Explanation:
Lodestone am iron were the only known magnetic materials in Gilbert's day, and his task was to investigate magnetism. Gilbert was so sure that the earth was a giant lodestone and used the earth as a primary reference, defining the north(magnetic) pole of a needle, or a a nail floating on a piece of cork, to be that which turns towards the Earth's north geographic pole. he wanted to prove this with a model Terella, using short pieces of iron.
what is the twisted ladder shape of the DNA called?
Answer:
Double helix
Explanation:
The Double helix is a DNA molecule. The two strands around the Double Helix is called the twisted ladder.
Answer:
double helix
Explanation:
A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m. It climbs a frictional incline of height 2.9 m inclined at an angle 16°, then moves along a second frictional surface of coefficient 0.1 before coming to rest.
The acceleration of gravity is 9.8 m/s^2. If the first frictional surface has a coefficient of 0.21 for a distance 1 m, how far does it slide along the second frictional region before coming to rest?
Answer:
[tex]D=99.4665307m \approx 99.5m[/tex]
Explanation:
From the question we are told that
Mass [tex]m=6kg[/tex]
Velocity of mass [tex]V_m=16[/tex]
Force of Tunnel [tex]F_t=8N[/tex]
Length of Tunnel [tex]L_t=1.6[/tex]
Height of frictional incline [tex]H_i=2.9[/tex]
Angle of inclination [tex]\angle =16 \textdegree[/tex]
Acceleration due to gravity [tex]g=9.8m/s^2[/tex]
First Frictional surface has a coefficient [tex]\alpha_1 =0.21\ for\ d_c=1[/tex]
Second Frictional surface has a coefficient [tex]\alpha _2=0.1[/tex]
Generally the initial Kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]K.E=\frac{1}{2}(6)(16)^2[/tex]
[tex]K.E=768[/tex]
Generally the work done by the Tunnel is mathematically given as
[tex]w_t=F_t*d_t[/tex]
[tex]w_t=8*1.6[/tex]
[tex]w_t=12.8J[/tex]
Therefore
[tex]Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t[/tex]
[tex]E_t=K.E+E_t\\E_t=768J+12.8J[/tex]
[tex]E_t=780.8J[/tex]
Generally the energy lost while climbing is mathematically given as
[tex]E_c=mgh[/tex]
[tex]E_c=(6)(9.8)(2.9)[/tex]
[tex]E_c=170.52J[/tex]
Generally the energy lost to friction is mathematically given as
[tex]E_f=\alpha *m*g*cos\textdegree*d_c[/tex]
[tex]E_f=0.21*6*9.8*cos16*1[/tex]
[tex]E_f=11.86965942 \approx 12J[/tex]
Generally the energy left in the form of mass [tex]Em[/tex] is mathematically given as
[tex]E_m=E_t+E_c+E_f[/tex]
[tex]E_m=(768J)-(170.52)-(12)[/tex]
[tex]E_m=585.48J[/tex]
Since
[tex]E_m=\alpha_2*g*m*d[/tex]
Therefore
It slide along the second frictional region
[tex]D=\frac{585.46}{0.1*9.81*6}[/tex]
[tex]D=99.4665307m \approx 99.5m[/tex]
Jogger A has a mass m and a speed v, jogger B has a mass m/2 and a speed 3v, jogger C has a mass 3m and a speed v/2, and jogger D has a mass 4m and a speed v/2. Rank the joggers in order of increasing kinetic energy. Indicate ties when appropriate. Show your work.
