The maximum height reached by the arrows is determined as 8.6 m.
What is the initial speed of the arrow?The initial velocity of the arrow is calculated by applying the principle of conservation of energy as shown below;
K.E = U
where;
K.E is the kinetic energy of the arrowU is the elastic potential energy of the bow¹/₂mv² = ¹/₂kx²
mv² = kx²
v² = kx²/m
v = √(kx²/m)
where;
k is spring constant of the bowm is the mass of the arrowx is the extension of the bowv = √(195 x 0.39²/0.088)
v = 18.36 m/s
The maximum height reached by the arrow is calculated as follows;
H = (v² sin²θ) / (2g)
where;
θ is angle of projection of the arrowg is acceleration due to gravityH = (18.36² (sin45)²) / (2 x 9.8)
H = 8.6 m
Thus, the height of the arrow above the ground when it reaches its peak is 8.6 m.
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A heart defibrillator passes 10.3 A through a patient's torso for 5.00 ms in an attempt to restore normal beating.(a) How much charge passed?(b) What voltage was applied if 492 J of energy was dissipated?KV(c) What was the path's resistance?ΚΩ(d) Find the temperature increase caused in the 8.00 kg of affected tissue. The specific heat of tissue is 3500 J/(kg. °C).°C
a) The formula for calculating the quantity of charge is expressed as
Q = IT
where
Q is the quantity of charge
I is the current
T is the time
From the information given,
I = 10.3
T = 5 ms = 5 x 10^-3 s
Q = 10.3 x 5 x 10^-3
Q = 51.5 x 10^- 5 C
The quantity of charge passed is 51.5 x 10^- 5 C
b) The formula for calculating the energy is expressed as
E = I^2RT
where
R is the resistance
E is the energy
From the information given,
E = 492 J
Thus,
492 = 10.3^2 x R x 5 x 10^-3
R = 492/(5 x 10^-3 x 10.3^2)
R = 927.514 ohms
Voltage, V = IR
Voltage = 10.3 x 927.514
Voltage = 9553.398 V
We would divide by 1000. It becomes
Voltage = 9.553 KV
c) From the calculations,
Resistance = 927.514 ohms
We would divide by 1000. It becomes
Resistance = 0.93 ΚΩ
d) Let the temperature increase be t
mass of tissue, m = 8 kg
Specific heat of tissue = 3500 J/(kg. °C).
°C
The formula for calculating the quantity of heat is
H = mcθ
where
H is the quantity of heat
From the informtaion given,
H = 492
θ = t
Thus,
492 = 8 x 3500 x t
t = 492/(8 x 3500)
t = 0.018
The temperature increase is 0.018 degrees
Consider the graph shown. Which of the motions is consistent with the graph?a) The object has a constant velocity in the negative direction.b) The object is moving in the negative direction with a changing speed.c) The object is moving in the positive direction and slowing down.
Given:
The graph of velocity vs time of an object
To find:
Which of the motions is consistent with the graph?
Explanation:
We see here that the object's initial velocity is positive, and the final velocity is zero. So, the object is slowing down. as the initial velocity is positive, the object's direction of movement is positive.
Hence, The object is moving in a positive direction and slowing down.
In a lightning discharge, 45 C of charge move through a potential difference of 1.0 x108 V in 0.030 s.A. What is the current of the lightning strike?B. How much energy is released by the lightning bolt?
Given:
Charge, Q = 45 C
Potential difference, V = 1.0 x 10⁸ V
Time, t = 0.030 s
Let's solve for the following:
• (A). What is the current of the lightning strike?
To find the current, apply the formula:
[tex]I=\frac{Q}{t}[/tex]Where:
I si the current
Q is the charge = 45 C
t is the time = 0.030 s
Thus, we have:
[tex]\begin{gathered} I=\frac{45}{0.030} \\ \\ I=1500\text{ A} \end{gathered}[/tex]Therefore, the current of the lightning strike is 1500 Amperes.
• (B). How much energy is released by the lightning bolt?
