1 kilo = 1000 gms
Therefore, 5 kilo = 5000 gms
Which conclusion can be made based on the information in the table?
Wave speed and wavelengths can vary inversely to produce the same frequency.
O Frequency and wave speed can vary directly to produce the same wavelength.
O Wavelengths and frequency can vary inversely to produce the same wave speed.
O Frequency and wavelengths can vary directly to produce the same wave speed.
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Answer:
The correct option is (b).
Explanation:
The relation between the wavelength and frequency is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
Where
v is the wave speed
f is the frequency of a wave
It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.
I need help with physics question.
(D)
Explanation:
Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is
F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)
= 2.0×10^-11 N
A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices
Answer:
Points away from the disk along the z-axis.
Explanation:
Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.
Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.
what units of measurement measures both velocity and speed
Answer:
[tex]metre \: per \: second[/tex]
Explanation:
Velocity is a derived quantity and the S.I unit is metre per second.Speed is also a derived quantity which is has the S.I unit to be metre per second.
Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?
Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.
Explanation:
It is known that formula for path difference is as follows.
[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex] ... (1)
where, n = 0, 1, 2, and so on
As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.
[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]
Hence, path difference is as follows.
[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]
For lowest frequency, the value of n = 0.
[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]
[tex]\lambda = 4 \Delta L[/tex]
where,
[tex]\lambda[/tex] = wavelength
The relation between wavelength, speed and frequency is as follows.
[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]
where,
[tex]\nu[/tex] = speed
f = frequency
Substitute the values into above formula as follows.
[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]
Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.
Problem
A charged particle is moving in the presence of uniform magnetic field. The mass of the particle
is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the
beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).
a.) How large will the x component of the velocity of the particle be in t = 2 s?
b.) Where will the particle be in t = 3.14 s?
c.) How large will the magnitude of the velocity be in t = 2.5 s?
Answer:
Answer is a I checked the work
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
Two point charges exert a 6.10 N force on each other. What will the force become if the distance between them is increased by a factor of 8
Answer:
0.0953125 N
Explanation:
Applying,
F = kq'q/r²................. Equation 1
Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.
From the equation,
If both charges remain constant,
Therefore,
F = C/r²
C = Constant = product of the two charge(q' and q) and k
Fr² = F'r'²................ Equation 2
From the question,
Given: F = 6.10 N
Assume: r = x m, r' = 8x
Substitute these value into equation 2
6.1(x²) = F'(8x)²
F' = 6.1/64
F' = 0.0953125 N
Hence the new force will become 0.0953125 N
If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?
Answer:
72 is the premimum of the insurance.
Explanation:
Below is the given values:
The loading = 0.4
Coinsurance rate = 0.2
Number of units = 100
Total number of units = 100 * 0.4 = 40
Remaining units = 60 * 0.2 = 12
Add the 60 and 12 values = 60 + 12 = 72
Thus, 72 is the premimum of the insurance.
You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.
Answer:
exactly 1273 N whether or not the box moves.
Explanation:
In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction
Therefore as per the given situation, the above represent the answer
assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K
Answer:
T = 2398 K
Explanation:
To calculate the emission of the light bulb we use the law is Stefan
P = σ A e T⁴
as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)
T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]
let's calculate
T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]
T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]
T = 2,398 10³ K
T = 2398 K
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?
Answer:
44.6 m
Explanation:
From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.
E = E'
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J
Substituting the values of the variables into the equation, we have
U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂
mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J
9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J
(735 kgm/s²)h + 75 kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J
(735 kgm/s²)h + 121.5 kgm²/s² = 29400 kgm²/s² + 3500 J
(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J
(735 kgm/s²)h + 121.5 J = 32900 J
(735 kgm/s²)h = 32900 J - 121.5 J
(735 kgm/s²)h = 32778.5 J
h = 32778.5 J/735 kgm/s²
h = 44.6 m
So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff?
Answer:
144 meters
Explanation:
it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .
so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144
which team won the champions league in 2020 2021
Answer:
Chelsea F.C
Explanation:
Chelsea F.C
Soccer
16. The sum of kinetic energies in an object.
17. The essential device in power plants that convert mechanical
energy to electricity.
18. The device that converts electricity back to mechanical energy
19. The only EM wave that is seen by naked eye.
20. A device that converts light to electricity.
A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction
Answer:
The kinetic energy and potential energy lost to friction is 2,420 J.
Explanation:
Given;
total mass, m = 40 kg
initial velocity of the girl, Vi = 5 m/s
hight of the hill, h = 10 m
length of the hill, L = 100 m
initial kinetic energy of the girl at the top hill:
[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]
initial potential energy of the girl at the top hill:
[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]
Total energy at the top of the hill:
E = 500 J + 3920 J
E = 4,420 J
At the bottom of the hill:
final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s
hight of the hill = 0
final kinetic energy of the girl at the bottom of the hill:
[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]
final potential energy of the girl at the bottom of the hill:
[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]
Based on the principle of conservation of energy;
the sum of the energy at the top hill = sum of the energy at the bottom hill
The energy at the bottom hill is less due to energy lost to friction.
