Answer:
6,000 kg
Explanation:
If the peak wavelength of a star at rest is 615 nm, then what peak wavelength is observed when the star is traveling 2,500,000 m/s toward the Earth.
Answer:
1612903nm
Explanation:
Doppler effect can be regarded as the change in frequency of a wave with respect toobserver that move relative to the wave source
We can expressed as
(λo - λs)/λs = v/s
λo= peak wavelength
λs= peak wavelength observed
C= speed of light
(λo -615×10^9 )/615×10^9 = 2,500,000/(3×10^8)
1.5375=3×10^8λo -184.5
1.5375+184.5=3×10^8λo
186=3×10^8λo
λo=1612903nm
Therefore the peak wavelength that is observed when the star is traveling away from the eart to the velocity given is 1612903nm
Describe why you are doing the experimen?
Answer:
An experiment is a procedure carried out to support, refute, or validate a hypothesis. Experiments provide insight into cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.
Explanation:
Answered: A 4 kg mass is attached to a horizontal spring with the spring constant of 600 N/m and rests on a frictionless surface on the ground. The spring is compressed 0.5 m past its equilibrium. What is the initial energy of the system.
Answer: 75 joules
Can someone help me with my physics
A pendulum is a body that is suspended from a fixed point so that it can swing back and forth through an exchange of kinetic energy and gravitational potential energy. Using 1–2 sentences, explain what happens to the kinetic energy and gravitational potential energy of the pendulum at the highest point and at the lowest point of its swing.
Answer:
Because mechanical energy (the sum of potential and kinetic energy) is conserved, as the kinetic energy increases, the potential energy decreases. The maximum kinetic energy is achieved when the pendulum passes through the lowest point, and the maximum potential energy is achieved at the highest point.
It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s).
Let B = 0.86 T , I = 2300 A , m = 20 kg , and L = 55 cm . For simplicity, assume the net force on the object is equal to the magnetic force, as in parts A and B, even though gravity plays an important role in an actual launch into space.
Express your answer using two significant figures.
Answer:
The distance of the bar D = 1153 km
Explanation:
The electric force is the one that takes place between electric charges.
The electric force with which two point charges are attracted or repelled at rest is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them.
Recall that:
Electrical force(F) = I*B*L
where;
I = the current,
B = the magnetic field strength,
L = the length of the bar
However;
From the second equation of motion,
F = Ma
Since; (F) = I*B*L
Then,
Ma = IBL,
where;
M is the mass;
a is the acceleration
Making the acceleration (a) the subject of the formula, we have
a = IBL/M
Similarly;
From the third equation of motion;
v^2= u^2+2as,
where v and u are the final velocity and the initial velocity respectively
Here u = 0
Also; let distance s = D
Then
v^2 = 2aD
where;
a = IBL/M
Making the distance D the subject of the formula, we get:
D = v^2/2a = v2*M/(2IBL)
D = 11200² × 20/(2×2300×0.86×0.55)
D = 1153047.155 m
D = 1153 km
In the video your blood is compared to a __________________ that delivers oxygen to your body and picks up CO2 to be released out when you breath.
Answer:
delivery truck
Explanation:
because i got it right
Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled eastward at 8 m/s while billiard ball B originally traveled westward at 2 m/s. Calculate the speed and direction of each ball after the collision.
Answer:
lucky mauld mauldgomary was an british poet...
3. Sarah drops a 10.0 kg objed from a 200 m high bridge over a river. a) What work is done by the objec as a result of its fall?
A train is traveling at 55m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed of9 m/s and it takes 49 s to properly slow down, what distance does the train travel while slowing down?
Answer:
x = 1127 [m]
Explanation:
In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.
[tex]v_{f} =v_{o} -a*t[/tex]
where:
Vf = final velocity = 9 [m/s]
Vo = initial velocity = 55 [m/s]
a = acceleration o desacceleration [m/s²]
t = time = 49 [s]
Now replacing:
9 = 55 - a*49
a*49 = 55 + 9
a = 1.306 [m/s²]
Note: The negative sign in the above equation means that the speed decreases.
Now using the second equation.
[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]
(9)² = (55)² - 2*(1.306)*x
2944 = 2.612*x
x = 1127 [m]
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is launched to a maximum height 50.2 cm. How much should the spring be compressed to send the ball twice as high?
We know, by conservation of energy :
[tex]\dfrac{kx^2}{2}=mgh[/tex]
Therefore,
[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}[/tex]
Putting given values, we get :
[tex]\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm[/tex]
Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.
Hence, this is the required solution.
A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How long until the bullet reaches the ground?
0.32 s
0.57 s
0.64 s
0.25 s
how does tom and jerry movie character influence your attitude
Answer:
it makes me wish I was a cartoon
Answer:
goofy and stupid and act like a kid
Explanation:
There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center) and repel each other with an electrostatic force of F1=0.0720 N . A thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed, the spheres still repel, but with a force of F2=0.115 N . The Coulomb force constant is k=1/(4π????0)=8.99×109 N⋅m2/C2 . Using this information, find the initial charge on each sphere, q1 and q2 , if q1 is initially less than q2 .
