Answer:
534.8 meters
Explanation:
Use T=(2*pi*r)/v
560=(2*pi*r)/6
3360=2*pi*r
1680=pi*r
534.8 meters=radius
It takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. The radius of a track is 534.7 m.
What is Distance?The distance covered by a body is equal to the sum of total path covered. It is equal to the total path traveled by an object during its entire journey.This quantity is always positive. It can't be 0 or a negative number.It is defined as a scalar quantity.
Mathematically, it can be calculated as follows :
distance = speed × time
The formula relating distance (d), speed (s), and time (t) is
d = st
First, Calculating the distance,
d = 560 s × 6 m·s⁻¹
= 3360 m
When, Calculating the track radius,
The distance travelled is the circumference of a circle,
C = 2пr
r = 3360/2п
= 534.7 m
The radius of the track is 534.7 m.
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a sharp image is formed when light reflects from a
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
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On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. The Earth's atmosphere slows down light. Assume the distance to the Moon is precisely 3.84×108 m, and Earth's atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n=1.000293. What is the difference in travel time for light that travels only through space to the moon and back and light that travels through the atmosphere and space?
Answer:
a) space only t = 1.28 s
b) space+ atmosphere t_ {total} = 1.28000003 s
Explanation:
The speed of light in each material medium is constant, which is why we can use the uniform motion relations
v= x / t
a) let's look for time when it only travels through space
t = x / c
t = 3.84 10⁸/3 10⁸
t = 1.28 s
b) we look for time when it travels part in space and part in the atmosphere
space
as it indicates that the atmosphere has a thickness of e = 30 10³ m
t₁ = (D-e) / c
t₁ = (3.84 10⁸ - 30.0 10³) / 3 10⁸
t₁ = 1.2799 s
atmosphere
we use the refractive index
n = c / v
v = c / n
we substitute in the equation of time
t₂ = e n / c
t₂ = 30 10³ 1,000293 /3 10⁸
t₂ = 1.000293 10⁻⁴ s
therefore the total travel time is
t_ {total} = t₁ + t₂
t_ {total} = 1.2799+ 1.000293 10⁻⁴
t_ {total} = 1.28000003 s
we can see that the time increase due to the atmosphere is very small
which one is odd copper,plastic,rubber
Answer:
It's plastic.
trust me it's plastic, i've rad it somewhere.
All of them have something that's not like the others.
-- Rubber is the only one on the list that has two repeated letters.
-- Plastic is the only one on the list thagt has no repeated letters.
-- Plastic is the only one on the list that has no 'r' in its name.
-- Copper is the only one on the list that is an element, not a compound.
-- Copper is the only good electrical conductor on the list.
-- Plastic is the only one on the list with more than six letters in its name.
-- Rubber is the only one on the list with no 'p' in its name.
-- Plastic is the only one on the list that doesn't end in "-er".
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
Mechanical energy is the most concentrated form of energy.
a. true
b. false
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?
___Hz
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
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A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False
If you could help me please.
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
2) How much does a 3-kg object weigh (on earth) in newtons?
3) How much does a 20-kg object weigh (on earth) in newton?
4) What must happen for the mass of an object to change?
5) What are 2 ways the weight of an object can change?
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
No. 1kg of mass does not weigh 9.8N on the moon.
Weight = (mass) x (gravity).
Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.
2) How much does a 3-kg object weigh (on earth) in newtons?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (3 kg) x (9.8 m/s² )
Weight = 29.4 N
3) How much does a 20-kg object weigh (on earth) in newton?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (20 kg) x (9.8 m/s² )
Weight = 196 N
4) What must happen for the mass of an object to change?
When an object moves, its mass increases. The faster it moves, the greater its mass gets. But this is all part of Einstein's "Relativity". The object has to move at a significant fraction of the speed of light before any change can be noticed or measured. So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.
5) What are 2 ways the weight of an object can change?
First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.
But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood. So the weight can change even though the mass doesn't.
The weight of an object changes if you take it to a place where gravity is stronger or weaker.
Let's say we have an object whose mass is 90.72 kilograms. Like me !
As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons (200 pounds).
. . . Fly me to the moon. Gravity = 1.62 m/s² Weight = 147 Newtons (33 lbs)
. . . Drag me to Jupiter. Gravity = 24.8 m/s² Weight = 2,249 N (506 pounds)
My mass never changed, but my weight sure did.
. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact
Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
Is there a way to see moon and the sun at once?
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
mark me brainliest
why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.
a.
3.81
b.
26.68
c.
9151.24
d.
1307.32
Hhhhhellllppp fastt
Answer:
a. 3.81
Explanation:
F = GMm/r^2
F = (6.67 x 10^-11 x 28 x 1) / 7^2
F = 3.81 x 10^-11 N
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
A hair dryer draws a current of 12.8 A.
(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.
(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.
Answer:
(a) 8.85×10⁻³ minutes
(b) 4.24×10¹⁹ electrons
Explanation:
(a) Using,
Q = it............................. Equation 1
Where Q = quantity of charge, i = current, t = time.
Make t the subject of the equation
t = Q/i............................. Equation 2
Given: Q = 6.8×10⁰ C, i = 12.8 A
Substitute these values into equation 2
t = 6.8×10⁰/12.8
t = 8.85×10⁻³ minutes
(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3
Where n = number of electrons.
Given: Q = 6.8×10⁰ C
Substitute into equation 2
n = 6.8×10⁰/1.602×10⁻¹⁹
n = 4.24×10¹⁹ electrons
(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b) Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
What will be the time of the charge and number of the electrons in the charge ?As we know Q = IT
Where Q = quantity of charge, i = current, T = time.
From the above equation
T= Q/I.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A
Substitute these values
T= [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8
T = [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes
Now the number of the electrons present in the charge will be
n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])
Where n = number of electrons.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C
Substitute Value of Q
n = [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]
n = [tex]4.24\times\d10^{19}[/tex] electrons
Thus
(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b)Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
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An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.
Answer:
B. in the direction of the force
Explanation:
Sana nakatulong
What is happening in the graph shown below?
A.
The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
B.
The object moves toward the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 2 m/s.
C.
The object moves toward the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 8 m/s.
D.
The object moves away from the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 8 m/s.
Answer:
D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
Explanation:
I just got it right lol
In this experiment, you will use a track and a toy car to explore the concept of movement. You will measure the time it takes the car to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.
Answer: if weight affects how fast they go?
Explanation:
Answer:
How can we change the speed of a toy car on a racetrack to describe the car’s motion?
Explanation:
thats the sample respond
A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3
Answer:
75.4
Explanation:
r= 2
h= 6
v= 22/7 *r*r*h
v= 75.42
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?
Answer:
work done= force × displacement
=800×5
=4000J
Explanation:
The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.
