Answers
Answer:
289.05N
Explanation:
The mass of the bullet is 4.2-g bullet, if we convert it to "kg" we have
4.2-g bullet/1000
= 0.0042kg
The speed of the bullet leaving the missle is 338 m/s.
The Lenght of the barrel= 0.83-m-long
From kinematics, v²= u² +2al
V= final speed when it leaves the barell=
338 m/s
U= intial speed=0
L= length of the barrel travelled by the bullet
V²= 2al( initial speed=0)
If we make "a" subject of the formula we have
a=v²/2l
a= (338)²/(2×0.83)
=68821.687 m/s²
We can now calculate the force acting on the body using the Expression below
F=ma
Where F= force acting on the body
m= mass
a= Acceleration
F= 0.0042kg×68821.687
= 289.05N
Therefore, the force exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle is 289.05N
Related Questions
Which trait do you think is most important for a boss or supervisor to have?
Answers
Answer:
A high EQ (emotional intelligence)
In today's transitioning workplace, having a high EQ is the most important trait of a good boss. Bosses must be able to discern between their own personal beliefs and the thoughts and beliefs of others, and other generations (boomers, Gen X, xennials, millennials and now Gen Z).
Explanation:
hope it helps you
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree. At some instant, it moves with a velocity of −1.25 m/s
. Then, 2.01 s
later, it moves with a velocity of 1.95 m/s
. What is the chipmunk's average acceleration during the 2.01 s
time interval?
Answers
The chipmunk's average acceleration during the 2.01-second time interval is approximately 1.59 m/s².
To find the average acceleration of the chipmunk during the 2.01-second time interval, we can use the formula for average acceleration:
Average acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = -1.25 m/s
Final velocity (v) = 1.95 m/s
Time interval (t) = 2.01 s
Plugging in the values into the formula
Average acceleration = (1.95 m/s - (-1.25 m/s)) / 2.01 s
Simplifying:
Average acceleration = (1.95 m/s + 1.25 m/s) / 2.01 s
Average acceleration = 3.2 m/s / 2.01 s
Average acceleration ≈ 1.59 m/s²
Therefore, the chipmunk's average acceleration during the 2.01-second time interval is approximately 1.59 m/s².
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in 1977 steve weldon ate 91 m of spaghetti in 29 s. at the same speed, how long would it take Mr. weldon to eat 5 m of spaghetti?
Answers
The time that would be taken is 1.6 s.
What is the speed?We know that the speed would be the ratio of the distance to the time that have been taken and that is how we would be able to solve the problem that we have in the case that is ahead of us in this question
We have that the speed is; Distance/Time
= 91 m/29s
= 3.1 m/s
At the same speed;
Time = Distance/Speed
= 5 m / 3.1 m/s
= 1.6 s
Hence we k now that the spaghetti would be able to take a time that is about 1.6 s
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A plank AB 3m long weighing 20kg and with center of gravity 2m from the end A carries a load of mass 10kg at the end A it rests on two supports CandD.
1, compute the values of the reaction forces R1 and R2 at C and D.
2, how far from D and on which side of it must a mass of 24kg be placed on the plank so as to make the reactions equal? What are their values.
3,without this 24kg what vertical force applied at B will just lift the plank clear of D? What is then the reaction at C.
Answers
The answers are 1) The value of R2 is not relevant as it implies a downward force on the plank, 2) The reactions at C and D are 66.3 N and 90 N, respectively, and 3) The vertical force at B to lift the plank clear of D is 686.4 N. The reaction at C is zero, and the reaction at D is 61.4 kg.
