Answer:
2.56 nC
Explanation:
By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:[tex]C =\frac{Q}{V} (1)[/tex]
For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:[tex]Q = \sigma * A (2)[/tex]
Assuming an uniform electric field E, the potential difference V can be expressed as follows:[tex]V = E*d (3)[/tex]
where d is the distance between plates.
Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:[tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]
Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:[tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]
Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:[tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]
Solving for Q, we get:[tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will ________ because the toy car ____
Answer:
If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration.
Explanation:
I hope this helped
Hypothesis: If the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car gains more potential energy at the higher starting point.
When a toy car moves along a sloped racetrack, it converts its potential energy (due to its height above the ground) into kinetic energy (energy of motion). The higher the starting height of the racetrack, the more potential energy the toy car possesses initially.
As the toy car moves down the sloped track, it will accelerate due to the force of gravity. The potential energy is converted into kinetic energy, and the car's speed increases. According to the law of conservation of energy, the total mechanical energy (sum of potential and kinetic energy) remains constant as long as no external forces, such as friction, act on the car.
Therefore, if the starting height of the racetrack is increased, the toy car will have more potential energy to start with. As it moves down the track, it will convert this increased potential energy into kinetic energy, resulting in a higher speed compared to when it starts from a lower height.
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2. An ambulance traveling at 20 m/s emits a sound at 500 Hz. What frequency does a person standing on the corner of a street detect?
A football player runs down the field at a speed of 8 m/s how long will it take him to run 20 m?
a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achieve this speed?
The time taken by the car to achieve the final speed is 6.25 seconds.
What is the equation of motion?The equations of motion can be defined as the equation that represents the relationship between the time, velocity, acceleration, and displacement of a moving object.
The mathematical expressions for the equations of motions can be written as:
[tex]v= u+at\\S=ut+(1/2)at^2\\v^2-u^2=2aS[/tex]
Given, the initial speed of the car, u = 15 m/s
The final speed of the given car, v = 25m/s
The distance covered by car, S = 125 m
From the third equation of motion: v² = u²+ 2aS
(25)² = (15)² + 2×a× 125
a = 1.6 m/s²
From the first equation of motion we can find the time to achieve the final speed:
v = u+ at
25 = 15 + (1.6) × t
t = 6.25 sec
Therefore, 6.25 seconds will be taken by the car to catch the final speed.
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what are the laws of newton
Answer:
Explanation:
These are the laws of Newton
Answer:
the first law, an object will not change its motion unless a force acts upon it. the 2nd one, the force of an object is equal to its mass times it acceleration. the 3rd one is when 2 objects interact, they apply forces to each other of equal magnitude and opposite direction.
2
10 points
Find the total displacement of each of the motions.
a) You walk 45 m W, then 34 mW
b) You drive 5 km N, then 7 km S
c) You cycle 350 m E, then 800 m W, then 200 m E
d) You fly 850 km N then 850 km S
Answer:
a) s = 79 m W
b) s = 2 km S
c) s = 250 m W
d) s = 0 km
Explanation:
We take the following sign convention for the directions:
North (N) ---> positive
South (S) ---> negative
East (E) ---> negative
West (W) ---> positive
a)
45 m W, 34 m W
s = 45 m + 34 m
s = 79 m W
b)
5 km N, 7 km S
s = 5 km - 7 km
s = - 2 km
s = 2 km S
c)
350 m E , 800 m W, 200 m E
s = -350 m + 800 m - 200 m
s = 250 m
s = 250 m W
d)
850 km N, 850 km S
s = 850 km - 850 km
s = 0 km
What is Energy in physics?
Answer:
Energy, in physics, the capacity for doing work. It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms. There are, moreover, heat and work—i.e., energy in the process of transfer from one body to another.
Explanation:
Hope this helps!
Which interaction contributes to the greenhouse effect?
A (Gases in the atmosphere absorb heat.
B (Dust particles scatter and reflect light from the Sun.
C (Green visible light is trapped in Earth’s atmosphere.
D (Light travels through a thick part of the atmosphere at sunrise and sunset.
Answer:
A
Explanation:
Answer:
A
Explanation:
Gases in the atmosphere absorb heat.
You know when you have a blanket around you and some heat gets trapped in but some still gets out. Thats basically what it is.
Plus I got it right in multiple questions including the test!
I hope that reassured you!
Have a good night!