To find the amount of energy released, apply the formula:
[tex]E=V\times Q[/tex]where:
E is the Energy released
V is the potential difference, V = 1.0 x 10⁸ V
Q is the charge = 45 C
Thus, we have:
[tex]\begin{gathered} E=1.0\times10^8\ast45 \\ \\ E=4.5\times10^9\text{ J} \end{gathered}[/tex]Therefore, the energy released is 4.5 x 10⁹ Joules.
ANSWER:
(a). 1500 A
(b). 4.5 x 10⁹ J
A NASA probe is moving horizontally above the surface of the moon at a constant speed to the right, as depicted in the diagram below. It releases an instrument package when it is directly above Point P. As seen from the lunar surface, which path would the package likely follow after the release and why?
B, because the gravity of the moon will pull the instrument to the ground with constant acceleration and the lack of an atmosphere allows the package to fall straight down
D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
C, because the reduced gravity of the moon pulls the package down vertically at a constant speed while the package travels horizontally at a constant speed resulting in a straight-line trajectory to the lunar surface.
A, because the gravity of the moon pulls the package down with constant acceleration, while the atmosphere of the moon creates horizontal drag on the package which reduces the horizontal component of the package’s velocity causing the package to be pulled backward as it falls
As seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant a, so that its vertical velocity increases.
A is incorrect because there is no wind pushing the package backwards. B is incorrect because the package has an initial velocity. C is incorrect because vertical velocity is not constant due to the presence of gravity. E and F are incorrect because gravity acts immediately after the package is dropped.
D is correct because the horizontal component remains constant because there is no horizontal force acting on the package. This is because in outer space there is no atmosphere, so there will be no air resistance. The vertical component increases with respect to time because of constant acceleration due to gravitational pull on the package.
Therefore, as seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
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A computer connected to a motion sensor creates velocity—time graph for a ball rolling down an incline. On the graph, the velocity increases by 0.3 m/s for every tenth of a second increment on the graph. What is the acceleration of the object? A. 0.3 m/s2. B. 1 m/s2. C. 3 m/s2. D. The acceleration cannot be determined from the given information.
We are asked to determine the acceleration of an object which speed changes by 0.3 m/s every 1/10 seconds. The definition of acceleration is the following:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]Where:
[tex]\begin{gathered} \Delta v=\text{ change in velocity} \\ \Delta t=\text{ change in time} \end{gathered}[/tex]Substituting the given values we get:
[tex]a=\frac{0.3\frac{m}{s}}{\frac{1}{10}s}[/tex]Solving the operations we get:
[tex]a=3\frac{m}{s^2}[/tex]Therefore, the acceleration is option C.
If the mass of the vase with flowers is 3.2 kg, what is the magnitude of the normal force?
The magnitude of normal force will be 31.36 N but in the opposite direction of the vase's weight.
We know that forces exist in pairs.
According to Newton's third law of motion,
" Every action has an equal and opposite reaction "
Also,
The second law of motion states, "The magnitude of a force is equal to the product of mass and acceleration acting upon it."
So, keeping in mind the above two laws:
The weight of the vase will be equal to:
W = mass of the vase × acceleration due to gravity
W = m × g
W = 3.2 kg × 9.8 m/s²
W = 31.36 N
So, the magnitude of normal will be the same as the Weight but the direction of Normal will be opposite to the direction of the weight.
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Ahmed wants to measure the weight of an object. What instrument should he use?
Answer:
a scale
Explanation:
you use it to weigh things
Answer:
its not a scale
Explanation:
An object has 25 J of kinetic energy and 35 J of potential energy. What isthe total energy possessed by the object? *
The total energy possessed by the object is 60 J
Given:
The kinetic energy of the object, K=25 J
The potential energy of the object, P=35 J
To find:
The total energy possessed by the object.
Explanation:
The kinetic energy of an object is the energy possessed by it due to its motion. The kinetic energy is directly proportional to the square of the velocity of the object.