Answer: B>A=D>C
Explanation:
Kinetic Energy is the product of mass and square of the velocity
For Jogger A
[tex]K.E._a=\frac{1}{2}mv^2[/tex]
For Jogger B
[tex]K.E._b=\frac{1}{2}\times\frac{m}{2}\times(3v)^2=\frac{9}{4}mv^2\\K.E._b=2.25mv^2[/tex]
For Jogger C
[tex]K.E._c=\frac{1}{2}\times3m\times(\frac{v}{2})^2=\frac{3}{8}mv^2\\K.E._c=0.375mv^2[/tex]
For Jogger D
[tex]K.E._d=\frac{1}{2}\times4m\times(\frac{v}{2})^2=\frac{1}{2}mv^2[/tex]
Kinetic Energy of Joggers in increasing order
B>A=D>C
Please help with an explanation for 30 points even if u only do one (but with good explanation)
Answer:
1- 4Ω
Explanation:
if 2Ω is 2 meters, than 4 metres is 4Ω
Which characterictic of motion could change without changing the velocity of an object
Answer:
The direction could change
A soccer ball, mass 3 kg, is kicked straight up with a velocity of 25 m/s. How high will the ball travel?
Answer:
h = 31.9 m
Explanation:
When the ball travels from the ground to the highest point, its kinetic energy is converted into potential energy. So by the law of conservation of energy:
[tex]Kinetic\ Energy\ Lost\ by\ ball = Potential\ Energy\ gained\ by\ ball\\\frac{1}{2}mv^{2} = mgh\\\\h = \frac{v^{2}}{2g}[/tex]
where,
h = height gained by ball = ?
v = speed of ball = 25 m/s
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(25\ m/s)^{2}}{2(9.8\ m/s^{2})}[/tex]
h = 31.9 m
The distance between a loudspeaker and the left ear of a listener is 2.70 m. (a) Calculate the time required for sound to travel this distance if the air temperature is 20 oc. (b) Assuming that the sound frequency is 523 Hz, how many wavelengths of sound are contained in this distance
Answer:
Explanation:
Distance covered by sound = 2.7 m
speed of sound at 20⁰C = 343 m /s
time take by sound to cover the distance = distance / speed
= 2.7 / 343
= 7.87 ms . ( millisecond )
b )
Wavelength of sound = speed / frequency
= 343 / 523 m
= .6558 m
= 65.58 cm
Distance = 2.70 m = 270 cm
No of wavelengths contained in the distance
= 270 / 65.58
= 4.11 or 4 wavelengths ( by rounding off to digit )
The time taken by sound and the wavelengths of sound is required.
Time taken is [tex]0.00787\ \text{s}[/tex]
The number of wavelenghts is 4.
s = Distance = 2.7 m
v = Speed of sound at [tex]20^{\circ}\text{C}[/tex] = 343 m/s
Time is given by
[tex]t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{2.7}{343}\\\Rightarrow t=0.00787\ \text{s}[/tex]
f = Frequency = 523 Hz
Wavelength is given by
[tex]\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{523}\\\Rightarrow \lambda=0.66\ \text{m}[/tex]
The wavelength is 0.66 m.
n = Number of wavelengths
[tex]n\lambda=s\\\Rightarrow n=\dfrac{s}{\lambda}\\\Rightarrow n=\dfrac{2.7}{0.66}\\\Rightarrow n\approx 4[/tex]
The number of wavelenghts is 4.
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A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0260 kg, and its speed is 42.8 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = [tex]p_{fx} - p_{ox}[/tex]
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m [tex]v_{fx}[/tex] - m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
[tex]\Delta p_{y}[/tex] = 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the charge Q is distributed uniformly, in the shell between r1 and r2, determine the electric field, magnitude and direction, in the following situations:______.
a. From r=0 to r=r1.
b. From r=r1 to r=r2.
c. Outside of r=r2.
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1250 kg when fired, had a muzzle velocity of 750 m/s. Assuming a constant force, determine the explosive force experienced by the shell inside the barrel. Start from a fundamental principle.
Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})[/tex] (1)
Where:
[tex]F[/tex] - Explosive force, measured in newtons.
[tex]\Delta s[/tex] - Barrel length, measured in meters.
[tex]m[/tex] - Mass of the shell, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the shell, measured in meters per second.