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 satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the speed and direction of the satellite after the crashes. Why would the crash affect the motion of the satellite, and which crash caused a greater change in motion for the satellite?
WILL GIVE BRAINLIEST
Answer:
The fastest satellite must change orbit
The most massive body (m₁) transfers more momentum to the satellite,
Explanation:
For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants
Most massive body (m1)
initial. Before the crash
p₀₁ = M v + m₁ v₁
after the crash
[tex]p_{f1}[/tex] = M v´ + m₁ v₁´
how momentum is conserved
p₀ = p_{f}
Lighter body (m2)
p₀₂ = M v + m₂ v₂
p_{f2} = M v´ + m₂ v₂´
Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.
We can see that p₀₁> p₀₂
Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.
The fastest satellite must change orbit, thus rotating at a different distance from Earth
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
Use the drop-down menus to complete each sentence.
A plant grows toward a sunny window. This response is an example of
✔ phototropism
.
Sometimes a plant grows around a tree for support. This response is an example of
✔ thigmotropism
.
The roots of a plant grow toward a water source. This response is an example of
✔ hydrotropism
.
The roots of a plant grow down into the soil. This response is an example of
✔ gravitropism
.
the answers are already there
Answer:
The correct answer is - phototropism, thigmotropism, hydrotropism, and gravitropism in order ( already match correctly).
Explanation:
phototropism is a phenomenon in which plants grow towards the light or sun which is accomplished by the hormone auxin in the cells far from the light.
Thigotropism is a type of plant growth that occurs around the tree to support itself which is a touch stimuli response.
The movement of the plant in the direction of the source of the water is known as hydrotropism. In which stimuli is humidity or the water concentration.
The movement of the plant or roots of the plants towards the soil or earth is known as gravitropism here gravity is the stimuli.
Answer:
Use the drop-down menus to complete each sentence.
A plant grows toward a sunny window. This response is an example of
phototropism
.
Sometimes a plant grows around a tree for support. This response is an example of
thigmotropism
.
The roots of a plant grow toward a water source. This response is an example of
hydrotropism
.
The roots of a plant grow down into the soil. This response is an example of
gravitropism
.
Explanation:
1. What does the pH scale measure?
Answer:
The pH scale measures of how acidic or basic water is.
The pH scale also measures whether there is more hydronium or hydroxide in a solution.
Explanation:
The range goes from 0-14, with 7 being neutral. Less than 7 indicates acidity and more than seven indicates the substance is a base.
A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself.
Answer:
m = 876.71 kg
Explanation:
This is an exercise of Archimedes' principle, which states that the thrust on a body is equal to the weight of the dislodged liquid
B = ρ g V
therefore the load that the balloon can lift is
B - W_structure - w_load = 0
w_load = B - W_structure
The volume of the balloon is
v = 4/3 π r³
let's substitute
w_carga = rho g 4/3 π r³ - m_structure g
the air density at T = 25ºc is ρ = 1.18 kg / m³
let's calculate
w_load = 1.18 9.8 4/3 π 7.15³ - 930 9.8
w_load = 17705,77 - 9114
w_ load = 8591.77 N
this corresponds to a mass of
w_load = m g
m = w_load / g
m = 8591.77 / 9.8
m = 876.71 kg
Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?
a. 250 kg
c. 10 Kg
b. .002 Kg
d. 45 Kg
Answer:
a. 250kg I think it's the right answer. hope it helps:)
Answer:
C.10
Explanation:
because when you divide 50 divided by 5 = 10
An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. All airplanes will start from rest(0m/s). Assuming this minimum acceleration, what is the minimum allowed length for the runway for take off?
Answer:
x = 704 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f} ^{2} =v_{o} ^{2} +2*a*x[/tex]
where:
Vf = final velocity = 65 [m/s]
Vo = initial velocity = 0 (starts from rest)
a = acceleration = 3 [m/s²]
x = distance [m]
Now replacing we have:
65² = 0 + 2*3*x
4225 = 6*x
x = 704 [m]
Select all correct answers....Covalent compounds
A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of the signals? а 7.15 m b 5.23 m C 2.93 m d 0.93 m
Answer:
2.93 m (which agrees with answer "C" on the list)
Explanation:
Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:
Wavelength = 2.997 10^8 / 1.023 10^8 = 2.93 m
A typical elevator car with people has a mass of 1500.0 kg. Elevators are currently approaching speeds of 20.0 m/s - faster than the speed.
Required:
What is the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s?