The potential energy of an object is the energy possessed by it due to its position. The gravitational potential energy of an object is directly proportioinal to the mass ofthe
Moving a charge from point A, where the potential is 369.53 V, to point B, where the potential is 217.85 V, takes 0.55 milliJ of work. What is the value of the charge, in micro-Coulombs?
Given:
The potential of point A is
[tex]V_A=369.53\text{ V}[/tex]The potential of point B is
[tex]V_B=\text{ 217.85 V}[/tex]The work done is W = 0.55 milli Joule.
To find the value of charge in micro Coulomb.
Explanation:
The charge can be calculated as
[tex]\begin{gathered} Q=\frac{W}{V_B-V_A} \\ =\frac{0.55\times10^3\text{ J}}{217.85-369.53} \\ =-3.63\text{ C} \\ =\text{ -3.63 C}\times\frac{10^6\text{ }\mu C}{1\text{ C}} \\ =-3.63\times10^6\text{ }\mu C\text{ } \end{gathered}[/tex]3. An object of mass 8 kg is sliding down a friction lessinclined plane of length 11 m that makes an angleof 70 deg with the horizontal. Calculate thework done by gravitational force as the objectslides from the top of the inclined plane to theground. (1 point)A. 099.045 JB. O1135.872 JC.810.391 JD.499.917
Given
Mass of the object, m=8 kg
Length of the inclined plane, l=11 m
Angle of inclination,
[tex]\theta=70^o[/tex]To find
Calculate the work done by gravitational force as the object slides from the top of the inclined plane to the
ground.
Explanation
The height of the ramp
[tex]h=lsin\theta=11sin70^o[/tex]The work done by gravitational force,
[tex]\begin{gathered} W=mgh \\ \Rightarrow W=8\times9.8\times11sin70^o \\ \Rightarrow W=810.390J \end{gathered}[/tex]Conclusion
The work done is C.810.391
What is the resistance in a circuit that has a current of 2.5A and a voltage of 40v
16 ohms
Explanation
Ohm's law relates the strength of a direct current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit,it is given by the expresssion
[tex]\begin{gathered} V=IR \\ if\text{ we isolate R} \\ R=\frac{V}{I} \end{gathered}[/tex]then
Step 1
a) Let
[tex]\begin{gathered} I=2.5\text{ A} \\ V=40\text{ V} \end{gathered}[/tex]b)replace in the formula
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{40V}{2.5A} \\ R=16\text{ ohms} \end{gathered}[/tex]therefore, the answer is 16 ohm
I hope this helps you
Imagine that you are on board a ship that was struck by a rogue wave.
Tell your story, from the calm before the wave hit to its aftermath.
The possibility of a wave toppling cruise liner is extremely low. They are made wide and have enough ballast on the lower decks to be heavy enough rogue waves. On the side, the crew's neglect also be necessary.
What are rogue waves?A wave that is double the region's major wave height is typically considered a rogue wave. The highest one-third of waves on average over a period of time make up the noteworthy wave height. Even the biggest ships and oil rigs can be rendered useless and sunk by rogue waves.
Have rogue waves ever struck a cruise ship?Rogue waves have occasionally hit cruise ships, although it is not frequently. Four cruise ships have collided with rogue waves since rogue wave records were first kept in 1995. All sustained damage, and some people reported injuries, but there have been no confirmed fatalities on cruise ships due to rogue waves.
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The acceleration of an object is ___proportional to the net force and ___ proportional to its mass? Can you tell me which ones it would be from these options?directly, directlydirectly, inverselyinversely, inverselyinversely, directly
Answer:
the acceleration of an object is directly proportional to the net force and inversely proportional to its mass.
Explanation:
when the net force on an object goes up, so does the acceleration, meaning one increases as the other increases
the acceleration of an object decreases as the mass increases, therefore this relationship is inversely proportional
A plane is traveling with a velocity of 70 miles/hr with a direction angle of 24 degrees. The wind is blowing at 25 miles/hr with a direction angle of 190 degrees. What is the vertical component of the wind velocity? Round your answer to the nearest whole number.
