km away from the Earth.Thus, from this we can conclude that,
The farthest is the star on far side of Andromeda Galaxy, then the star on near side of Andromeda Galaxy, then comes the star on far side of Milky Way Galaxy, then the star near center of Milky Way Galaxy, then it is the Alpha Centauri, then the Orion Nebula and then it is the Pluto with the Sun being the nearest one.
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8. What would the shape of the graph be if the speed of the object decreased from 50.0 km/hr at 20 s to 30 km/hr at 40 s?9. What is the acceleration in Problem 8?
We will have the following:
8. The graph will be the following:
9. The acceleration of part 8 will be determined as follows:
First, we determine the equivalency of the velocities, that is
[tex]\begin{gathered} \frac{50km}{h}\ast\frac{1000m}{1km}\ast\frac{1h}{3600s}=\frac{125}{9}m/s \\ \\ \frac{30km}{h}\ast\frac{1000m}{1km}\ast\frac{1h}{3600s}=\frac{25}{3}m/s \end{gathered}[/tex]Then, the acceleration will be:
[tex]a=\frac{(25/3\text{ }m/s)-(125/9\text{ }m/s)}{40s-20s}\Rightarrow a=-\frac{5}{18}m/s^2[/tex]So, the acceleration was of -5/18 m/s^2.
How to do Projectile Motion?
A thrown ball undergoes projectile motion so throwing a ball in the air is an example of projectile motion.
What is Projectile Motion?Projectile motion is the motion of an object pitched (projected) into the air. After the starting force that launches the object, the only occurrence of the force of gravity in the object is called a projectile motion, and its path is called its trajectory. Projectile motion is a form of motion in which an object or particle ( called a projectile, is thrown near the earth's surface and moves along a curved path under the action of gravity only. Throwing a ball or a cannonball. The motion of a billiard ball on the billiard table. t. The motion of the earth around the un-projectile motion is a special case of two-dimensional motion. A particle in motion at a vertical level with an initial velocity and experiencing a free-fall (downward) acceleration, displays projectile motion.
So we can conclude that Projectile motion is applicable in both throwing and hitting. A thrown ball undergoes projectile motion.
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An object is projected or flung into the air, and only gravity's acceleration affects the object's velocity. A projectile is what it is, and its trajectory is what it took to get there.
What is Projectile motion?An item or particle that is projected toward the surface of the Earth and moves along a curved path only under the influence of gravity is said to be experiencing projectile motion. Galileo demonstrated that this curving path was a parabola, however it can also be a straight line in the unique situation where it is hurled straight up.
Ballistics is the study of such motions, and this trajectory is a ballistic trajectory. Gravity, which works downward and gives the item a downward acceleration toward the Earth's center of mass, is the sole force of mathematical significance that is actively acting on it.
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A 0.95 kg stone attached to a string is whirled in a horizontal circle of radius 38 cm as a conical pendulum. The string makes an angle of 40° with the vertical. (a) Find the speed of the stone. (b) Find the tension in the string.
The vertical component of the tension will be responsible for the weight of the stone while the horizontal component will be responsible for centripetal force.
a.) Speed v = 1.8 m/s
b.) Tension T = 12.2 N
Types of Circular MotionThere are motion of a body in a vertical circle and motion of a body in horizontal circle in which a conical pendulum is a good example.
Given that a 0.95 kg stone attached to a string is whirled in a horizontal circle of radius 38 cm as a conical pendulum. The string makes an angle of 40° with the vertical.
The parameters given are;
mass m = 0.95kgRadius r = 38 cm = 0.38mAngle Ф = 40°Speed v = ?Tension T = ?To find the speed of the stone, we will use the formula
TsinФ = mv²/r ...... (1)
Let us first find the tension by using the formula
TcosФ = mg
Tcos40 = 0.95 × 9.8
Tcos40 = 9.31
T = 9.31/cos40
T = 12.15 N
Substitute T in equation 1 to find the speed v
12.15sin40 = 0.95v²/0.38
2.97 = 0.95v²
v² = 2.97/0.95
v² = 3.125
v = √3.125
v = 1.77 m/s
Therefore, the tension in the string is 12.15 N and the speed of the stone is 1.8 m/s approximately
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The owner of a recycling company wants to reduce his electrical consumption and costs. The electromagnet used in his operation uses 12 A of current, has 7000 loops and a lifting force of 9800 N. If the lifting force needs to remain the same but the owner would like to reduce the current to only 5 A, how many loops would the electromagnet have?
