Answer:
are nearby and bright i hope this helps
You desire to observe details of the Statue of Freedom, the sculpture by Thomas Crawford that is the crowning feature of the dome of the United States Capitol in Washington, D.C. For this purpose, you construct a refracting telescope, using as its objective a lens with focal length 86.3 cm. In order to acheive an angular magnification of magnitude 5.01, what focal length fe should the eyepiece have?
Answer:
the focal length of the eyepiece is 17.23 cm
Explanation:
The computation of the focal length of the eyepiece is shown below:
= Focal length of objective lens ÷ angular magnification magnitude
= 86.3 ÷ -5.01
= 17.23 cm
Hence, the focal length of the eyepiece is 17.23 cm
We simply divided the angular magnification magnitude from the focal length of objective lens so that the focal length of the eyepiece could come
Which sentence best explains the law of conservation of mass as applied to
chemical reactions?
A. The amount of mass changes only slightly during a chemical
reaction.
B. The volumes of the reactants and products are equal during a
chemical reaction.
C. The types of atoms can change during a chemical reaction, but
their masses cannot.
D. In a chemical reaction, the total mass of the reactants equals the
total mass of the products.
Answer:
A. The amount of mass changes only slightly during a chemical
reaction.
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How many radians did the disk turn while stopping ? C. how many revolutions?
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
[tex]\omega_f = \omega_i + \alpha t[/tex]
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
[tex]0\ rad/s = 3.14\ rad/s + \alpha(3\ s)[/tex]
α = - 1.047 rad/s²
B.
We can use the second equation of motion to find the angular distance:
[tex]\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2[/tex]
θ = 14.1 rad
C.
θ = (14.1 rad)(1 rev/2π rad)
θ = 2.24 rev
When you are high up in the air you
have
greater potential energy
less potential energy
Answer:
Its Greater potential energy because the air is high up and that makes high energy power
Explanation:
Particles q1 = -53.0 uc, q2 = +105 uc, and
q3 = -88.0 uc are in a line. Particles qı and q2 are
separated by 0.50 m and particles q2 and q3 are
separated by 0.95 m. What is the net force on
particle qı?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-53.0 μC
-88.0 C
+105 με
+92
91
93
K 0.50 m
0.95 m
Enter
no
Answer:
[tex]-180.38\ \text{N}[/tex]
Explanation:
[tex]q_1=-53\ \mu\text{C}[/tex]
[tex]q_2=105\ \mu\text{C}[/tex]
[tex]q_3=-88\ \mu\text{C}[/tex]
r = Distance between the charges
[tex]r_{12}=0.5\ \text{m}[/tex]
[tex]r_{23}=0.95\ \text{m}[/tex]
[tex]r_{13}=1.45\ \text{m}[/tex]
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Net force is given by
[tex]F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}[/tex]
The force on the particle [tex]q_1[/tex] is [tex]-180.38\ \text{N}[/tex].
Answer:
The answer sir would be 180.38
Explanation:
Put in 180.38 trust
d. What is the net force on the bowling ball rolling lane
Answer:
Friction.
Explanation:
form
bonds with each other.
There are many kinds of mixtures. Some mixtures are
chunky like a mixture of peanuts and raisins. These
mixtures are called
I
mixtures.
Answer:
Homogeneous mixtures
Explanation:
I think so because homogeneous means mixed mixtures
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.67 10-6 W/m2 at a distance of 233 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
the distance is 315.3696 m
Explanation:
The computation of the distance is given below:
Given that
Sound intensity = 1.67 × 10^-6 W/m^2
And, the distance = 233 m
Now as we know that
Power = Intensity × surface area
1.67 × 10^-6 × 4π(233)^2 = 1.67 × 10^-6 ÷ 2× 4π × d^2
d^2 = 2 × (223)^2
= √2 × 223
= 315.3696 m
Hence, the distance is 315.3696 m
A 1.00 x 109 kg object is raised vertically at a
constant velocity of 4.00 m/s by a crane. What
is the power output of the crane?