Wind velocity:
25 m/h with a direction angle of 190°.
Vertical component:
25 sin 190 = -4.34 m/s = - 4 m/s
Tritium has a half-life of 12.3 years. How many years will have elapsed when the radioactivity of a tritium sample has decreased to 10 percent of its original value?
Given
Half life of tritium is
[tex]t_{\frac{1}{2}}=12.3\text{ years}[/tex]The sample is reduced to 10% of its original value.
To find
How many years will have elapsed when the radioactivity of a tritium sample has decreased to 10 percent of its original value?
Explanation
The activity is given by
[tex]\begin{gathered} A=A_oe^{-\lambda t} \\ \Rightarrow0.1A_o=A_oe^{-\frac{0.693}{t_{\frac{1}{2}}}t} \\ \Rightarrow ln(0.1)=-\frac{0.693}{1.23}t \\ \Rightarrow-2.302=-\frac{0.693}{1.23}t \\ \Rightarrow t=4.08 \end{gathered}[/tex]Conclusion
The time taken is 4.08 years
A 6 kg object is being pulled by a horizontal force F=120 N on a friction-less horizontal surface. It moved a distance of 18 m. If its initial kinetic energy was 100 Joules, what is the final kinetic energy in Joules?
The final kinetic energy is 120m.
What is Work-Energy Theorem?
The Work-Energy Theorem states that the work done is equal to the change in the K.E. i.e Kinetic Energy of the object.
W = Δ(K.E.)
In the given question we had,
Mass = 6 kg,
Force = 120 N,
Distance = 18 m,
Initial Kinetic Energy ( KE1 ) = 100 Joules
According to Work-Energy Theorem,
W = Δ(K.E.)
W = KE2-KE1
F x S = KE2 - 100
120 x 18 = KE2 - 100
2160 + 100 = KE2
2260J = KE2
So, the final kinetic energy is 2260J.
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A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)
Given:
Initial velocity, vi = 41.0 m/s
Mass of ball, m = 195 g = 0.195 kg
Final velocity, vf = 37.0 m/s
Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:
• (A). What is the impulse delivered to the ball by the bat?
To find the impulse, apply the change in momentum formula:
[tex]\Delta p=p_f-p_i[/tex]Where:
pi is the initial momentum = -mvi
pf is the final momentum = mvf
Thus, we have:
[tex]\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Impulse can be said to equal change in momentum.
Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.
• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?
Apply the formula:
[tex]\text{ Impulse = Force }\ast\text{ time}[/tex]Rewrite the formula for force:
[tex]\text{ Force=}\frac{impulse}{time}[/tex]Where:
time = 3.00 m/s
impulse = 15.21 kg.m/s
Hence, we have:
[tex]\begin{gathered} \text{ F=}\frac{15.21}{3} \\ \\ F\text{ = 5.07 kN} \end{gathered}[/tex]Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.
ANSWER:
(A). 15.21 kg.m/s away from the bat
(B). 5.07 kN.
I need help on this science homework I forgot on what to do through number 4-7 including a,b,c,d, and e.
Answer:
[tex]\begin{gathered} (a)\Rightarrow v=2ms^{-1} \\ (b)\Rightarrow v=1ms^{-1} \\ (c)\Rightarrow v=6.67ms^{-1}_{} \\ (d)\Rightarrow v=1.2ms^{-1} \\ (e)\Rightarrow v=0ms^{-1} \end{gathered}[/tex]Explanation: We need to calculate the speed on intervals a b c d and e, the speed can be calculated with the following formula:
[tex]v=\frac{\Delta S}{\Delta t}\Rightarrow(1)[/tex](a) 0-5 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(10m-0m)}{(5s-0s)}=\frac{10m}{5s}=2ms^{-1} \\ v=2ms^{-1} \end{gathered}[/tex](b) 5-15 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(20m-10m)}{(15s-5s)}=\frac{10m}{10s}=1ms^{-1} \\ v=1ms^{-1} \end{gathered}[/tex](c) 15-18 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(40m-20m)}{(18s-15s)}=\frac{20m}{3s}=6.67ms^{-1} \\ v=6.67ms^{-1} \end{gathered}[/tex](d) 18-23 seconds Interval:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(46m-40m)}{(23s-18s)}=\frac{6m}{5s}=1.2ms^{-1} \\ v=1.2ms^{-1} \end{gathered}[/tex](e) 23-25 seconds Intervals:
[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(40m-40m)}{(25s-23s)}=\frac{0m}{2s}=0ms^{-1} \\ v=0ms^{-1} \end{gathered}[/tex]A duck is paddling due East at 1.1 m/s across the river while it flows due South at 0.55 m/s What is her resultant velocity?
