The car and the delivery truck both start from rest and accelerate at the same rate. So, the final speed of the car as compared to the truck is four times as much. So, option D is correct.
What is acceleration?Acceleration is the term used in mechanics to describe how quickly an object's velocity changes in relation to time. The magnitude of accelerations as a vector. An object will accelerate in the direction that it is being pulled in by the net force.
Acceleration is a vector quantity since it has a magnitude and a direction. Velocity is another aspect of vessel quantities. The change in the vector of velocity over a time interval divided by the time interval is the definition of acceleration.
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a projectile starting from ground hits a target on the ground located at the distance of 1000m after 40 sec
A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) The size of the angle θ is = 83°
b)The initial velocity was the projectile launched is (u)=205.13 m/s.
What is velocity?The velocity is a physical term that is refer to how much the object has covered the distance in a given time. It can be measured in m/s and cm/s.
How can we calculate the velocity?a) To calculate the angle we are using two formulas, they are
r= u cos(θ)T
T= 2u sin (θ)/g
Here we are given,
r= The distance covered in the motion = 1000m.
T = The time covered in the motion = 40 Seconds.
g= The acceleration due to gravity = 9.8 m/s²
We have calculate the values of angle = θ
Now we put the values in above equation we get,
1000= u cosθ*40...(1)
40*g= 2u sinθ.....(2)
Equation(2) divided by equation(1) we get,
2tanθ=40*40*g/1000
Or, tanθ=7.84
Or, θ= 82.7°≈83°
From the above calculation we can say that, The size of the angle θ is = 83°
b) The initial velocity was the projectile launched is = u
As we know, r= u cos(θ)T
Now we put the values in the equation we get,
1000= u cos(83)*40
Or, u=205.13 m/s
From the above calculation we can say that, The initial velocity was the projectile launched is (u)=205.13 m/s.
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A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) What is the size of the angle θ?
b) At what initial velocity was the projectile launched?
Moments after making the dreaded decision to jump out the door of the airplane, Darin's 75.5-kg body experiences +128 N of air resistance upward. Determine Darin's acceleration (in m/s^2) at this instant in time.Hint: Your answer will be negative since he is falling (i.e., his acceleration is still in the down direction). Answer: ________ m/s^2
Given:
The mass of the Darin is m = 75.5 kg
The air resistance is
[tex]F_a=128\text{ N}[/tex]Required: Darin's acceleration.
Explanation:
According to Newton's second law, the downward force will be
[tex]F_g=\text{ mg}[/tex]Here, g = -9.8 m/s^2 is the acceleration due to gravity.
On substituting the values, the downward force will be
[tex]\begin{gathered} F_g=75.5\times(-9.8) \\ =\text{ -739.9 N} \end{gathered}[/tex]The net force will be
[tex]\begin{gathered} F_{net}=\text{ F}_g+F_a \\ =-739.9+128 \\ =-611.9\text{ N} \end{gathered}[/tex]Final Answer: Darin's acceleration is -611.9 N
If a car has a momentum of 2315Ns and its mass is 382kg, how fast is it moving?
ANSWER:
6.06 m/s
STEP-BY-STEP EXPLANATION:
Given:
Momentum = 2315 N*s
Mass = 382 kg
We can calculate the speed as follows:
[tex]\begin{gathered} p=m\cdot v \\ v=\frac{p}{m} \\ \text{ we replacing} \\ v=\frac{2315}{382} \\ v=6.06\text{ m/s} \end{gathered}[/tex]It moves at a speed of 6.06 m/s
A car traveling at 11.6 meters per second crashes into a barrier and stops in 0.287 meters. What force must be exerted on a child of mass 21.2 kilograms to stop him or her in the same time as the car? Answer must be in 3 significant digits.
The equation to obtain the final speed of car is,
[tex]v^2=u^2+2as[/tex]Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(11.6m/s)^2+2a(0.287\text{ m)} \\ a=\frac{-134.56m^2s^{-2}}{2(0.287\text{ m)}} \\ \approx-234.4m/s^2 \end{gathered}[/tex]The negative sign of acceleration indicates that the car is deaccelerating.