If we know that [tex]m = 1250\,kg[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 750\,\frac{m}{s}[/tex] and [tex]\Delta s = 15\,m[/tex], then the explosive force experienced by the shell inside the barrel is:
[tex]F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}[/tex]
[tex]F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}[/tex]
[tex]F = 23437500\,N[/tex]
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.4 m/s in the positive x direction and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time
Answer:
33.5J
Explanation:
Given:
Mass of the canister= 3.1 kg
Initial velocity of canister= v(i)= 4.4i m/s
Final velocity of canister= v(f)= 6.4j m/s
Force magnitude( xy plane)= 5 N
The magnitude of vector V'= Vxi + Vyj + Vzk
|V|= √( Vx^2 + Vy^2 + Vz^2
From Kinectic energy and work theorem.
Net work = Kinectic energy of the canister
ΔK= W
(Kf - Ki)= W
Where Kf= final Kinectic energy
= 1/2 mv^2
If we input the given values we have,
= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J
Ki= initial Kinectic energy
= 1/2 mv^2
If we substitute the given values we have
=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J
Work done by canister = (final Kinectic energy - initial energy)
= 63.5- 30
=33.5J
Hence, work done on the canister 33.5J
Which is the best tool to use when measuring mass
Beam balance is used to determine the mass of an object.
How long has a victim been dead if his body temperature was 89.2°F?
An RLC parallel circuit has an applied voltage of 240 volts. R= 60 ohms, XL = 20 ohms, and Xc =36 ohms. What is the capacitor current?
Answer:
6.67A
Explanation:
The voltage across the capacitor formula is expressed as;
VL = IXL
VL is the voltage across the capacitor = 240volts (since it is a parallel connection, all the elements will have the same voltage)
I is the capacitor current
XL is the capacitive reactance = 36 ohms
Recall from the formula:
VL = IXL
I = VL/XL
I = 240/36
I = 6.67A
Hence the capacitor current is 6.67A
A spring has a spring constant of 200 N/m. How much elastic potential energy is stored in the spring when it is stretched by 0.1 meters? A:2000 J B:20 J C:0 J D:1 J
Answer:
D, 1J
Explanation:
PE=1/2kx^2 and plug in the variables.
PE=1/2 x 200 x 0.1^2= 1J
A space station consists of three modules, connected to form an equilateral triangle of side length 82.0 m. Suppose 100 people, with an average mass of 75.0 kg each, live in each capsule and the mass of the modules is negligible compared to the mass of the people. At the current rotational rate the effective acceleration of gravity is g/2. (a) What angular momentum of the system
Answer:
Angular momentum of the system is 16221465.4617 kgm²/s
Explanation:
Given that;
length of the side of the triangle L = 82 M
m = 75.0 kg × 100 = 7500 kg
distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m
effective acceleration a = 9.8 / 2 = 4.9 m/s²
we know that; effective acceleration is being provided by centripetal acceleration.
so
a = R × w²
rate of rotation w = √( a / R) = √( 4.9 / 47.34) = 0.3217 rad/seconds
Moment of Inertia I = 3mR²
we substitute
I = 3 × 7500 × (47.34)²
Also, Angular momentum L is expressed as;
L = I × w
so
L = 3 × 7500 × (47.34)² × 0.3217
L = 16221465.4617 kgm²/s
Therefore, Angular momentum of the system is 16221465.4617 kgm²/s
The Angular momentum of the system is 16221465.4617 kgm²/s
The calculation is as follows:Given that;
Length of the side of the triangle L = 82 M
[tex]m = 75.0 kg \times 100 = 7500 kg[/tex]
Now
distance of each vertex from center R = L/√3
= 82/√3
= 47.34 m
Now
effective acceleration [tex]a = 9.8 \div 2[/tex] = 4.9 m/s²
So,
a = R × w²
Now
rate of rotation [tex]w = \sqrt( a \div R) = \sqrt( 4.9 \div 47.34)[/tex] = 0.3217 rad/seconds
Moment of Inertia I = 3mR²
Now
[tex]I = 3 \times 7500 \times (47.34)^2[/tex]
Also, Angular momentum L is expressed as;
L = I × w
so
L = 3 × 7500 × (47.34)² × 0.3217
= 16221465.4617 kgm²/s
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A uniform ladder whose length is 5.2 m and whose weight is 400 N leans against a frictionless vertical wall. The coefficient of static friction between the level ground and the foot of the ladder is 0.35. What is the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping
Answer:
the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping is 3.3424 m
Explanation:
Given the data in the question and as illustrated in the images below;
without the ladder immediately slipping, the net torque and the net force mus all balance out.