The diagram representing this scenario is shown below
A right angle triangle is formed. R represents the resultant velocity. To find R, we would apply the pythagorean theorem which is expressed as
hypotenuse^2 = one leg^2 + other leg^2
From the diagram,
hypotenuse = R
one leg = 1.1
other leg = 0.55
Thus,
R^2 = 1.1^2 + 0.55^2 = 1.5125
Taking square root of both sides,
R = square root of 1.5125
R = 1.23
The resultant velocity is 1.23 m/s
The crazed physic's student's lab partner decides to throw another pumpkin off the
roof with an initial velocity of 19.4 m/s. What is the velocity when the pumpkin
strikes the ground if it takes 3.2 seconds for it to fall?
Please Help :(
The velocity when the pumpkin strikes the ground if it takes 3.2 seconds for it to fall: 50.78 m/s
From the definition of velocity, we can find the velocity of a falling object is:
v = v₀ + gt
Here
v₀ - 19.4 m/s , t - 3.2 seconds , g - 9.80665 m/[tex]s^2\\[/tex]
v = 19.4 + (9.86 x 3.2)
v = 50.78 m/s
What is velocity?
Velocity and speed describe how quickly or slowly an object is moving. We frequently encounter circumstances when we must determine which of two or more moving objects is going faster. If the two are travelling on the same route in the same direction, it is simple to determine which is quicker. It is challenging to identify who is moving the fastest when their motion is in the other direction. The concept of velocity is useful in these circumstances. Learn about the definition of velocity in this article as well as the distinction between speed and velocity.
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A kid is on a stationary sled, onsnowy ground with fls 0.105.It takes 71.2 N of force to setthe sled moving. How muchnormal force is acting?(Unit = N)
ANSWER
[tex]678.10N[/tex]EXPLANATION
Parameters given:
Coefficient of static friction, μs = 0.105
Force, F = 71.2 N
The minimum force required to set the sled moving must be equal to the frictional force acting on the sled.
The frictional force is given mathematically as:
[tex]F_f=\mu_sN[/tex]where N = normal force
Since the force required is equal to the frictional force, we have that:
[tex]F=F_f[/tex]Therefore:
[tex]F=\mu_sN[/tex]Solve for N:
[tex]\begin{gathered} N=\frac{F}{\mu_s} \\ N=\frac{71.2}{0.105} \\ N=678.10N \end{gathered}[/tex]That is the normal force.
state the dimension of energy in physics
hint: Energy = force × distance
Force = Mass * Acceleration = kg* m/s^2= MLT^-2
Distance = metres= L
Energy = MLT^-2 * L =ML^2T^-2
Two long parallel wires 0.552 meter apart are each carrying 1.75 amperes of current, as shown(a) Find the magnitude and direction of the magnetic field at point A due to the current in the top wire. (b) Find the magnitude and direction of the force per unit length this field exerts on the bottom wire.
,Given,
Distance between two-wire, d=0.552 m
Current through the wires, I=1.75 A
(a) The magnetic field at a point on the bottom wire due to top wire is given by,
[tex]B_a=\frac{\mu_0I_1}{2\pi d}[/tex]Where μ₀ is the permeability of the free space.
The direction is given by the right-hand thumb rule. According to this, the direction of the magnetic field produced by top wire will into the plane of the two wires.