Given:
Current, I = 12 A
Loops, B = 7000
Force, F = 9800 N
Let's determine the loops if the force remains the same but the current redudces to 5A.
Apply the formula:
[tex]F=\frac{I\times N}{L}[/tex]Let's solve for L.
[tex]\begin{gathered} L=\frac{I\times N}{F} \\ \\ L=\frac{12\times7000}{9800} \\ \\ L=8.57\text{ m} \end{gathered}[/tex]If the current reduces to 5 A, we have:
[tex]\begin{gathered} N=\frac{F\times L}{I} \\ \\ \text{Where I = 5 A} \\ \\ N=\frac{9800\times8.57}{5} \\ \\ N=16800\text{ } \end{gathered}[/tex]The number of loops the electromagnet would have is 16800 loops.
If a hammer is dropped from a height of 52 m and there is no air resistance, what is the acceleration the hammer experiences while it is falling towards the ground?
The hammer is fallingtowards the ground under the action of only one force which is gravity.
Thus, the acceleration of the hammer while falling towards the ground is the acceleration due to gravity.
Hence, the acceleration of the hammer in the given case is,
[tex]g=9.81ms^{-2\text{ }}[/tex]Jake did an experiment in his science class. He found two of the same kind of beaker, and he filled them with different liquids. Then, he measured the volume and mass of each. The image below shows the two beakers.Based on the image, which of the following is the best conclusion? A. The liquids have the same volume and different masses. B. The liquids have different volumes and the same mass. C. The liquids have different volumes and different masses. D. The liquids have the same volume and the same mass.
Yellow liquid:
Mass = 470 g
Volume = 450ml
Pink liquid
Mass= 570 g
Volume= 450ml
Volumes are equal (450ml) and masses are different (470g and 570g)
A. The liquids have the same volume and different masses.
One mole of an ideal gas at 1.00 atm and 0.00°C occupies 22.4 L. How many molecules of an ideal gas are in one cm^3 under these conditions?a. 28.9 b. 22 400 c. 2.69 × 1019 d. 6.02 × 1023
Given:
Atm = 1.00 atm
Temperature = 0.00°C
Amount of gas = 1 mole which occupues 22.4 L
Let;s find the number of molecules of an ideal gas are in one cm^3 under these conditions.
We have:
1 mole of ideal gas = 22.4 L
This is called the molar volume of gas.
To find the amount of molecules, apply the avogrado's constant:
Number of molecules in 1 mol = 6.023 x 10²³
Hence, for 1 cm³, we have:
[tex]\text{ No. of molecules in 1 cm}^3=\frac{6.023\times10^{23}}{22.4\times10^3}[/tex]Solving further:
[tex]\text{ No. of molecules in 1 cm}^3=\frac{6.023\times10^{23}}{22.4\times10^3}=2.69\times10^{19}mol/cm^3[/tex]Therefore, the amount of molecules of an ideal gas in one cm^3 under these conditions is:
2.69 x 10¹⁹ mol/cm^3
ANSWER:
C. 2.69 x 10¹⁹
A 2.0 microF capacitor is connected across a 60 Hz voltage source, and a current of 2.0 mA is measured on a VOM. What is the capacitive reactance of the circuit?
Let's write down and name the variables we know.
C: capacitance; C = 2 μF = 2*10^-6 F
f: frequency of voltage source; f = 60 Hz
Xc: capacitive reactance of circuit (we are solving for this)
We also know that ω = 2πf = 120π.
From this information, we can use the following equation:
Xc = 1/(ωC)
And we can solve for Xc.
Xc = 1/(120π*2*10^-6)
Xc = 1326.291 Ω
Using an electric current, you can split liquid water to form two new substances, hydrogen and oxygen gases. Is this a change in state. Explain your answer.
This process (called electrolysis) is not have a change of state. This comes from the fact that the substance (in this case water) does not keep its identity during the change, this means that we don't end up with the same substance; rather we have two new ones, hydrogen and oxygen.