1) R1 and R2 at C and D respectively are given by the equation R2 = (m1 + m2)g - R1, where m1 and m2 are the masses of the plank and load, respectively, and g is the acceleration due to gravity. Hence, substituting values R2 = (20 + 10) × 9.81 - R1 = 294.3 - R1. Now, taking moments about D, the following equation can be obtained: (20 × 1 + 10 × 3)g = R1 × 2 + R2 × 3 = 2R1 + 3 × (294.3 - R1) = 882.9 - R1, from which R1 = 343.7 N and R2 = 294.3 - 343.7 = -49.4 N. Since the support at D can only push the plank upwards and cannot pull it downwards, a negative value for R2 implies that the plank is actually being pulled downwards by an external force. Therefore, the value of R2 is not relevant. 2) The total weight of the plank and the load acting at the end A is 20 + 10 = 30 kg. For the reactions at C and D to be equal, the 24 kg mass must be placed at a distance x from D such that x × 30 = 24 × 6, from which x = 12/5 = 2.4 m. Since the 24 kg mass is being placed to the left of the plank, it will cause the reaction at C to decrease and that at D to increase. Thus, if R is the vertical force applied at B, then taking moments about D gives 20g × 1 - 10g × 3 + R × 6 = 0, from which R = 90 N. Taking moments about C gives R × 3 - 10g × 2 = 0, from which R = 66.3 N. 3) The vertical force applied at B that will just lift the plank clear of D is the weight of the plank and the load acting at the end A plus the weight of the part of the plank that is to the right of D. The weight of the plank and the load acting at the end A is 20 + 10 = 30 kg, and the weight of the part of the plank that is to the right of D is 24 × 1.6 = 38.4 kg. Therefore, the vertical force applied at B that will just lift the plank clear of D is (20 + 10 + 38.4)g = 686.4 N. The reaction at C is zero because the plank is not being supported there anymore. The reaction at D is the same as the weight of the plank and the load acting at the end A plus the weight of the part of the plank that is to the right of D, which is 20 + 10 + 24 × 1.6 = 61.4 kg.For more questions on acceleration
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Which statement can be made about the amplitude of any transverse wave?
It is the length from the midpoint to the crest.
It is half the length from the midpoint to the trough.
It is the length of the wavelength.
It is half the length of the wavelength.
Answers
Answer:
A) It is the length from the midpoint to the crest.
Explanation:
The correct statement about the amplitude of any transverse wave is: It is the length from the midpoint to the crest. The correct option is A.
What is a transverse wave?A transverse wave is a type of wave in which the displacement of the medium is perpendicular to the direction of wave propagation.
Examples of transverse waves include light waves and water waves in which the surface of the water oscillates up and down while the wave moves horizontally.
The properties of a transverse wave include:
Amplitude: The maximum displacement of the medium from its rest position. This represents the intensity or strength of the wave.
Wavelength: The distance between two adjacent points in the wave that are in phase with each other, for example, between two consecutive crests or troughs. It is typically represented by the symbol λ.
Frequency: The number of complete wave cycles that pass a point in a given amount of time, usually measured in hertz (Hz). The frequency is inversely proportional to the wavelength and is represented by the symbol f.
Period: The time taken for one complete wave cycle to pass a given point, usually represented by the symbol T. The period is directly proportional to the wavelength and inversely proportional to the frequency.
Speed: The speed at which the wave propagates through the medium, usually represented by the symbol v. The speed is directly proportional to the frequency and wavelength.
These properties are related to each other by the wave equation:
v = fλ,
Where v = the speed of the wave,
f = the frequency,
and λ = the wavelength.
Therefore, The correct statement about the amplitude of any transverse wave is A. It is the length from the midpoint to the crest.
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If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answers
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
\(Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\\)
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
\(W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)\)
W = 52.92 J + 588.6 J
W = 641.52 J
Three different groups each measured the diameter of a CD three times (the actual diameter of a CD is 12.00cm). The class data is shown in the table. Which group(s) was the most accurate in their measurements? Group 1 only Group 1 only Group 2 only Group 2 only Group 3 only Group 3 only Groups 1 & 3 were equally accurate
Answers
1. Which of the following does not affect the resistance of a wire?
a) Length
b) Temperature
c) Usage time
d) Cross-sectional area
2. If a 12V battery is passing current through a resistor with a current of 2A, what is the value of the resistor?
a 24resistance
b) 14resistance
c) 10resistance
d) 6resistance
3. Describe the differences between series and parallel circuits.
4. A circuit contains resistors of 8resistance and 4resistance，what is combined resistance if the resistors are combined:
a) In series
b) In parallel
5. A 0.5A current is passing across three resistors of 8resistance, 4resistance and 12resistance that are linked in series.
