The electric potential is given by:
[tex]\begin{gathered} V=\frac{Kq}{r} \\ \end{gathered}[/tex]Let's find r first:
[tex]\begin{gathered} r=\frac{Kq}{V}=\frac{8.988\times10^9\cdot2.42\times10^{-9}}{293.97} \\ r\approx0.074m \end{gathered}[/tex]Now we can find the radius of the new drop:
[tex]r_t=2(r)=2(0.074)=0.148[/tex]So:
[tex]\begin{gathered} V=\frac{K(2q)}{r_t}=\frac{8.988\times10^9\cdot2(2.42\times10^{-9})}{(0.148)} \\ V=293.93V \end{gathered}[/tex]After reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain. In memory, this process is called
In memory, the process in which after reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain is called storage.
In memory there are three phases. They are:
Encoding StorageRetrievalIn the given scenario, the process in which converting the bold, key words to coded representations is encoding process. The process which maintains the coded representations in a network of neurons in your brain is storage. The process of remembering the information stored when needed is retrieval.
Therefore, in memory, this process is called Storage
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Block 1 has a mass of 12 kg is moving to the right on a levelsurface at a speed of 2 m/s. Block 2 has a mass of 2.5 kg andis at rest on the surface. Block 1 collides with block 2, causingblock 2 to move to the right with a speed of 4 m/s. How fast,and in what direction, is block 1 moving after the collision?
Given:
The mass of block 1, m₁=12 kg
The mass of block 2, m₂=2.5 kg
The velocity of block 1 before the collision, u=2 m/s
The velocity of block 2 after the collision, v₂=4 m/s
To find:
The velocity of block 1 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of the blocks before the collision will be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]v_1=\frac{m_1u-m_2v_2}{m_1}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{12\times2-2.5\times4}{12} \\ =1.17\text{ m/s} \end{gathered}[/tex]The positive sign of the velocity indicates that block 1 will continue to move to the right.
Final answer:
The velocity of block 1 after the collision will be 1.17 m/s and its direction is to the right.
Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child hasthe larger linear displacement?Select one:a. There is not enough information given to answer the question.b. Child Ac. They have the same non-zero linear displacementd. Child Be. They have the same zero linear displacement.
The linear displacement in a rotation is given by:
[tex]s=r\theta[/tex]where r is the distance from the axis of rotation and theta is the angular displacement.
Since the linear displacement is proportional to the radius we conclude that child A has a larger linear displacement.
A block of mass 0.500 kg slides on a flat smooth surface with a speed of 2.80 m/s. It then slides over a rough surface with μk and slows to a halt. While the block is slowing, (a) What is the frictional force on the block? (b) What is the magnitude of the block’s acceleration? (c) How far does the block slide on the rough part before it comes to a halt?
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
m = 0.5 kg
v = 2.8 m / s
Since there is no vertical motion,
∑ [tex]F_{y}[/tex] = 0
N - mg = 0
N = 0.5 * 9.8
N = 4.9 N
μ = 0.3
[tex]f_{k}[/tex] = μ N
[tex]f_{k}[/tex] = 0.3 * 4.9
[tex]f_{k}[/tex] = 1.47 N
The net force acting on the block is due to friction,
F = [tex]f_{k}[/tex] = 1.47 N
F = m a
1.47 = 0.5 * a
a = 2.94 m / s²
Since, acceleration is towards the opposite of motion,
a = - 2.94 m / s²
v² = u² + 2 a s
0 = 2.8² + ( 2 * - 2.94 * s )
s = 1.33 m
Therefore,
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
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2. A 14000 kg air jet accelerates from rest to 70 m/s before it takes off. What is the changein momentum of the jet?
Answer:
980,000 kg m/s
Explanation:
The change in momentum can be calculated as
[tex]\begin{gathered} \Delta p=m\Delta v \\ \Delta p=m(v_f-v_i) \end{gathered}[/tex]Where m is the mass, vf is the final velocity and vi is the initial velocity. Replacing m = 14000 kg, vf = 70 m/s and vi = 0 m/s, we get
[tex]\begin{gathered} \Delta p=14000\text{ kg \lparen}70\text{ m/s - 0 m/s\rparen} \\ \Delta p=14000\text{ kg \lparen70 m/s\rparen} \\ \Delta p=980000\text{ kg m/s} \end{gathered}[/tex]Therefore, the change in momentum of the jet is 980,000 kg m/s
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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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 fires a cannonball vertically upwards. The cannonball returns to the ground after a 3.80s flight. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the cannonball. Use regular metric units (ie. meters).How long did the cannonball rise? What was the cannonball's initial speed? What was the cannonball's maximum height?