Answer:
1500NExplanation:
Force = mass * acceleration
Given
Mass = 1500kg
Get the acceleration using the equation of motion;
v² = u²+2aS
20² = 0+2s(200)
400 = 400a
a = 400/400
a = 1m/s²
Get the upward force required
F = 1500 * 1
F = 1500N
Hence the upward force required if the elevator moves upward 200.0 meters before reaching 20.0 m/s is 1500N
If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the internal energy of the gas?
Answer:
The change in the internal energy of the gas 1,595 J
Explanation:
The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)
This is summarized in the expression:
ΔU= Q - W
where the heat absorbed and the work done by the system on the environment are considered positive.
Taking these considerations into account, in this case:
Q= 500 cal= 2,092 J (being 1 cal=4.184 J) W=500 JReplacing:
ΔU= 2,092 J - 500 J
ΔU= 1,592 J whose closest answer is 1,595 J
The change in the internal energy of the gas 1,595 J
Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat of 1,700 J/(kg°C).
Answer:
Q = 425 kJ
Explanation:
Given that,
Mass, m = 25 kg
The clown doll head that heats up from 25°C to 35°C
The specific heat is 1700 J/kg°C
We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :
[tex]Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ[/tex]
So, 425 kJ of thermal energy is severed.
that delivers oxygen to your body and In the video your blood is compared to a picks up CO2 to be released out when you breath. PLEASE I NEED A ANSWER
What specific changes in two climate variables are expected to lead to major decreases in soil moisture southern Africa and the Mediterranean region?
Answer:
Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.
Explanation:
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.
What is drought?
Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.
Droughts can last months or years, although they can be proclaimed in as little as 15 days.
It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.
Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.
Hence Precipitation and droughts are the specific changes in two climate variables.
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Which possible component of initial energy is caused by molecular motion within a material?
Answer: thermal energy
Answer:
Thermal energy
Explanation:
The internal energy of a system is widely known as thermal energy. Now, thermal energy is also called heat energy and it is an internal energy of a component which is produced when an increase in temperature causes atoms and molecules within the component to move faster and start colliding with one other.
Therefore, the more heat the is applied to the component, the hotter the substance and the more its particles move which in turn leads to a higher thermal energy.
15 points.
An object of mass 100 kg is observed to accelerate at a rate of 15
m/s/s. Calculate the force required to produce this acceleration.
Answer:
its 0.5 for all i beleive
Explanation:
A long wire carries a current 5 A from west to east. A magnetic compass pointing North is placed underneath the wire at a distance of 2 mm. What is the deflection of the compass when it is placed under the wire?
Answer:
no deflection
Explanation:
current is flowing from west to east. As the magnetic field of a long wire carrying current is circular, its direction will be north below the wire and south above the wire (according to the right hand rule). So, when the compass is placed underneath the wire, it will still point towards the north direction.
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel?A) The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.B) Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.C) Pressure fluctuations travel along the direction of propagation of the sound wave.D) Propagation of energy that passes through empty spaces between the particles that comprise the mediumDoes air play a role in the propagation of the human voice from one end of a lecture hall to the other?a) yesb) no
Answer:
None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave
Explanation:
At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coefficient of friction was estimated to be 0.38. a) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.b) Why does the car's mass not matter?
Answer:
a. V = 19.1m/s
b. The mass of the car does not matter
Explanation:
A.
KE = 1/2mv² = fd --------(1)
Fd = umgd ---------(2)
Therefore,
1/2mv² = umgd ---------(3)
M will cancel itself out from both sides of equation 3.
Then we will have:
1/2v² = ugd
Then we cross multiply to make v² the subject of the formula
V² = 2ugd
V = √2ugd -------(4)
U = 0.38
g = 9.81
d = 98
When we input these values into equation 4, we will have:
V = √2x0.38x9.81x98
V = √730.6488
V = 27.03m/s
B.
The mass of the car does not actually matter as the mass was cancelled out on the both sides of equation 3
Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds
Answer:
6.56km
Explanation:
Given parameters:
Speed = 11.5km/hr
Time = 2060s
Unknown:
Distance covered = ?
Solution:
Speed is distance divided by the time taken.
Speed = [tex]\frac{distance}{time}[/tex]
Distance = Speed x time
Let us convert the seconds to hours;
3600s = 1hr
2060s = [tex]\frac{2060}{3600}[/tex] = 0.57hr
Now
Distance = 11.5km/hr x 0.57hr = 6.56km