The force required to stop the car is,
[tex]F=ma[/tex]Substitute the magnitude of known values,
[tex]\begin{gathered} F=(21.2kg)(234.4m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =4969.28\text{ N} \\ \approx4970\text{ N} \end{gathered}[/tex]Thus, the force required to stop the car is 4970 N.
If you see an object following a curve path, you can safely assume:There is a net force being applied The net force is zero
If an object is following a curve, it means that it is changing direction. This change in direction will cause a change in velocity. Change in velocity causes acceleration. Since the velocity keeps changing, the acceleration is not zero. This also means that the net force is not zero. Therefore, you can safely assume:
There is a net force being applied
a power transmission line is used to transmit 100 kw of power at a voltage of 10 kv with a loss of 1 kw. if the voltage is increased to 200 kv, what is power loss (in watts) to transmit the same amount of power?
To transmit the same amount of electricity, there is a 2.5 W power loss.
What is power loss?A power loss is the difference in power between a device, piece of equipment, pump set, or process' input and output (Pv). Pumps, equipment, and procedures, as well as electrical and electronic devices, turn this undesirable loss into heat.
Consumers receive electric energy in different units than what power plants create. A fraction of the units is lost in the distribution network. This difference between the distributed and created units is referred to as transmission and distribution loss.
According to Ohm's law, power can be defined as
P = VI
Here,
V = Voltage
I = Current
replacing
100kW = 10WI
I = [tex]\frac{100}{10}[/tex]
I = 10 Amp
Now we can find the resistance
R = [tex]\frac{P_{LOSS} }{I^{2} }[/tex]
Replacing,
R = [tex]\frac{1000}{10^{2} }[/tex]
R = 10Ω
In the second state when the voltage is 200kv, we have,
I = [tex]\frac{100*10^{3} }{200*10^{3} }[/tex]
I= [tex]\frac{100000}{200000}[/tex]
I= 0.5 A
Now power loss,
[tex]P_{t}[/tex] = [tex]I^{2}[/tex]R
[tex]P_{t}[/tex] = ([tex]0.5^{2}[/tex]) (10)
[tex]P_{t}[/tex] = 0.25 *10
[tex]P_{t}[/tex] =2.5 W
As a result, 2.5W is needed to transport the same amount of electricity.
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What letter from the picture below represents the position of the maximum kinetic energy?
Given that a pendulum has a mean position as C, and two extreme points A and E.
We have to find the position of kinetic energy.
Here, the total energy is conserved. So, the sum of potential energy and kinetic energy is constant.
Potential energy increases with the increase in height.
At the extreme positions, A and E, potential energy are maximum and potential energy is zero at point C.
Also, Kinetic energy is zero is at points A and E.
As energy is conserved, Kinetic energy is maximum at point C and potential energy is zero.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." Explain the error in the student's claim. Provide an example of each quantity to support your answer.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." The error here is that acceleration is said to be done when either speed of that object changes or direction of that object changes. Hence , acceleration is not a scaler quantity.
Scalar quantities are quantities that only have a magnitude and do not have any direction
A vector quantity is defined as the physical quantity that has magnitude as well as directions associated to it.
Acceleration is said to be occurred in two cases :
when the object changes its speed
or when the object changes its direction
since , acceleration depends upon both direction as well as magnitude ,hence it is a vector quantity not a scaler quantity.
for example : a stone attached to a string moving in a circular motion at a constant speed will be considered in accelerated motion because it is constantly changing its direction. Here we can see speed is constant hence magnitude (value of speed) is not changing but direction of the stone is changing . since , direction is changing the object is said to be in accelerated motion.
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Use the data that you have for the air-filled tube. The tube is closed at one end.• Make a drawing of the standing wave in the tube when you have two, three, and four nodes or peaks. These will correspond to the first, second, and third overtones.• For each standing wave, label the nodes and the antinodes.• Label and record the distance in meters of the half wavelength (λ/2) for each standing wave.• Explain how the standing wave occurs.