What is the potential difference of the circuit?
6. Wire A has a resistance of 24resistance. If wire B is double the length and has a diameter four times as large as wire A, what is the resistance of wire B?
Answers
After watching this video, Blake, a student in an introductory physics class, makes the following claim: The acceleration and velocity of the glider are both momentarily zero when the glider changes direction. The velocity of the glider must be zero for an instant when the glider changes direction. Because the velocity is zero, the acceleration must also be zero. 1. Respond to Blake's claim. Which parts, if any do you agree with, and which parts do you not agree with
Answers
Answer:
Please see below as the answer is self- explanatory.
Explanation:
Any time that an object changes direction (from leftward to rightwward, or from upward to downward) the velocity must be zero just for one instant, when is on the verge of changing the direction.This is needed because velocity changes as a continuous function of time, so it needs to cross the t-axis when passing from positive to negative or vice versa.However, the claim that in the moment that velocity is zero, the acceleration is also zero, is false.Due to acceleration is the rate of change of velocity, and velocity is a vector, this means at any time there is a change of direction, there is an acceleration that is non-zero.For example, when an object that has been thrown upward, reaches to its maximum height, just one instant before starting to fall, the velocity becomes zero, but the acceleration (which causes the object to fall) is non-zero, due to it's the acceleration due to gravity.Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I predicted. This is because I used the wrong mass for the puck when I did all my calculations. I accidentally used the mass of the smaller puck rather than the mass of the larger puck in my video." Is this a plausible explanation? Would the using the wrong mass for the puck during the calculations mean the puck would reach a greater height? Explain your reasoning.
Answers
Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".
When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.
So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.
The representation is as followed:
\(K_{i}+P_{i}=K_{f}+P_{f}\)
\(K_{i}+0=0+P_{f}\)
\(\frac{1}{2}mv^{2} = mgh\)
As we noticed, mass of the object can be cancelled from the equation, making height be:
\(h=\frac{v^{2}}{2g}\)
So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.
I don’t understand what it’s asking me to do
Answers
Because the mass and displacement are already given in Kg and m, respectively, in the first part of your question, there is no need to convert them. However, in the second part of your question, you must use the given equation to calculate the spring constant.
if the table data is given in grams and cm you have to convert it using the following conversion,
1. To convert grams to kilograms, we divide the mass values by 1000.
2. To convert centimeters to meters, we divide the displacement values by 100.
But here in the given table it's already given the mass in kg and the displacement in meters (m). so no need to convert it.
Now comes the second part of your question,
To calculate the spring constants for the given data, we can use the equation:
k = -mg/Δx
where:
k is the spring constant (in N/m),
m is the mass (in kg), and
Δx is the displacement of the spring (in m).
Let's calculate the spring constants using the provided data:
Mass (kg): 0.05 0.1 0.2 0.3 0.4 0.5 0.6
Displacement of Spring (m): 0.012 0.027 0.065 0.1 0.135 0.17 0.199
Using the equation
k = -mg/Δx,
we can calculate the spring constant for each data point:
For the first data point (m = 0.05 kg, Δx = 0.012 m):
k = -0.05 kg * 9.8 m/s² / 0.012 m
k ≈ -40.833 N/m
Similarly, we can calculate the spring constants for the other data points:
For the mass of 0.05 kg, the spring constant is approximately -40.833 N/m.
For the mass of 0.1 kg, the spring constant is approximately -18.519 N/m.
For the mass of 0.2 kg, the spring constant is approximately -6.154 N/m.
For the mass of 0.3 kg, the spring constant is approximately -3.267 N/m.
For the mass of 0.4 kg, the spring constant is approximately -2.222 N/m.
For the mass of 0.5 kg, the spring constant is approximately -1.716 N/m.
For the mass of 0.6 kg, the spring constant is approximately -1.449 N/m.
Therefore, In the first part of the question, there is no need to convert the mass into kg and the displacement cm into m because it is already given in kg and m respectively, and in the second part question you have to calculate the spring constant using the given equation.
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Introduction to Simple Machines
This activity will help you meet this educational goal:
You will compare and contrast information from a video with information from a text.