[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]
Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.
What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.
Answer:
[tex]V=8.08m/s[/tex]
Explanation:
From the question we are told that:
Height[tex]h=5.00m[/tex]
Mass [tex]m=0.750kg[/tex]
Radius [tex]r=4.00cm=>0.04m[/tex]
Generally the equation for Total energy is mathematically given by
[tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]
Therefore
[tex]V=\sqrt{\frac{4gh}{3}}[/tex]
[tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]
[tex]V=8.08m/s[/tex]
Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence
Answer:
potential energy is a type of energy an object has because of it's position
Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Force and power. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hphp
Answer:
The number of crates is 84580.
Explanation:
mass, m = 30 kg
height, h = 0.9 mm
Power, P = 0.5 hp = 0.5 x 746 W = 373 W
time, t = 1 minute = 60 s
Let the number of crates is n.
Power is given by the rate of doing work.
[tex]P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580[/tex]
A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s
Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *
Answer:
you drove 50km
Explanation:
10×5 hope this helps
Answer:
50 Km
Explanation:
This is how far you have got on your journey if traveling like this.
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Hope this Helps
7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
driver of the car applies the brakes after a reaction time of 0.5 s and the car slows down at a rate of
10 ms. What happens?
Answer: Car collide with man
Explanation:
Given
Speed of car is [tex]u=30\ m/s[/tex]
Distance of the man from the car is [tex]s=55\ m[/tex]
Reaction time [tex]t_r=0.5\ s[/tex]
Rate of deceleration [tex]a_d=-10\ m/s^2[/tex]
Distance traveled in the reaction time [tex]d_o=30\times 0.5=15\ m[/tex]
Net effective distance to cover [tex]d=55-15=40\ m[/tex]
Distance required to stop the car
[tex]\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m[/tex]
Require distance is more than that of net effective distance. Hence, car collides with the man.
Given that o.2i+bj+o.4k is a unit vector,what is the value of b?
Answer:
b = 0.89
Explanation:
The given vector is, [tex]A=0.2i+bj+0.4k[/tex]
A is a unit vector
We need to find the value of b.
For a unit vector, |A| = 1
So,
[tex]0.2^2+b^2+0.4^2=1\\\\0.04+b^2+0.16=1\\\\0.2+b^2=1\\\\b^2=1-0.2\\\\b=0.89[/tex]
So, th value of b is 0.89.
A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)
Answer:
The force between the 10 th car and the 11 th car is 13636.4 N.
Explanation:
Force, F = 150 kN
acceleration, a = 2 m/s^2
Let the mass of each car is m. \Total numbers of cars = 11
F = n m a
150000 = 11 x m x 2
m = 6818.18 kg
The force between the 10 th and 11 th car is
T = ma = 6818.18 x 2 = 13636.4 N
PLEASE ANSWR 1ST AND I WILL MARK U BRAINLIEST
Two statements are given- one labeled Assertion (A) and the other labeled Reason ®. Select
the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
a. Both A and R are true, and R is correct explanation of the assertion.
b. Both A and R are true, but R is not the correct explanation of the assertion.
c. A is true, but R is false.
d. A is false, but R is true.
Assertion: An object has a negative acceleration.
Reason: The velocity of an object decreases in the same direction.
Answer:
Where is the R statement?
Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W
Compute the work done on the table:
W = Fd = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W
Please help I need this done within 30 mins
It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.
A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .
Required:
How far from the second lens is the final image of an object infinitely far from the first lens?
Answer:
the required distance is 6 cm
Explanation:
Given the data in the question;
f₁ = 15 cm
f₂ = 5.0 cm
d = 45 cm
Now, for first lens object distance s = ∝
1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'
Now, image distance of first lens s' = 15cm
object distance of second lens s₂ will be;
s₂ = 45 - 15 = 30 cm
so
1/f₂ = 1/s₂ + 1/s'₂
1/5 = 1/30 + 1/s'₂
1/s'₂ = 1/5 - 1/30
1/s'₂ = 1 / 6
s'₂ = 6 cm
Hence, the required distance is 6 cm
The distance of the final image from the first lens will be is 6 cm.
What is mirror equation?The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).
The given data in the problem is;
f₁ is the focal length of lens 1= 15 cm
f₂ s the focal length of lens 2= 5.0 cm
d is the distance between the lenses = 45 cm
From the mirror equation;
[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]
If f₁ is the focal length of lens 1 is 15 cm then;
[tex]s'=15 cm[/tex]
f₂ s the focal length of lens 2= 5.0 cm
s₂ = 45 - 15 = 30 cm
From the mirror equation;
[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]
Hence the distance of the final image from the first lens will be is 6 cm.
To learn more about the mirror equation refer to the link;
https://brainly.com/question/3229491