from the first image;
In the x, the force is;
[tex]F_{f}[/tex] = N₂
mg = N₁
the torque about the ground contact point gives the following equation
N₂Lsin∅ = mgcos∅[tex]\frac{L}{2}[/tex]
solving for ∅
tan∅ = mg / 2N₂ 55
∅ = tan⁻¹ ( mg / 2N₂ )
we already know that N₂ = [tex]F_{f}[/tex] = μN₁ = μmg
so,
∅ = tan⁻¹ ( mg / 2μmg )
∅ = tan⁻¹ ( 1 / 2μ )
given that; The coefficient of static friction between the level ground and the foot of the ladder μ = 0.35
we substitute
∅ = tan⁻¹ ( 1 / (2×0.35 ) )
∅ = tan⁻¹ ( 1.42857 )
∅ = 55°
now to get the required distance;
from the second image; cos∅ = d / L
d = Lcos∅
given that; length of the ladder = 5.2 m
we substitute
d = 5.2cos(50)
d = 3.3424 m
Therefore, the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping is 3.3424 m
Suppose you have two metal spheres that are exactly the same size, separated by a very large distance. Sphere A carries an excess of eight negative charges while sphere B carries an excess of two positive charges. Sphere A is momentarily connected to Sphere B using a metal wire and then the wire is removed. Draw charge diagrams of the two spheres for the two cases below before and after the connecting wire is used.
1. Before the spheres are connected using the wire .
2. After the wire has been removed .
Answer:
a) phere A has 8 negative charges and sphere B has 2 positive charges
b) each one has then 3 negative charges
Explanation:
In this case, it is asked to determine the charge of the spheres in two conditions
a) Before connecting the cable, sphere A has 8 negative charges and sphere B has 2 positive charges
b) After connecting the cable, as the spheres are metallic, the load is distributed, we have
q = 8 -2 = 6 negative charges
these charges are distributed between the two spheres, each one has then 3 negative charges
calculated the pressure exerted when a force of 100N is applied on a area of 25m2
Answer:
P = 4[Pa]
Explanation:
Pressure is defined as the relationship between Force over area. So we can use the following equation.
[tex]P=F/A[/tex]
where:
P = pressure [Pa] (units of Pascals)
F = force = 100 [N]
A = area = 25 [m²]
Now replacing:
[tex]P=100/25\\P=4[Pa][/tex]
[tex]\\ \rm\longmapsto P=\dfrac{F}{A}[/tex]
[tex]\\ \rm\longmapsto P=\dfrac{100}{25}[/tex]
[tex]\\ \rm\longmapsto P=4Pa[/tex]
What is Nature? How does it explain the human condition?
34000×0.4
----------------
0.02×200
In scientific notation
Answer:
the simplified expression is written as 3.4 x 10³
Explanation:
Given expression;
[tex]\frac{34000\times 0.4}{0.02 \times 200}[/tex]
in scientific notation, the expression is simplified as;
[tex]\frac{34000\times 0.4}{0.02 \times 200} = \frac{13600}{4} = 3400 = 3.4 \times 10^3[/tex]
Therefore, in scientific notation, the simplified expression is written as 3.4 x 10³
A boy throws a stone straight upward with an initial
speed of 17.0 m/s.