Directions
Read the instructions for this self-checked activity. Type in your response to each question, and check your answers. At the end of the activity, write a brief evaluation of your work.
Activity
Watch this video and then answer the following questions based on what you learned.
Part A
How does a bicycle make work easier?
Part B
Which two examples of levers are mentioned in the video?
The picture shows a bicycle’s pedals. Look at the shaft that the pedals are attached to. Do you think the shaft is a lever? Why or why not?
Answers
Answer:
word for word answers!
Explanation:
1) Part A: By pedaling a bicycle lightly, the rider can go a long way
2) Part B: The two examples mentioned in the video are the handlebars and the brakes
3) Yes, it’s a type of lever because the two pedals rotate around a fixed point
Einstein’s principle of equivalence states:
A) the gravitational constant is the same everywhere in the universe
B) it is impossible to tell the difference between gravitational force and the normal force
C) every mass exerts a gravitational force on every other mass
D) gravitational mass and inertial mass are the same
E)the laws of physics are the same in all inertial reference frames
Answers
Hope this helps you have a great night!!
Good luck
ED. 05.39 m/sE. O 0.388 m/s10. An object of mass 16 kg moving with a speedof 26 m/s to the right collides with an objectof mass 9 kg at rest. If the collision is completely inelastic,calculate the kinetic energy lost during the collision. (1 point)A. O-3302.995 JB. O-1703.969 JC. O-1205.975 JD. O-3546.668 JE. O-1946.88 JSubmit Query
Answers
Given:
The mass of one object is m1 = 16 kg
The initial speed of the object is
\(v_i=\text{ 26 m/s}\)The mass of another object is m2 = 9 kg.
The speed of the object will be zero as it is at rest.
To calculate the kinetic energy lost.
Explanation:
The collision is inelastic.
According to the conservation of momentum, the speed after the collision can be calculated as
\(\begin{gathered} m1v_i+m2\times0=(m1+m2)v_f \\ v_f=\frac{m1v_i}{(m1+m2)} \\ =\frac{16\times26}{16+9} \\ =\text{ 16.64 m/s} \end{gathered}\)The loss in kinetic energy during a collision can be calculated as
\(\begin{gathered} \Delta K.E\text{ =K.E.}_f-K.E._i \\ =\frac{1}{2}(m1+m2)v_f-\frac{1}{2}m1v_i^2 \\ =\frac{1}{2}\times25\times(16.64)^2-\frac{1}{2}\times16\times(26)^2 \\ =3461.12-5408 \\ =-1946.88\text{ J} \end{gathered}\)Thus, the correct choice is E
A ball of mass 0.900
kg is carefully balanced on a shelf that is 2.00
m above the ground. What is its gravitational potential energy?
Answers
The gravitational potential energy of the ball is 17.64 J. The gravitational potential energy (U) of an object at a height (h) above the ground is given by U = mgh
What does gravitational potential energy refer to?The energy that an object has due to its location in a gravitational field is known as gravitational potential energy. For an object close to the surface of the Earth, when the gravitational acceleration can be assumed to remain constant at roughly 9.8 m/s, gravitational potential energy is most frequently used.
How does gravitational potential energy come about?The position of an object in relation to its surroundings within the Earth-object system determines its gravitational potential.
\(U=mgh\\U = (0.900 kg) *(9.81 m/s^2) * (2.00 m)\\U = 17.64 J\)
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It is necessary to to secure an inflated balloon tightly give reason
Answers
Two hot air balloons with the same mass and amount of helium put inside of them if one is a rigid material and the other expands which one would be the highest?
Answers
Answer:
One is that atmospheric pressure is dramatically reduced at high altitudes, so a helium balloon expands as it rises and eventually explodes. If you inflate a balloon beyond its limits at room temperature, it will break into small pieces up to about ten centimetres long
Explanation:
Scientists might make a computer model of volcanic eruptions. What is the
biggest benefit of this model?
Answers
Answer:
Computer Model May Help to More Accurately Predict Volcano Eruptions. Scientists at the GFZ German Research Center in Potsdam, Germany, have developed a computer model which they say boosts the accuracy of volcanic eruption prediction.