Part 1)
When a standing wave oscilates in an air-filled tube closed at one end and open at the other, the closed end is a node and the open end is an antinode. Then, the diagram of the wave inside the tube looks like the following:
Part 2)
The nodes are the points where the two curves intersect (darker regions) and the antinodes are the clearer regions.
Part 3)
In each wave on the diagram, the distance between two consecutive nodes or two consecutive antinodes is the same as half the wavelength.
Part 4)
The standing waves arise from the combination of reflection and interference of waves inside the air column when the wavelengths match the length of the tube in such a way that the open end is an antinode and the closed end is a node.
Apassenger in a drops a and it freely from rest to the groundWhat is the speed of the ball a distance of 20 massuming the acceleration due to gravity equal to 9.01 m/s ^ 2
Given data
*The given initial speed is u = 0 m/s
*The given distance is s = 20 m
*The value of the acceleration due to gravity is g = 9.01 m/s^2
The formula for the final speed of the ball at a distance of 20 m is given by the kinematic equation of motion as
[tex]v^2=u^2+2gs[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v^2=(0)^2+2(9.01)(20) \\ v^2=\text{36}0.04 \\ v=18.98\text{ m/s} \end{gathered}[/tex]Hence, the speed of the ball at a distance of 20 m is v = 18.98 m/s
A fuzzy die that has a weight of 1.70N hangs from the ceiling of a car by a massless string. The car travels on a horizontal road and has an acceleration of 2.70m/s^2 to the left. The string makes an angle theta with respect to the vertical, as shown in the figure below. 1) What is the angle theta?
First we calculate the mass of the fuzzy die
[tex]m=\frac{W}{g}[/tex]m is the mass, W is the weight, and g is the gravity
m=?
W=1.70N
g=9.8 m/s^2
we susbtitute
[tex]m=\frac{1.70}{9.80}=0.17\text{ kg}[/tex]Then we calculate the force of x
[tex]Fx=0.17(2.70)=0.468\text{ N}[/tex][tex]Fy=1.70N[/tex]Then we have the next diagram
Therefore for the angle
[tex]\theta=\arctan (\frac{F_x}{F_y})=arctan(\frac{0.468}{1.70})=15.4\text{ \degree}[/tex]ANSWER
The angle is 15.4°
Two toy cars collide in an inelastic collision. The first car has a mass of 10 kg
and a velocity of 4 m/s to the right. The second car has a mass of 8 kg and a
velocity of 6 m/s to the left.
The velocity of the cars after the collision is ____________ m/s.
The velocity of the cars after the inelastic collision is 0.44 m/s.
What is inelastic collision ?
When some of the kinetic energy of a colliding object or system is wasted, the collision is said to be inelastic. In a perfectly inelastic collision, the colliding particles stay together and the most kinetic energy is lost. Such situations involve the use of wasted kinetic energy to bind the two bodies together. The conservation of momentum and energy is typically used to tackle collision-related problems.
m 1 = 10kg
m 2 = 8kg
u 1 = 4m/s
u 2 = 6m/s
Total mass of the combined system,
M=m 1+m 2 = 18kg
Let the velocity of the combined system after the collision be v
Applying conservation of momentum before and after the collision :
m 1 x u 1 − m 2 x u 2 =Mv
10 x 4 - 8 x 6 = 18v
v = 8/18 = 0.44 m/s
The velocity of the cars after the inelastic collision is 0.44 m/s.
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Which optical instrument produces a magnified, virtual, and inverted image of small objects?1) a refracting telescope2) a single lens reflex camera3) a microscope4) a pair of binoculars
Microscope
Explanations:Microscopes are known for magnifying tiny objects
The images produced by a microscope are:
virtual (formed behind the screen)
Inverted
Magnified or enlarged
Therefore, the optical instrument which produces a magnified, virtual, and inverted image is the microscope
2. During which Epoch did humans first appear?Holocene3. How many million years ago did humans first appear?2.8 million years ago4. What are the three periods of the Mesozoic Era?Cretaceous, Jurassic, Triassic5. When did dinosaurs first appear?Triassic period6. In what period did birds first appear?Jurassic7. During what period did the dome-like uplift of the Adirondack region begin?