On substituting the known values in the above equation,
[tex]B_a=\frac{4\pi\times10^{-7}\times1.75}{2\pi\times0.552}=6.34\times10^{-7}\text{ T}[/tex]The force per unit length is given by,
[tex]F=I_2\times B_a[/tex]The direction is given by the right-hand rule. According to this the force is directed towards the top wire.
On substituting the known values in the above equation,
[tex]F=1.75\times6.34\times10^{-7}=1.11\times10^{-6}\text{ N}[/tex]Therefore the magnetic field acting on the bottom wire due to the current in the top wire is 6.34×10⁻⁷ T and the magnetic force due to this field is 1.11×10⁻⁶ N
which are neutrally charged, are found in thenucleus of the atom.
Given:
Nucleus of the atom
Required:
Neutra
I got first part correct but dont know how to solve second part: Two new particles with identical positive charge 3 are placed the same 0.0809 m apart. The force between them is measured to be the same as that between the original particles. What is 3 ?
Answer:
5.92 *10^-6 C
Explanation:
For the two charges q3 the force between them is given by
[tex]F=k\frac{q_3\times q_3}{d^2}[/tex]Now we know that
F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives
[tex]48.1=(8.99\times10^9)\frac{q_3\times q_3}{(0.0809)^2}[/tex][tex]\Rightarrow48.1=(8.99\times10^9)\frac{(q_3)^2}{(0.0809)^2}[/tex]Now we solve for q_3.
Dividing both sides by 8.99 * 10^9 gives
[tex]\frac{48.1}{(8.99\times10^9)}=\frac{(q_3)^2}{(0.0809)^2}[/tex]multiplying both sides by (0.0809)^2 gives
[tex]\frac{48.1}{(8.99\times10^9)}\times\mleft(0.0809\mright)^2=(q_3)^2[/tex]finally, taking the square root of both sides gives
[tex]\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}=\sqrt{(q_3)^2}[/tex][tex]q_3=\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}[/tex]Evaluating the right-hand side gives
[tex]\boxed{q_3=_{}5.92\times10^{-6}C\text{.}}[/tex]Hence, the charge q_3 is 5.92 x 10^-6 C.
Three vectors are shown in this figure. Their respective moduli are A = 4.00m.B = 3, 20m and C = 2.70mCalculate 2.00 A - B + 1.30 CExpress your answer according toa) Unit vectorsb) The modulus and orientation with respect to the positive part of the x-axis
Given that,
Modulus of vector A=4.00
The angle made by the vector A with the y axis, θ₁=33.0°
The modulus of vector B=3.20 m
The angle made by the vector B with the x-axis is θ₂=40.0+90.0=130°
The modulus of the vector C=2.70 m
The angle made by the vector C with x-axis θ₃=-90°
The x and y components of the vector can be written as
[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]Where r is the magnitude (or modulus) of the vector and θ is the angle made by the vector.
Or a vector, in cartesian coordinates, can be written as,
[tex]R=r\cos \theta\hat{\text{i}}+r\sin \theta\hat{j}[/tex]Therefore, vector A is cartesian coordinates is
[tex]\begin{gathered} \vec{A}=4.00\cos 33^{\circ}\hat{i}+4.00\sin 33.0^{\circ}\hat{j} \\ =3.35\hat{i}+2.18\hat{j} \end{gathered}[/tex]And the vector B is
[tex]\begin{gathered} \vec{B}=3.20\cos (130^{\circ})\hat{i}+3.20\sin (130^{\circ})\hat{j} \\ =-2.06\hat{i}+2.45\hat{j} \end{gathered}[/tex]And vector C is given by,
[tex]\begin{gathered} \vec{C}=2.70\cos (-90^{\circ})\hat{i}+2.70\sin (-90^{\circ})\hat{j} \\ =-2.7\hat{j} \end{gathered}[/tex]The given equation is
[tex]2.00\vec{A}-\vec{B}+1.30\vec{C}[/tex]Let this represents a vector V
On substituting the known values,
[tex]\begin{gathered} \vec{V}=2.00\vec{A}-\vec{B}+1.30\vec{C} \\ =2.00\times(3.35\hat{i}+2.18\hat{j})-(-2.06\hat{i}+2.45\hat{j)}+1.30(-2.7\hat{j}) \\ =8.76\hat{i}-1.6\hat{j} \end{gathered}[/tex](a) This is the representation with the unit vectors, where i and j are the unit vectors along the x-axis and y-axis respectively.