Answer:
P = 3.92 10¹⁰ W
Explanation:
The power is data by the expression
P = W / t
the work of a force is
W = F. y
the bold ones represent vectors. In this case the displacement is vertical upwards and the vertical forces upwards, therefore the angle is zero and the cos 0 = 1
W = F y
we substitute
P = F y / t
P = F v
as the body rises at constant speed the acceleration is zero and from the equilibrium condition
F -W = 0
F = mg
we substitute
P = m g v
let's calculate
P 1.00 10⁹ 9.8 4
P = 3.92 10¹⁰ W
What has to happen for a star to join the main sequence ?
1) Nuclear Fusion
2) Shell Heating
3) Use up most of its available fuel
4) Hydrostatic Equilibrium
Answer:
Nuclear fusion
Explanation:
This is because main sequence of star is powered by fusion of hydrogen to helium atoms together and this process releases energy. The energy released when the gas collapse into a protostar make the center of the protostar to be extremely hot. When the core becomes very hot, nuclear fusion can start.
Question /
Which object is shown below?
A. Convex mirror
B. Convex lens
C. Concave lens
D. Concave mirror
SLIDNAT
6. Applying Explain how scientists see what early galaxies looked like..
Answer: The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.
Explanation:
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm10nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10−5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.
1. What is the magnitude of the electric field between the membranes?
A. 1×10^−15N/C
B. 5×10^−5N/C
C. 1×10^6N/C
D. 9×10^−2N/C
2. What is the magnitude of the force on a Ca++ ion between the cell walls?
A. 4×10^−13N
B. 4×10^−12N
C. 2×10^−12N
D. 2×10^−11N
3. What is the potential difference between the cell walls?
A. 1×10^7V
B. 1×10^−2V
C. 6×10^−3V
D. 10V
4. What is the direction of the electric field between the walls?
A. Toward the outer wall.
B. Parallel to the walls.
C. Toward the inner wall.
D. There is no electric field.
5. If released from the inner wall, what would be the kinetic energy of a 3fC charge at the outer wall? 1fC=10^−15C.
A. 3×10^−14J
B. 3×10^−17J
C. 3×10^−8J
D. 3×10^−2J
Answer:
the correct answue are B, A, C, C, B
Explanation:
1) The electric field is requested, let's approximate the membrane by a parallel plate with surface charge density
E = [tex]\frac{\sigma }{2 \epsilon_o }[/tex]
E = [tex]\frac{ 10^{-5}}{2 \ 8.85 \ 10^{-12}}[/tex]
E = 5.65 10⁵ N / C
the correct answer is B
2) A calcium ion has two positive charges, so the force applied by each side of the membrane (plate)
F = q E
F = 2 1.6 10⁻¹⁹ 5.65 10⁵
F = 1.8 10⁻¹³ N
the total force is the sum of the force of each membrane and the two forces go to the same side
F = total = 2 F
F_total = 3.6 10⁻¹³ N
the correct answer is A
3) the field and the electric potential are related
ΔV = - E s
ΔV = - 5.65 10⁵ 10 10⁻⁹
ΔV = - 5.65 10⁻³ V
the correct answer is C
4) In the exercise they indicate that the outer wall has a positive charge, therefore, as they indicate that we approximate the system to a capacitor, the inner wall must be negatively charged.
The electric field goes from the positive to the negative charge, which is why it goes from the outer wall to the inner wall
the correct answer is C
5) For this part we use conservation of energy
starting point. On the inside wall, brown
Em₀ = U = qV
final point. On the outside
Em_f = K
energy is conserved
Em₀ = Em_f
q V = K
K = 3 10⁻¹⁵ 5.65 10⁻³
K = 1.7 10⁻¹⁷ J
the correct answer is B
Your friend says an appliance uses energy. How would you correct his statement?