(1)
The three eras within the Phanerozoic eon are
1) The Paleozoic era (541 million to 252 million years ago)
2) The Mesozoic era (252 million to 66 million years ago)
3) The Cenozoic era (66 million years ago to the present)
. Chloe performed an experiment on the amount of a certain gas that can be dissolved in water at different temperatures. Her data from the experiment is shown on the graph below. What type of function would best fit this set of data?
The values on the curve corresponding to the y axis are the y values
The values on the curve corresponding to the x axis are the x values
We would pick corresponding x and y values on the graph and compare them
On the graph, each small square represents 1 unit on both axes.
If x = 1, y = 1
If x = 4, y = 2
if x = 9, y = 3
We know that the square root of 1 is 1, the square root of 4 is 2 and the square root 9 is 3. This means that
y is the square root of x
The function is
[tex]y\text{ = }\sqrt[]{x}[/tex]Thus, it is a square root function
The first option is correct
An x-ray with a wavelength of 3.5 × 10^-9 m travels with a speed of 3.0 × 10^8 m/s. What is the frequency of this electromagnetic wave? A.9.52 × 10^-1 Hz B.8.57 × 10^16 Hz C.1.17 × 10^-17 Hz D.1.05 Hz
Answer. B
Determine the maximum static friction when the normal force is 890 N and the coefficient of static friction is 0.55.
Answer:
489.5 N
Explanation:
The maximum static friction can be calculated as
[tex]F_f=\mu_sF_N[/tex]Where μ is the static friction and Fn is the normal force.
So, replacing the values, we get:
[tex]F_f=0.55(890N)=489.5N[/tex]Therefore, the maximum static friction is 489.5 N
Point charges 88μC,-55μC and 70 μC are placed in a straight line. The central one is 0.75m from each of the others. Calculate the net force on each due to the other two.
The net force on the charges is 139.04 N.
What is the net force between the charges?
The net force between the charges is calculated by applying Coulomb's law as follows;
F = kq₁q₂/r²
where;
k is Coulomb's constantq₁ is the first chargeq₂ is the second charger is the distance between the chargesF(12) = (9 x 10⁹ x 88 x 10⁻⁶ x 55 x 10⁻⁶) / (0.75²)
F(12) = 77.44 N
F(23) = (9 x 10⁹ x 55 x 10⁻⁶ x 70 x 10⁻⁶) / (0.75²)
F(23) = 61.6 N
The net force on the charges is calculated as follows;
F(net) = 77.44 N + 61.6 N
F(net) = 139.04 N
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There are three point charges placed in a straight line, so the net force on each due to the other two is 139.04 N.
What is a charge?Charged matter experiences a force when it comes into contact with an electromagnetic field because electric charge is a property of matter. An electric field can have either a positive or negative charge. Charges that are similar to one another and dissimilar to one another are drawn to one another.
Given information in the question,
Charge, q₁ = 88 μC
Charge, q₂ = -55 μC
Charge, q₃ = 70 μC
Distance, r = 0.75 meters.
F = kq₁q₂/r²
Put the values in the above formula,
F₁₂ = (9 x 10⁹ x 88 x 10⁻⁶ x 55 x 10⁻⁶) / (0.75²)
F₁₂ = 77.44 N
F₂₃ = (9 x 10⁹ x 55 x 10⁻⁶ x 70 x 10⁻⁶) / (0.75²)
F₂₃ = 61.6 N
Now, calculate the net force :
F(net) = 77.44 N + 61.6 N
F(net) = 139.04 N
Hence, the net force due to the other two is 139.04 N.
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A force of 10 Newtons is the only force exerted on a block, and the acceleration of the block is measured. When the same force is the only force exerted on a second block, the acceleration is three times as large. What can you conclude about the masses of the two blocks?
When the acceleration is three times as large, the mass of the new block will be one-third the initial mass.
What is the acceleration of the block?
The acceleration of the block is the rate of change of velocity of the block with time.