Pls Mark As Brainliest
What is the formula for Ba+2 and F-?
Answers
Answer:
BaF2
Explanation:
since you got the valence numbers just do the scissors move where you:
give the F the 2 and give the Ba the 1 so it be like
BaF2 here is the chemical compound
A ball is dropped from the top of a building.
When does the ball have the least potential energy?
O after it has hit the ground
O half way through the fall
O as it is released
O just before it hits the ground
Answers
Answer:
after it has hit the ground
.........
Answer: A. after it has hit the ground
plzzz help will give brainlist
Below are views related to education.
---------------------------
Civics should be taught in public schools.
Good citizenship is learned in communities.
---------------------------
Which conclusion can be drawn from these views?
A. Civic involvement is important to building strong societies.
B. Civic participation should be learned at home.
C. Political conflict is a threat to strong societies.
D. Political conflict should be reduced through legislation.
Answers
Answer:
A.
Explanation:
this would make sense but it seems to be more like they both sound different
Two boys searching for buried treasure are standing underneath the same tree. One boy walks 18 m east and then 18 m north. The other boy walks 16 m west and then 11 m north. Find the scalar product of their net displacements from the tree.
Answers
NEED HELP When a guitar string is plucked, what part of the standing wave is found at the fixed ends of the string?(1 point)
the nodes
the resonance
the antinodes
the harmonics
Answers
Explain the advantages and disadvantages of having a racecar with a small mass in reference to Newton’s 1st law of motion. Use the term inertia in your answer.
Answers
Answer:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This tendency to resist changes in a state of motion is inertia.
Explanation:
At time t = 0 s, a puck is sliding on a horizontal table with a velocity 3.60 m/s, 35.0° above the +x axis. As the puck slides, a constant acceleration acts on it that has the following components: a = - 2 0.360 m/s and da, = -0.980 m/s. What is the velocity of the puck at time t = 1.50 s? 2 0
Answers
The velocity of the punk at time t = 1.5 s is equal to 2.47 m/s.
What are the components of velocity?The object will have two velocities during the projectile motion, one is horizontal velocity and the other is vertical velocity. The horizontal velocity of the object is constant throughout the projectile motion but the vertical velocity is changed in the projectile motion.
Given that the velocity vector is making an angle θ with a horizontal table, θ = 35°
The vertical component of the velocity of the puck can be determined as:\(u_x = u Sin 35^o = 3.60 \times 0.573 =2.06 m/s\)
The horizontal component of the velocity of the puck can be determined as:
\(u_y = u cos 35^o = 3.60 \times 0.82=2.95 m/s\)
Given the acceleration aₓ = -0.360 m/s² and a y = -0.980 m/s²
The final vertical components of the velocity of the punk are:
\(v_x = u_x + a_xt\)
\(v_x = 2.95 - (0.360) (1.5)\\v_x = 2.41 m/s\)
and, the horizontal velocity of the punk is:
\(v_y = u_y +a_yt\\v_y = 2.06 - (0.980) (1.5)\\v_y = 0.59 m/s\)
The velocity of the punk can be determined as:
\(v = \sqrt{v_x^2 + v_y ^2}\\ v =\sqrt{(5.80)^2 + (0.35)^2}\\ v = 2.47 m/s\)
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a single conservative force Fx= (2x+7) N acts on a particle of mass 6 kg as the particle moves along the X-axis from X1 = 1 m to X 2 = 5m. calculate the work done by this force
Answers
Answer:
To calculate the work done by a force, we can use the formula:
Work = ∫F dx
In this case, the force is given by Fx = (2x + 7) N, and the particle moves along the X-axis from X1 = 1 m to X2 = 5 m. Let's calculate the work done.
Work = ∫(2x + 7) dx
Integrating the function (2x + 7) with respect to x, we get:
Work = (x^2 + 7x) evaluated from X1 to X2
Plugging in the values X2 = 5 and X1 = 1 into the expression, we have:
Work = (5^2 + 7 * 5) - (1^2 + 7 * 1)
= (25 + 35) - (1 + 7)
= 60 - 8
= 52
Therefore, the work done by the force is 52 Joules (J).