Answer:
Not all appliances run on energy. Some of them run on gas. Some both. It just depends on the age of the appliance, the make of the appliance, and the company who made it.
A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two pulses of sound, one that travels through the air and a longitudinal wave that travels through the rail. (The speeds of sound in air and in iron are 343 m/s and 5950 m/s, respectively.)
Required:
Find the separation in time between the arrivals of the two pulses.
Answer:
ΔT = 0.02412 s
Explanation:
We will simply calculate the time for both the waves to travel through rail distance.
FOR THE TRAVELING THROUGH RAIL:
[tex]T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s[/tex]
FOR THE WAVE TRAVELING THROUGH AIR:
[tex]T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s[/tex]
The separation in time between two pulses can now be given as follows:
[tex]\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\[/tex]
ΔT = 0.02412 s
Water has a heat capacity of 4.184 J/g °C. If 50 g of water has a temperature of 30ºC and a piece of hot copper is added to the water causing the temperature to increase to 70ºC. What is the amount of heat absorbed by the water?
The amount of heat absorbed by the water will be 8368 J.
What are heat gain and heat loss?Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.
It is given by the formula as ;
[tex]\rm Q= mcdt[/tex]
The given data in the problem is;
Equilibrium temperature = 30°C.
mass of water = 50 g ,
Temperature change = 70ºC
The specific heat of water =4.184 J//g °C
The amount of heat absorbed by the water is;
[tex]\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J[/tex]
Hence, the amount of heat absorbed by the water will be 8368 J.
To learn more about the heat gain refer to the link;
https://brainly.com/question/26268921
#SPJ2
Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Answer:
The potential difference between the ends of a wire is 60 volts.
Explanation:
It is given that,
Resistance, R = 5 ohms
Charge, q = 720 C
Time, t = 1 min = 60 s
We know that the charge flowing per unit charge is called current in the circuit. It is given by :
I = 12 A
Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :
V = IR
V = 60 Volts
So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.
which process of the method a neutral object obtains an. electrical charge
The gauge pressure in your car tires is 3.00 ✕ 10^5 N/m2 at a temperature of 35.0°C when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to
−42.0°C?
Assume the tires have not gained or lost any air.
Explanation:
Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.
So, we have:
P1/T1=P2/T2
Solving for P2, we have:
P2=(P1*T2)/T1
NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!
Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:
P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]
P2=228930.7156 N/m^2
Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2
Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!
Hope this helps!
Mark as brainlist...
5. What is the period of a vertical mass-spring system that has an amplitude of
71.3 cm and maximum speed of 7.02 m/s? The spring constant is 12.07 N/m.
The period of the vertical mass-spring is 0.64 s.
The given parameters:Amplitude of the spring, A = 71.3 cm Maximum speed of the spring, V = 7.02 m/sSpring constant, k = 12.07 N/mThe angular speed of the vertical mass-spring is calculated as follows;
[tex]V_{max} = A \omega\\\\\omega = \frac{V_{max}}{A} \\\\\omega = \frac{7.02}{0.713} \\\\\omega = 9.85 \ rad/s[/tex]
The period of the vertical mass-spring is calculated as follows;
[tex]f = \frac{\omega }{2\pi} \\\\T = \frac{1}{f} \\\\T = \frac{2 \pi}{\omega } \\\\T = \frac{2\pi }{9.85} \\\\T = 0.64 \ s[/tex]
Thus, the period of the vertical mass-spring is 0.64 s.
Learn more about period of oscillation here: https://brainly.com/question/20070798
PLEASE HELPPPPPP <333
Answer:
Explanation:
The answer is c. I am very sure
Answer:
i think its b
Explanation:
im not very sure
What does the outer part of the disk turn into?
1) Planets and Moons
2) Interstellar Cloud
3) Planetary Nebula
4) It gets sucked into the star
Answer:
what does the outer part of the disk turn into
Explanation:
4) it gets sucked into the star
The Image shows a magnetic field around the poles of a magnet. Identify the areas where the magnetic force is the strongest.
N
Answer:
strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.
Explanation:
The lines of magnetic force are drawn so that the density of lines is proportional to the intensity of the magnetic field.
Therefore, the sections where the magnetic field is strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.
Although the use of absorbances at 450 nm provided you with maximum sensitivity, the absorbances at, say, 400 nm or 500 nm are not zero and could have been used throughout this experiment. Would the same value of K be obtained at one of these wavelengths
Answer:
Yes, the value will be the same.
Explanation:
Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.
During a soccer game, a player grabs and holds an opponent's shirt outside of the penalty box. After the foul is called, what kick is awarded to put the ball back into play?
a
Penalty Kick
b
Indirect Free Kick
c
Kickoff
d
Direct Free Kick
Question below in photo!! Please answer! Will mark BRAINLIEST! ⬇⬇⬇⬇⬇⬇⬇
its wave length
its wave lenght because how its measure
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a coordinate system, unable to move at all times. The second object is initially placed 3.00 cm along the positive x-axis and is free to move. The moment the second object is released at x = 3.00 cm, what is the acceleration of this second object? This experiment is done far away from other massive objects, in outer space.
Answer:
a = 640 m/s²
Explanation:
From work-kinetic energy principles,
The net force acting on the second object is the gravitational force and the electric force due to the first object.
So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²
Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²
The net Force F = ma
So, -F₁ + F₂ = F (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)
-Gm²/r² + kq²/r² = ma
ma = -Gm²/r² + kq²/r²
a = (-Gm²/r² + kq²/r²)/m
a = (-G + kq²/m²)m/r²
Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.
Substituting the variables into the equation, we have
a = (m/r²)(-G + k(q/m)²)]
a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg² ]
a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)
a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²
a = 0.064 × 10⁴N/kg
a = 64 × 10 N/kg)
a = 640 m/s²
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process. Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m3 to 0.026 m3 while doing work on a piston.
This question is incomplete, the complete question is;
The entropy of an ?-ideal gas changes in the following way as a function of temperature and volume:
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.
Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.
1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K
2) For this adiabatic expansion, what is the final temperature? T[tex]_f[/tex] = ? K
Answer:
1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) the final temperature is 158.66 K
Explanation:
Given the data in the question;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
P[tex]_i[/tex] = 100 kPa = 100000 Pa
V[tex]_i[/tex] = 0.01 m³
V[tex]_f[/tex] = 0.026 m³
T[tex]_i[/tex] = 300 K
1) the change in entropy due to the volume change alone
from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
so change in entropy due to the volume change alone is;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])
we know that, from ideal gas law; PV = nRT
so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] ---- let this be equation 1
∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])
we substitute
ΔS = [( 100000 Pa × 0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )
ΔS = 3.185 J/K
Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) Final temperature
we know that, in an adiabatic expansion;
[tex]PV^Y[/tex] = K
where Y = 5/3
so
[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]
[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]
we substitute
[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]
[tex]P_f[/tex] = 20341.255 Pa
Also from ideal gas law;
PV = nRT
T = PV / nR
so
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
but from equation 1; nR = PV/T
so
T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )
T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )
we substitute
T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ × 300 K) / 100000 Pa × 0.01 m³ )
T[tex]_f[/tex] = 158.66 K
Therefore, the final temperature is 158.66 K
Franny drew a diagram to compare images produced by concave and convex lenses.
2 overlapping circles, the left circle labeled Concave lenses and the right circle labeled Convex lenses. An X in the overlap.
Which belongs in the area marked X?
Answer:
Virtual
Explanation:
Answer:
B. Virtual
Good Luck!
The type of brightness in which all
stars being observed are the same
distance from Earth is known as
which type of brightness?
A. absolute brightness.
B. apparent brightness.
C. obvious brightness.
D. compositional brightness.