[tex]\vec{V}=8.76\hat{i}-1.6\hat{j}[/tex]b) The modulus of any vector is the square root of the sum of the squares of its components.
That is, the magnitude of the vector V is
[tex]\begin{gathered} V=\sqrt[]{8.76^2+(-1.6)^2} \\ =8.90\text{ m} \end{gathered}[/tex]The angle of this vector with the x-axis is given by
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{-1.6}{8.75}) \\ =-10.36^{\circ}^{} \end{gathered}[/tex]The negative sign indicates that the vector is below the positive x-axis
Therefore the modulus of the resultant of the above equation is 8.90 m and its angle with the positive x-axis is -10.36°
A 6.5 kg lump of clay is sliding to the right on a fricitonless surface with a speed of 23 m/s. It collides head-on and sticks to a 2 kg metal sphere that is sliding to the left with a speed of -7 m/s. What is the kinetic energy of the combined objects after the collision?
The kinetic energy of the combined objects after the collision = 1768.25 Joules
Explanation:The mass of the lump of clay, m₁ = 6.5 kg
The speed of the lump of clay, v₁ = 23 m/s
The mass of the metal sphere, m₂ = 2 kg
The speed of the metal sphere, v₂ = -7 m/s
The Kinetic Energy (KE) of the combined objects after collision is calculated as shown below:
[tex]KE=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2^{}[/tex][tex]\begin{gathered} KE=\frac{1}{2}(6.5)(23^2)+\frac{1}{2}(2)(-7)^2 \\ KE=1719.25+49 \\ KE\text{ = }1768.25J \end{gathered}[/tex]The kinetic energy of the combined objects after the collision = 1768.25 Joules
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The time it takes for the moon to rotate once on its axis is the same amount of time it takes for
the moon to revolve around the Earth once. The result is that the SAME SIDE OF THE MOON
ALWAYS FACES THE EARTH.
The amount of time it takes for the moon to complete one full rotation around the Earth is the same as its period of one full rotation on its axis. Due to this, the EARTH ALWAYS SEE THE SAME SIDE OF THE MOON.
What does the earth's rotation entail?With an inclination of 23.45 degrees from the plane of its orbit around the sun, the Earth revolves on its axis in relation to the sun every 24.0 hours mean solar time. The differences brought on by the Earth's are averaged out to create mean solar time.
What occurs when the Earth rotates?Rotation causes the day-night cycle, which in turn induces a cycle of temperature and humidity. As the world spins, the sea level rises and falls twice each day. The tidal range is determined by the gravitational pull of the sun and moon together
To know more about rotation around the earth visit:
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Sports biomechanics looks at athletes to better understand how the body:
A. grows.
B. creates energy.
C. moves.
D. resists disease.
Answer:
The ans is C
Explanation:
In a sporting context, biomechanics examines an athlete in relation to his or her environment and equipment. Forces acting on the body (kinetics) and movements of the body (kinematics) are analysed prior to mapping out an exercise, performance or recovery plan.
Whats the percent of 10 of 20
In order to determine the associated percent of 10 related to 20, proceed as follow:
If x is the percentage, then, you can write:
[tex]\frac{x}{100}\cdot20=10[/tex]which means that x percentage of 20 is equal to 10. By solving for x, you get:
[tex]\begin{gathered} x=\frac{10}{20}\cdot100 \\ x=50 \end{gathered}[/tex]Hence, 10 is the 50% of 20
Answer:2
Explanation:multiply 0.20 times 10 you get it