The time taken by ball to rise is given as,
[tex]t=3.80\text{ s}[/tex]Therefore, the total time taken to move the ball in the upward direction is calculated as,
[tex]\begin{gathered} t_0=\frac{3.80\text{ s}}{2} \\ =1.90\text{ s} \end{gathered}[/tex]Therefore, the ball rise for 1.90 s.
The final speed of the ball can be given as,
[tex]v=u+gt[/tex]At the maximum height the final speed of ball is zero.
Substitute the known values,
[tex]\begin{gathered} 0m/s=u+(9.8m/s^2)(1.90\text{ s)} \\ u=-(9.8m/s^2)(1.90\text{ s)} \\ =-18.62\text{ m/s} \end{gathered}[/tex]Therefore, the initial speed of the ball is -18.62 m/s where negative sign indicates the direction of ball.
The final speed of the ball can be given as,
[tex]v^2=u^2-2gh_m[/tex]At the maximum height the final speed is zero. Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(-18.62m/s)^2-2(9.8m/s^2)h_m \\ h_m=\frac{(-18.62m/s)^2}{2(9.8m/s)^2} \\ =17.7\text{ m} \end{gathered}[/tex]Therefore, the maximum height of the canon is 17.7 m.
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.
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°
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
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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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.
Jacob Grena raises a spoon 0.210 m above a table . If the spoon and its contents have a mass of 30.0 g, what is the gravitational potential energy associated with the spoon at that height relative to the table's surface?
Explanation:
The spoon is raised so it gains Gravitational potential energy. Formula to find Gravitational potential energy;
Gravitational potential energy = mass × Gravitational field strength × height of the body from the surface (table in this scenario)
In symbols; E = m×g×h
Substitute values:
m = 30g = 0.03kg (don't forget to convert grams to kg)
g = 10N/kg
h = 0.210 m
So it's;
0.03kg × 10N/kg × 0.210m = 0.063 Joules
SI Unit of energy is joules
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
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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Which shape fits a position vs. time graph of an object that is slowing down? Which shape fits a position vs. time graph of an object that is speeding up?
1.
In a position x time graph, if the velocity is constant, so the position increases at the same rate over the time, so we have a linear relation between the position and the time.
Therefore the shape that represents this relation is C.
2.
In a velocity x time graph, if the velocity is constant, its value is always the same over the time, it doesn't change. That is represented graphically by a horizontal line, therefore the shape that represents this relation is 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
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 penguin runs 29,000 m/s how far will it travel in 10 seconds
Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.
[tex]s=\frac{d}{t}[/tex]s = speed
d = distance
t = time
[tex]\begin{gathered} d=st \\ d=29000\text{ m/s}\cdot10s \\ d=290000\text{ m} \end{gathered}[/tex]The distance would be 290,000 m
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 the wavelength of the sound produced by a bat if the frequency of the sound is 90 kHz on a night when the air temperature is 22°C?
We know that the wavelength is related to the speed of the wave by:
[tex]v=\lambda f[/tex]where f is the frequency.
The speed of sound on air at a given temperature is given by:
[tex]v=331\sqrt[]{1+\frac{T}{273}}[/tex]so in this case the speed is:
[tex]v=331\sqrt[]{1+\frac{22}{273}}=344.08[/tex]Plugging this and the frequency in the first expression above we have:
[tex]\begin{gathered} 344.08=90\times10^3\lambda \\ \lambda=\frac{344.08}{90\times10^3} \\ \lambda=3.82\times10^{-3} \end{gathered}[/tex]Therefore the wavelength is:
[tex]3.82\times10^{-3}\text{ m}[/tex]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]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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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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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
carts, bricks, and bands
8. What acceleration results when 2 rubber bands stretched to 20 cm are used to pull a cart with three bricks?
a. About 0.25 m/s2
b. About 0.33 m/s2
c. About 0.50 m/s2
d. About 1.00 m/s2
A. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with three bricks is 0.25 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with three bricks is 0.25 m/s².
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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
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?