The magnitude of the acceleration of each block can be obtained by applying Newton's second law of motion as follows;
F = ma
m = F/a
where;
m is the mass of each blockF is the applied force = 10 Na is the acceleration of each block.When the acceleration is three times as large, the mass of the new block is calculated as;
m = F/3a
m = 1/3 (F/a)
new mass = one-third the initial mass
Thus, we can conclude that when the acceleration is three times as large, the mass of the new block will be one-third the initial mass.
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A 22-pound force that makes an angle of 12° with an inclined plane is pulling a box up the plane. The inclined plane makes a 25° angle with the horizontal. What is the magnitude of the effective force pulling the box up the plane?
We will ave that the effective force is:
[tex]\begin{gathered} F=(22Lb)cos(12)\Rightarrow F=21.51924722...Lb \\ \\ \Rightarrow F\approx21.52Lb \end{gathered}[/tex]So, the effective force is 21.52 Pounds.
A ball is thrown directly downward with an initial speed of 8.45 m/s, from a height of 29.9 m. After what time interval does it strike the ground? s
Given:
The initial speed of the ball is: u = 8.45 m/s.
The ball is thrown from the height: h = 29.9 m
To find:
The time ball takes to strike the ground.
Explanation:
The time taken by the ball to strike the ground can be determined by using the following equation.
[tex]x=ut+\frac{1}{2}at^2[/tex]Here, x = -29.9 m, u = -8.45 m/s and a = -9.8 m/s^2. The negative sign indicates that the ball is falling in the downward direction.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} -29.9=-8.45t-\frac{1}{2}\times9.8t^2 \\ \\ 4.9t^2+8.45t-29.9=0 \end{gathered}[/tex]Solving the above quadratic equation, we get:
t = 1.754 s and t = -3.478 s
But the time is never negative, thus t = 1.754 s.
Final answer:
The ball takes 1.754 seconds to strike the ground.
A car with a mass of 1180 kilograms is being driven at 18.2 meters per second when it runs into a tree. What is the change in kinetic energy of the car? Include units in your answer. Answer must be in 3 significant digits.
The kinetic energy formula is
[tex]K=\frac{1}{2}mv^2[/tex]Where m = 1180 kg and v = 18.2 m/s.
[tex]K=\frac{1}{2}\cdot1180\operatorname{kg}\cdot(18.2\cdot(\frac{m}{s}))^2=195,431.6J[/tex]Therefore, the kinetic energy is 195,431.6 Joules.
A car initially traveling at 3.2 m/s accelerated uniformly to a speed of 14.9 m/s over a distance of 60 meters. How much time does it take for the car to reach 9 m/s? Report your answer to 1 decimal place. Do not type in the unit or the computer will mark your answer incorrect.
Given
Initial velocity, u=3.2 m/s
Final speed , v=14.9 m/s
Distance travelled, s=60 m
To find
Time taken to reach 9 m/s
Explanation
By equation of kinematics,
[tex]\begin{gathered} v^2=u^2+2as \\ \Rightarrow14.9^2=3.2^2+2a\times60 \\ \Rightarrow222.01=10.24+12a \\ \Rightarrow a=17.65\text{ m/s}^2 \end{gathered}[/tex]Let the time taken to reach 9 m/s be t
Thus
taking v=9 m/s here we have
[tex]\begin{gathered} v=u+at \\ \Rightarrow9=3.2+17.65t \\ \Rightarrow t=0.3\text{ s} \end{gathered}[/tex]Conclusion
The required time is 0.3 s
Two traveling pulses on a rope move toward each other at a speed of 1.0 m/s. The waves have the same amplitude. The drawing shows the position of the waves at time t = 0 s. Which one of the following drawings depicts the waves on the rope at t = 4.0 s?
ANSWER:
STEP-BY-STEP EXPLANATION:
We know the graph when t = 0. Therefore, at t = 4 pulse at "A" will reach to "B" and pulse at "B" will reach to "A".