Part A
What maximum height will the stone reach before falling back down?
Answer:
14.8m
Explanation:
Given parameters:
Initial speed = 17m/s
Unknown:
Maximum height = ?
Solution:
At the maximum height, the final speed will be 0m/s;
We use of the kinematics equation to solve this problem.
V² = U² - 2gH
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height
0² = 17² - (2 x 9.8 x h )
0 = 289 - (9.6h)
-289 = -19.6h
h = 14.8m
Bruce pulls a spring with a spring constant k=100 Nmk=100\, \dfrac{\text N}{\text m}k=100mNk, equals, 100, start fraction, start text, N, end text, divided by, start text, m, end text, end fraction, stretching it from its rest length of 0.20 m0.20\,\text m0.20m0, point, 20, start text, m, end text to 0.40 m0.40\,\text m0.40m0, point, 40, start text, m, end text.What is the elastic potential energy stored in the spring?
Answer:
K_{e} = 2.0 J
Explanation:
In this exercise you are asked to calculate the elastic potential energy of a spring
[tex]K_{e}[/tex] = ½ k x²
where k is the spring constant and x is the displacement from equilibrium position
In this exercise, indicate that the spring constant is k = 100 N/m, the length at rest is x₀ = 20 cm = 0.20 m, up to the position x₁ = 40 cm = 0.40 m, therefore the elongation
Δx = x₁ - x₀
Δx = 0.40 - 0.20
Δx = 0.20 m
let's calculate the elastic potential energy
K_{e} = ½ 100 0.20²
K_{e} = 2.0 J
A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 193.0 km/h. The conditions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.80 m/s2. What is the minimum required length of the runway
Answer:
513 m
Explanation:
We have;
final speed of the airplane = 193.0 km/h * 1000/3600 = 53.6 m/s
acceleration of the air plane = 2.80 m/s2
initial velocity of the airplane = 0 m/s
length of the runway = distance covered
v^2 = u^2 + 2as
v^2 - u^2 = 2as
s = v^2 - u^2/2a
s = (53.6)^2 - 0^2/ 2 * 2.80
s = 2872.96/ 5.6
s = 513 m
3N
3 N
What is the net force of the box?
A mimibus drives with a constant speed of 39 km/h. how far can it travel in 1.94 hours?
Answer:
The minibus traveled 75.66 km
Explanation:
Motion with Constant Speed
An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.
The formula to calculate the speed is:
[tex]\displaystyle v=\frac{d}{t}[/tex]
Where
v = Speed of the object
d = Distance traveled
t = Time taken to travel d.
From the equation above, we solve for d:
d = v . t
The minibus has a constant speed of v=39 km/h and it's required to find the distance it travels in t=1.94 hours.
Calculating the distance:
d = 39 km/h * 1.94 h
d = 75.66 km
The minibus traveled 75.66 km
1.
Atennis ball is shot straight up with an initial velocity of 34 m/s. What is its velocity two seconds after launch?
Answer:
The speed (magnitude of the velocity) is 14.4 m/s
Explanation:
Vertical Launch Upwards
It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the speed vf at any time is calculated by:
[tex]v_f=v_o-g.t[/tex]
A tennis ball is launched vertically up with an initial speed of vo=34 m/s. At time t=2 s, its speed is:
[tex]v_f=34-9.8*2[/tex]
[tex]v_f=34-19.6[/tex]
[tex]v_f=14.4\ m/s[/tex]
The speed (magnitude of the velocity) is 14.4 m/s
____ and____ are 2 major atmospheric gases
Answer:
Nitrogen and Oxygen are the two major atmospheric gases.
A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision. What is the final velocity of the two combined vehicles?
Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
[tex](m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}[/tex]
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
[tex](1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s][/tex]
Please help me......