Explanation:
Question 4 A car of mass 820 kg has a maximum power or 30 kW and moves against a constant resistance of motion to 910 N. Calculate the maximum speed of the car in the following situation On ground level ?
\(p = fv \\ v = p \div f \\ 30 \div 910 = \)
Answers
Explanation:
power = force × velocity
velocity=power/force
=(30×1000)/910
=32.97m/s
The correct answer is 32.97m/s.
How to calculate the maximum speed?By using the formula:
power = force × velocity
velocity = power / force
= (30×1000)/910
= 32.97m/s
The answer is 32.97m/s.
What is maximum speed physics?
Maximum speed is the highest rate of speed an athlete can attain. Acceleration refers to velocity, and because velocity has both a magnitude and direction associated with it, acceleration changes when athletes change the magnitude of their motion (how fast they are running), the direction of their motion, or both.
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Trevor is charting the motion of his pet tortoise across the floor. The diagram below shows its position every 60 seconds.
A.The tortoise is moving at a constant speed
B. The tortoise’s speed is decreasing through time
C. The tortoise’s speed is increasing through time
D. The tortoise’s position is not changing through time
Answers
Answer:
A.The tortoise is moving at a constant speed
Explanation:
The chart shows the tortoise covering a distance of 0.5 meters every 60.0 seconds. Since this rate of motion is not changing, the tortoise is moving at a constant speed.
The diagram shows the tortoise moving at a constant speed. Thus the correct option is option A.
What is speed?Speed is defined as how far the object travels. It is also defined as the rate of change of distance per unit of time. Speed is the scalar quantity and the unit of speed is m/s.
Speed is denoted by the letter v. v = s/t, where s is the distance traveled along the path and t is the time taken by the object to move along the path. The SI unit of distance is meter and the unit of time is seconds.
When the object covers an equal distance at equal intervals of time is called uniform speed. When the object covers unequal distance with equal intervals of time or equal distance with unequal intervals of time is called Non-uniform speed.
The graph shows the Tortoise covers 0.5 m in 60 seconds is called uniform speed. Thus, option A is correct.
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What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg.m/s of
momentum?
Answers
Answer:
The product of the mass and the volume is known as momentum.
According to the law of momentum, it is stated that the two or more bodies remain in a constant state unless an external force is applied in an isolated room.
Momentum depends on the following:-
Mass
Velocity
Momentum = MASS \ X \ VOLUMEMomentum=MASS X VOLUME
\begin{gathered}momentum = 3000kgm/s\\\\mass = 2000kg\\\\velocity =\frac{momentum}{mass}\\\\v= \frac{3000}{2000} \\\\v= 1.5m/s\end{gathered}
momentum=3000kgm/s
mass=2000kg
velocity=
mass
momentum
v=
2000
3000
v=1.5m/s
You stand on top of a cliff facing a perfectly flat valley below. In one hand is a stone and in the other a stopwatch. You throw the stone at 30.0 m/s and an angle of 30.0° above the horizontal. You time the stone with your stopwatch and find that it lands 6.91 seconds after you threw it.
A)How high above the valley floor was the stone when you threw it?
y0 = ____m
B)At what horizontal distance (x) from the cliff did the stone land?
x= ___ m
C)What was the magnitude of the stone's velocity just before it hit the ground?
v=___ m/s
D)Graph the the x-velocity vx versus time for the motion of the stone. Then choose the correct graph below.
E)Graph the the y-velocity vy versus time for the motion of the stone. Then choose the correct graph below.
Answers
Answer:
Height of cliff = 147.43 m
Explanation:
We have equation of motion s = ut + 0.5 at²
Consider the vertical motion of stone and up as positive,
Initial velocity, u = 15sin30=7.5m/s
Time taken, t = 6.30 s
Acceleration, a = -9.81 m/s²
Substituting
s = ut + 0.5 at²
s = 7.5 x 6.30 + 0.5 x (-9.81) x 6.30²
s = -147.43 m
So the stone goes below 147.43 m below the initial postilion
Height of cliff = 147.43 m