This means that the graph is contrary, the only one that fulfills this is the graph (e).
what is electric power
Answer:
Definition- Electric power is the rate at which electrical energy is transferred by an electric circuit.
Answer:
the rate at which electrical energy is transferred by an electric circuit.
Explanation:
Popeye the Sailor man, who has a mass of 85 kg, ran ( at a constant rate ) up a flight of stairs that are 3.55 m high in 6 seconds. How many watts of power did he generate during his run ?
Given data
*The given mass of the Sailorman is m = 85 kg
*The given height is h = 3.55 m
*The given time is t = 6 s
*The value of the acceleration due to gravity is
[tex]g=9.8m/s^2[/tex]The formula for the power generated by the Sailorman during his run is given as
[tex]\begin{gathered} P=\frac{W}{t} \\ =\frac{\text{mgh}}{t} \end{gathered}[/tex]*Here W is the work done
[tex]\begin{gathered} P=\frac{85\times9.8\times3.55}{6} \\ =492.85\text{ W} \end{gathered}[/tex]how a baby monster could have one eye even if both parents had two eyes.
Answer:
Both parents could be heterogeneous: Hh.
Explanation:
Both parents could be heterogeneous.
Both parents are Hh.
H = 2 eyes , dominant trait
h = one eye, recessive trait
The childs genotype is hh.
Find the force (N) on a 7.36cm radius piston on the other end of a hydraulic system that is driven by 23.37N of force from a 2.98cm radius piston.
The force (N) on a 7.36cm radius piston would be 142.55 Newtons, if on the other end of a hydraulic system that is driven by 23.37 N of force from a 2.98cm radius piston.
What is pressure?The total applied force per unit of area is known as the pressure.
The pressure depends both on externally applied force as well the area on which it is applied.
Pressure = Force / Area
As given in the problem we have to find out the force (N) on a 7.36cm radius piston on the other end of a hydraulic system that is driven by 23.37N of force from a 2.98cm radius piston,
By using the Pascal Law,
F₁ / A ₁ = F₂ / A₂
23.37 / 2.98² = F₂ / 7.36²
F₂ = 23.37 × 7.36² / 2.98 ²
= 142.55 Newtons
Thus, the force (N) on a 7.36cm radius piston would be 142.55 Newtons ,
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A 61 −kg ice skater coasts with no effort for 50 m until she stops. If the coefficient of kinetic friction between her skates and the ice is μk=0.10 , how fast was she moving at the start of her coast?
The 61 Kg ice skater was moving with a velocity of 9.9 m/s at the start of her coast.
How do I determine the initial velocity ?We'll begin by calculating the force. This can be obatined as follow:
Mass (m) = 61 KgAcceleration due to gravity (g) = 9.81 m/s²Normal reaction (N) = mg = 61 × 9.8 = 597.8 NCoefficient of kinetic friction (μK) = 0.10Force (F) = ?F = μKN
F = 0.1 × 597.8
F = 59.78 N
Next, we shall obtain the deceleration. This can be obtained as follow:
Force (F) = 59.78 NMass (m) = 61Deceleration (a) = ?a = -F / m (since the skater is coming to rest)
a = -59.78 / 61
a = -0.98 m/s²
Finally, we shall determine the initial velocity. This can be obtained as follow:
Distance (s) = 50 mFinal velocity (v) = 0 m/sDeceleration (a) = -0.98 m/s²Initial velocity (u) = ?v² = u² + 2as
0² = u² + (2 × -0.98 × 50)
0 = u² - 98
Collect like terms
u² = 0 + 98
u² = 98
Take the square root of both sides
u = √98
u = 9.9 m/s
Thus, the skater was moving with a velocity of 9.9 m/s
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How much work is done when a 25.0 kg object is lifted 3.00 m?
The amount of work done is 735J.
We all know that the potential energy U is equal to the work we must do against the force for moving an object from the reference point to the exact position.
The work done is equal to the potential energy through the work energy theorem.
W = mgh
here,
m = mass
g = gravitational acceleration
h = height
W = work done
On solving the above equation
W = 25x9.8x3
W = 75x9.8
W = 735 J
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