Answer:
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex]
Explanation:
[tex]F_1=0.072\ \text{N}[/tex]
[tex]F_2=0.115\ \text{N}[/tex]
r = Distance between shells = 40.4 cm
[tex]q_1[/tex] and [tex]q_2[/tex] are the charges
[tex]k[/tex] = Coulomb constant = [tex]8.99\times10^{9}\ \text{Nm}^2/\text{C}^2[/tex]
Force is given by
[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}[/tex]
[tex]F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}[/tex]
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}[/tex]
Substituting the above value of [tex]q_1[/tex] we get
[tex]\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}[/tex]
Since we know [tex]q_1<q_2[/tex]
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex].
Suppose two parallel-plate capacitors have the same charge Q, but the area of capacitor 1 is A and the area of capacitor 2 is 2 A.If the spacing between the plates, d, is the same in both capacitors, and the voltage across capacitor 1 is V, what is the voltage across capacitor 2?Express your answer in terms of V but do not type in the symbol "V"
Answer:
V' = V/2
Explanation:
The voltage across a parallel plate capacitor is given as follows:
V = Q/C
where,
V = Voltage across capacitor
Q = Charge on Capacitor
C = Capacitance of Capacitor = A∈₀/d
Therefore,
V = Qd/A∈₀
where,
A = Area of plate
d = distance between plates
∈₀ = permittivity of free space
FOR CAPACITOR 1:
Q = Q
d = d
A = A
V = V
Therefore,
V = Qd/A∈₀ --------------- equation (1)
FOR CAPACITOR 2:
V' = ?
Q' = Q
d' = d
A' = 2A
Therefore,
V' = Q'd'/A'∈₀
V' = Qd/2A∈₀
V' = (1/2)(Qd/A∈₀)
using equation (1):
V' = V/2
Please help!!! I will give brainliest,
Answer:
C. a liter of salt water.
Explanation:
Defination of Solution =>
a liquid mixture in which the minor component (the solute) is uniformly distributed within the major component (the solvent).
What ate the two safety precautions that should be taken before driving your car?
Answer:
When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.
Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.
define these terms about speed and state their units
speed
distance covered
time taken
Explanation:
Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.
Speed = [tex]\frac{distance}{time}[/tex]
The unit is m/s or km/hr or mile/hr
Distance covered is simply the length of the path traveled.
The unit is m or km or miles
Time taken is the duration of an event.
The unit is seconds or minutes or hour.
HELP ASAP!
What happens to a circuit's resistance (R), voltage (V), and current (I) when you increase the length of the wire in the circuit?
Answer:
B R is constant V increases /increases
If one increase the length of the wire in the circuit, then, R increases, V decreases, I decreases. The correct option is D.
The resistance (R) of a wire increases as its length is increased in a circuit. This is because longer wires have more resistance because resistance increases with length.
Ohm's Law states that the current (I) in the circuit will drop if the resistance (R) rises while the voltage (V) stays constant. This is due to the inverse proportionality between current and resistance.
Thus, the correct option is D.
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During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in the cannon,determine the average net force exerted on him in thebarrel of the cannon.
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!
A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?
Answer:
The number is [tex]N = 300[/tex]
Explanation:
From the question we are told that
The net charge is [tex]Q = -4.8 *10^{-17 } \ C[/tex]
Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]
Generally the number of excess electrons is mathematically represented as
[tex]N = \frac{Q}{e}[/tex]
=> [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]
=> [tex]N = 300[/tex]
How do lenses and mirrors compare in their interactions with light?
A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.
Answer:
C. lenses refract light; mirrors do not
This question involves the concepts of reflection and refraction.
The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".
LENSES AND MIRRORSWhen it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the bending of light rays, while passing through a medium, without any rebound or absorption.
Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.
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A box of mass 7.0 kg is accelerated from rest across a floor at a rate of 2.0 m/s2 for 9.0 s .Find the net work done on the box. Express your answer to two significant figures and include the appropriate units.
Answer:
Explanation:
Step one:
given data
mass = 7kg
acceleration =2m/s^2
time= 9seconds
acceleration = velocity/time
velocity= acceleration *time
velocity=2*9
velocity= 18m/s
distance moved= velocity* time
distance= 18*9
distance=162m
we also know that the force on impulse is given as
Ft=mv
F=mv/t
F=7*18/9
F=126/9
F=14N
work done = Force* distance
work done=14*162
work=2268Joules
work= 2.27kJ
What is the result of increasing the speed at which a magnet moves in and
out of a wire coil?
A. The current in the wire increases.
B. The magnetic field around the magnet decreases.
C. The current in the wire decreases.
D. The magnetic field around the magnet increases.
Answer:
A. The current in the wire increases.
Explanation:
Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.
This phenomenon shows the inter-relationship between electricity and magnetic fields.
Magnetic fields are induced by passage of electric current. Also, electric current can be produce by magnetic fields. When the speed at which a magnet moves in and out of a wire coil increases, the current also increases.Zinc has a work function of 4.3 eV. a. What is the longest wavelength of light that will release an electron from a zinc surface? b. A 4.7 eV photon strikes the surface and an electron is emitted. What is the maximum possible speed of the electron?
Answer:
a
[tex]\lambda_{long} = 288.5 \ nm[/tex]
b
The velocity is [tex]v = 3.7 *0^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The work function of Zinc is [tex]W = 4.3 eV[/tex]
Generally the work function can be mathematically represented as
[tex]E_o = \frac{hc}{\lambda_{long}}[/tex]
=> [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]
Here h is the Planck constant with the value [tex]h = 4.1357 * 10^{-15} eV s[/tex]
and c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]
=> [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]
=> [tex]\lambda_{long} = 288.5 \ nm[/tex]
Generally the kinetic energy of the emitted electron is mathematically represented as
[tex]K = E -E_o[/tex]
Here E is the energy of the photon that strikes the surface
So
[tex]E- E_o = \frac{1}{2} m * v^2[/tex]
Here m is the mass of electron with value [tex]m = 9.11*10^{-31 } \ kg[/tex]
Generally [tex]1 ev = 1.60 *10^{-19} \ J[/tex]
=> [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]
=> [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]
=> [tex]v = 3.7 *0^{5} \ m/s[/tex]
Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
branches of sicence
Answer: Natural science can be divided into two main branches
Explanation:
life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.
A 37.0-kg child swings in a swing supported by two chains, each 3.06 m long. The tension in each chain at the lowest point is 410 N. (a) Find the child's speed at the lowest point.______m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.)_______ N(upword)
Answer:
1. 6.15m/s
2. 820N
Explanation:
The total upward force
= 410x2
= 820
g = 9.81
a = v²/r
= 2xT - msg = m x v²/r
= 820-37*9.81 = 37v²/3.06
= 820-362.97 = 37v²/3.06
= 457.03 = 12.09v²
To get v²
V² = 457.03/12.09
V² = 37.8
V = √37.8
V = 6.15m/s
B. We already have the answer to this question
The force exerted is simply gotten by this calculation
2x410
= 820N
A seated musician plays a C4 note at 262 Hz . How much time Δ does it take for 346 air pressure maxima to pass a stationary listener?
Answer:
t = 1.32 s
Explanation:
We are given;. Frequency of C4 note; F_c = 262 Hz
In conversions, we know that 1 Hz = 1 cycle/s
Thus, F_c = 262 cycles/s
Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.
346 air pressure maxima denotes that the air pressure maxima is 346 cycles.
Thus, time will be;
t = 346 cycles/262 cycles/s
t = 1.32 s
The time taken for the musical note to pass the stationary listener is 1.32 s.
The given parameters:
frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346The frequency of a sound wave is defined as the number of cycles completed per second by the wave.
[tex]F = \frac{n}{t}[/tex]
where;
t is the time to compete the maximum cycleThe time taken for the musical note to pass the stationary listener is calculated as follows;
[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]
Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.
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A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The other end of the spring is fixed to a wall. The spring is compressed by 10.0 cm from its equilibrium position and released from rest.
1) What is the speed of the object when it is 8.00 cm from equilibrium? (Express your answer to three significant figures.)
2) What is the speed when the object is 5.00 cm from equilibrium? (Express your answer to three significant figures.)
3) What is the speed when the object is at the equilibrium position? (Express your answer to three significant figures.)
Answer:
1) v = 0.45 m/s
2) v = 0.65 m/s
3) v = 0.75 m/s
Explanation:
1) We can find the speed of the object by conservation of energy:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
Where:
k: is the spring constant = 280 N/m
v: is the speed of the object =?
m: is the mass of the object = 5.00 kg
x: is the displacement of the spring
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]
2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]
3) When the object is at the equilibrium position, the speed of the object is:
[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]
[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]
[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]
I hope it helps you!
(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s
(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s
(3) the speed of the object when the spring is at equilibrium is 0.748 m/s
Compression of spring and conservation of energy:
Given that the mass of the object, m = 5 kg
spring constant, k = 280 N/m
compression of the spring , x = 10 cm = 0.1m
(i) the spring compression is at d = 8cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]
v = 0.449 m/s
(ii) the spring compression is at d = 5cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]
v = 0.648 m/s
(iii) the spring is at equilibrium so compression is at d = 0cm
according to the conservation of energy:
[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]
where v is the speed at the given compression of the spring.
[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]
v = 0.748 m/s
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7. What does the changing colour perceived by the person as the filter changes indicate to you
about white light?
Answer:
lo
Explanation: