Answer:
t1 = t2 + 3.02 V = 41.5
V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2
Both stones reach the same height after the specified times
V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)
2 V / g = t1 + t2 = 2t1 + 3.02
t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s
t2 = t1 + 3.02 = 5.74 sec
Check:
41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m
41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m
Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s
Which trait do you think is most important for a boss or supervisor to have?
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 mechanical lift is used to pull an engine out of a car. The engine is attached
to the lift with chains and lifted straight up. The free-body diagram below
shows the engine when it is suspended in the air.
Force 1
Force 2
What is force 2 in this diagram?
O A. Weight
B. Tension
C. Normal force
D. Friction
Answer:
the answer is weight
Explanation:
Answer:
Weight
Explanation: Just took the quiz
A carousel is (more or less) a disk of mass, 15,000 kg, with a radius of 6.14. What torque must be applied to create an angular acceleration of 0.0500 rad/s^2?round to 3 significant figures
(Plssss help me im suffering from severe brainrot)
To calculate the torque required to create an angular acceleration, we can use the formula:
Torque = Moment of Inertia × Angular Acceleration
The moment of inertia of a disk can be calculated using the formula:
Moment of Inertia = (1/2) × Mass × Radius^2
Given:
Mass = 15,000 kg
Radius = 6.14 m
Angular Acceleration = 0.0500 rad/s^2
First, calculate the moment of inertia:
Moment of Inertia = (1/2) × Mass × Radius^2
Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2
Next, calculate the torque:
Torque = Moment of Inertia × Angular Acceleration
Torque = Moment of Inertia × 0.0500 rad/s^2
Now, let's plug in the values and calculate:
Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2
Moment of Inertia ≈ 283,594.13 kg·m^2
Torque = 283,594.13 kg·m^2 × 0.0500 rad/s^2
Torque ≈ 14,179.71 N·m
Rounding to three significant figures, the torque required to create an angular acceleration of 0.0500 rad/s^2 is approximately 14,180 N·m.
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To better understand crash dynamics we have to look at "__________."
A. the law of gravity
B. Bernoulli's principle
C. the laws of motion
D. Archimedes' principle
To better understand crash dynamics we have to look at "the laws of motion."
The laws of motion
The laws of motion were introduced by Sir Isaac Newton in 1687 in his book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), which defined the laws of motion, or three fundamental laws that govern the movement of bodies. The laws of motion, according to Newton, govern the motion of an object or a system of objects that interact.
It defines the concepts of force and mass, and the fundamental dynamics of motion.The following are the laws of motion: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. The velocity of an object changes proportional to the force applied to it, and the acceleration of an object is proportional to both its force and its mass. For every action, there is an equal and opposite reaction.
Therefore, these laws are necessary to fully grasp crash dynamics because they explain how objects respond to outside forces that cause them to accelerate or decelerate.
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A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?
Explanation:
The energy of the system before the collision must equal the energy after the collision.
After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is
\(\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J\)
Before the collision only the bullet has kinetic energy.
So we can now determine the speed of the bullet using
\(\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s\)
I NEED WITH THIS HELP ASAP PLS
which of the following statements about ocean waves a true?
A. they travel on the surface of the water
B. they travel deep underwater
C. they are secondary waves
D. they are primary waves
The assertion that ocean waves move on the water's surface is accurate. Therefore, choice A is right.
What is the ocean waves?A wind wave, water wave, or tornado water wave is a surface wave that develops on a body of water's free surface as a result of the air flowing over the ocean's surface. The fetch is the distance at which two objects come into touch when facing the wind.
The wind is the most frequent source of waves. The friction between the wind and the surface of the water produces wind-driven waves, also known as surface waves. A wave crest is produced when wind continuously disturbs the water's surface in an ocean or lake.
The statements about ocean waves a true will be they travel on the surface of the water.
Thus, option A is correct.
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When a skater pulls her arms in, it
reduces her moment of inertia from
2.12 kg m² to 0.699 kg-m². If she was
initially spinning 3.25 rad/s, what is
her final angular velocity?
The skater's final angular velocity is approximately 9.86 rad/s.
The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, the skater has an angular momentum of:
L_initial = I_initial * ω_initial
Substituting the given values:
L_initial = 2.12 kg m² * 3.25 rad/s
The skater's final angular momentum remains the same, as angular momentum is conserved:
L_final = L_initial
The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:
L_final = I_final * ω_final
0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s
Solving for ω_final:
ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²
Hence, the skater's final angular velocity is approximately 9.86 rad/s.
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Calculate the work done to push a 200-N object 5-meters
What is the answer to question 4
In order to make it all the way from the opening to the detector, a wave has to travel through air, glass, water, plastic, and vacuum.
-- The siren and the tuning fork are sounds.
-- Sound cannot travel through vacuum.
-- That knocks out choices B, C, and D.
The answer is A.
Note: Making a big exception here. We don't do test questions on Brainly. That would be cheating. Don't let me catch you doing it again.
What are the symmetry operations of molecule AB4, where the molecule b lies at the center of the square and A lies at the center of the square and is not coplaner with B atoms. How to find the multiplication table
The symmetry operations of molecule AB₄ is tetrahedral and they are :
E : Identity4C₃ : axis of rotation 3C₂ : axis of rotation3S₄ : Rotation-reflection axis 6бdTo find the multiplication table we have to apply the multiplication table for Td symmetry ( attached below )
Given that molecule B lies at the corners of the square and molecule A lie at the center and is not coplanar with molecule B the symmetry operations of the molecules AB₄ will belong to a tetrahedral symmetry group which contains :
E : Identity4C₃ : axis of rotation 3C₂ : axis of rotation3S₄ : Rotation-reflection axis 6бdLearn more about tetrahedral symmetry group : https://brainly.com/question/1968705
A stone of mass 1kg is thrown at 10m/s upwards making an angle of 37°with the horizontal from a building that is 20m high. Using the law of conservation of energy calculate the speed wjen the stone hits the ground.
Answer:
31.68 m/s
Explanation:
The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.
As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:
\(H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21\)
Now we just add the time accelerating:
\(Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s\)
Help! Il give brainlest to who answers first
Answer:
1. The density of the cube is 1.03 g/mL.
2. Dish soap
Explanation:
1. Determination of the density of the cube.
From the question given above, the following data were obtained:
Mass (m) of cube = 21.7 g
Volume (V) of cube = 21 mL
Density (D) of cube =?
The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Thus, density can be expressed mathematically as:
Density (D) = mass (m) / volume (V)
D = m / V
With the above formula, we can obtain the density of the cube as follow:
Mass (m) of cube = 21.7 g
Volume (V) of cube = 21 mL
Density (D) of cube =?
D = m / V
D = 21.7 / 21
D = 1.03 g/mL
Thus, the density of the cube is 1.03 g/mL.
2. Determination of the layer of density the cube will settle in.
From the question given above,
Subtance >>>>>>>> Density
Vegetable oil >>>>> 0.91 g/mL
Grape juice >>>>>> 0.97 m/L
Water >>>>>>>>>>> 1 g/mL
Dish soap >>>>>>>> 1.03 g/mL
Maple syrup >>>>>> 1.37 g/mL
Comparing the density of the cube (i.e 1.03 g/mL) with those in the table able, we can conclude that the cube will settle in the DISH SOAP layer since they both have the same density.
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks that have lost their brakes on mountain grades. Typically, such a lane is horizontal (if possible) and about 36.0 m36.0 m long. Think of the ground as exerting a frictional drag force on the truck. A truck enters a typical runaway lane with a speed of 50.5 mph50.5 mph ( 22.6 m/s22.6 m/s ). Use the work-energy theorem to find the minimum coefficient of kinetic friction between the truck and the lane to be able to stop the truck.
Answer:
The coefficient of kinetic friction \(\mu_k = 0.724\)
Explanation:
From the question we are told that
The length of the lane is \(l = 36.0 \ m\)
The speed of the truck is \(v = 22.6\ m/s\)
Generally from the work-energy theorem we have that
\(\Delta KE = N * \mu_k * l\)
Here N is the normal force acting on the truck which is mathematically represented as
\(\Delta KE\) is the change in kinetic energy which is mathematically represented as
\(\Delta KE = \frac{1}{2} * m * v^2 \)
=> \(\Delta KE = 0.5 * m * 22.6^2 \)
=> \(\Delta KE = 255.38m \)
\( 255.38m = m * 9.8 * \mu_k * 36.0 \)
=> \( 255.38 = 352.8 * \mu_k \)
=> \(\mu_k = 0.724\)
earth energy budget is the relationship between how much energy the earth _______ and energy the earth _________
earth energy budget is the relationship between how much energy the earth receive from the sun and energy the earth radiates out.
What is energy?Energy is described as the quantitative property that is displaced to a body or to a physical system, recognizable in the performance of work and in the form of heat and light.
The term earth's energy budget is also described as the balance between of the amount of energy, that gets to the earth. from the Sun and the energy that leaves Earth and returns to the universe.
The earth's energy budget was mainly three types as shown:
shortwave radiation, longwave radiation, and internal heat sources.Learn more about energy at:
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You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water.
(a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?
Answer:
a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º
Explanation:
We can solve this exercise using the projectile launch ratios
a) Let's find the time it takes for the bullet to reach the water level
y = y₀ + v_{oy} t - ½ g t²
when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t =\(\sqrt{2y_o/g}\)
t = \(\sqrt{2 \ 7 /9.8}\)
t = 1,195 s
now we can calculate the speed with the horizontal movement
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 75.0 / 1.195
v₀ₓ = 62.76 m / s
b) if the speed of the bullets is half of that found
v₀ = 62.76 / 2 = 31.38 m / s
let's write the expressions for the distance
x = v₀ cos θ t
y = y₀ + v_{oy} sin θ t - ½ g t²
t = \(\frac{x}{v_o \ cos \theta}\)
we substitute
\(0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2\)
\(0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}\)
let's use the identified trigonometry
sec² θ = 1 + tan² θ
sec θ = 1 / cos θ
we substitute
\(0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)\)
\(\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0\)
we change variable
tan θ = H
\(\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0\)
we subtitle the values
\(\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0\)
27.99 H² - 75 H + 20.99 = 0
H² - 2.679 H + 0.75 = 0
we solve the quadratic equation
H = [2.679 ± \(\sqrt{2.679^2 - 4 0.75}\)] / 2
H = [2,679 ± 2,044] / 2
H₁ = 0.3175
H₂ = 2.3615
now we can find the angles
H₁ = tan θ₁
θ₁ = tan⁻¹ H₁
θ₁ = tan⁻¹ 0.3175
θ₁ = 17.6º
θ₂ = 67.0º
for these two angles the bullet hits the boat
Etienne asks Hakim to write Newton's second law in the horizontal direction for the block-spring system when it is displaced a distance x from the equilibrium position. Which equation is correct?
Answer:
Explanation: The correct equation for Newton's second law in the horizontal direction for the block-spring system when it is displaced a distance x from the equilibrium position is:
m(d^2x/dt^2) + kx = 0
where m is the mass of the block, k is the spring constant, and x is the displacement from the equilibrium position. This equation represents the balance of forces acting on the block-spring system, with the restoring force of the spring (proportional to x) opposing the force required to accelerate the mass (proportional to d^2x/dt^2).
un auto si muove lungo una strada rettilinea
Answer:
Un'auto si muove lungo un percorso rettilineo con velocità variabile come mostrato in figura. Quando l'auto è in possesso di A, la sua velocità è 10 ms-1 e quando è in posizione B, la sua velocità è 20 ms-1. Se l'auto impiega 5 secondi per spostarsi da A a B, trova l'accelerazione dell'auto.
Explanation:
PLEASE HELP!! WILL MARK BRAINLEST!!! What are 3 types of waves?
Answer:
transverse waves, longitudinal waves, and surface waves.
a question was asked by a teacher to a student. She gave the student a jumbled word and told him to make words out of it. The jumbled word is gzeysktqix. Now you know what to do. see ya!
When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.
When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.
This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.
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Which of the following fields can only exist in one direction?
A small block, with a mass of 250 g, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, around the loop, and then onto the final level section on the right. The maximum height of the incline is 80 cm, and the radius of the loop is 15 cm.
a.) Find the initial energy of the block.
b.) Find the velocity of the block at the bottom of the loop.
c.) Find the velocity of the block at the top of the loop.
(a) The initial energy of the block due to its position is 1.96 J.
(b) The velocity of the block at the bottom of the loop is 3.96 m/s.
(c) the velocity of the block at the top of the loop is 3.13 m/s.
Initial energy of the blockThe initial energy of the block due to its position is calculated as follows;
P.E = mgh
P.E = 0.25 X 9.8 X 0.8
P.E = 1.96 J
Conversation of the energyThe velocity of the block at the bottom of the loop is determined by applying the principle of conservation of energy as shown below;
P.Ei + P.Ef = K.Ei + K.Ef
1.96 + 0 = 0 + ¹/₂mvf²
vf² = 2(1.96)/m
vf² = (2 x 1.96) / (0.25)
vf² = 15.68
vf = √15.68
vf = 3.96 m/s
Velocity of the block at top of the loopThe velocity of the block at the top is calculated by applying principle of conservation of energy,
P.Ei + P.Ef = K.Ei + K.Ef
1.96 = mghf + ¹/₂mvf²
where;
hf is the position of the ball at the top of the loop = 2r = 2 x 15 cm = 30 cm = 0.31.96 = 0.25 x 9.8 x 0.3 + 0.5 x 0.25vf²
1.225 = 0.125vf²
vf² = 1.225/0.125
vf² = 9.8
vf = 3.13 m/s
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Why losing is an important aspects of sports
Answer:
It reminds us that we need to work harder. It allows us to make adjustments in the way and manner in which we train and practice. In a loss, we are able to identify our vulnerabilities and weaknesses, and work to improve.
Explanation:
newtons first law 1 to 5.
What is each of the net force for all of the 5 questions?
Answer:
1. 65 N.
2. 160 N.
3. 0 N.
4. 210 N.
5. 90 N.
Explanation:
1. Determination of the net force.
Force applied to the right (Fᵣ) = 80 N
Force applied to the left (Fₗ) = 145 N
Net force (Fₙ) =?
Fₙ = Fₗ – Fᵣ
Fₙ = 145 – 80
Fₙ = 65 N
Thus, the net force is 65 N
2. Determination of the net force.
Force 1 applied to the left (F₁) = 35 N
Force 2 applied to the left (F₂) = 125 N
Net force (Fₙ) =?
Fₙ = F₁ + F₂
Fₙ = 35 + 125
Fₙ = 160 N
Thus, the net force is 160 N.
3. Determination of the net force.
Force applied to the right (Fᵣ) = 75 N
Force applied to the left (Fₗ) = 75 N
Net force (Fₙ) =?
Fₙ = Fₗ – Fᵣ
Fₙ = 75 – 75
Fₙ = 0
Thus, the net force is 0 N
4. Determination of the net force.
Force 1 applied to the right (F₁) = 150 N
Force 2 applied to the right (F₂) = 60 N
Net force (Fₙ) =?
Fₙ = F₁ + F₂
Fₙ = 150 + 60
Fₙ = 210 N
Thus, the net force is 210 N.
5. Determination of the net force.
Force applied to the right (Fᵣ) = 115 N
Force applied to the left (Fₗ) = 25 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 115 – 25
Fₙ = 90 N
Thus, the net force is 90 N
The largest flowers in the world are the Rafflesia Arnoldii, found in Malaysia. A single flower is almost a meter across and has a mass up to 11.0 kg. Suppose you cut off a single flower and drag it along the flat ground. If the coefficient of kinetic friction between the flower and the ground is 0.39, what is the magnitude of the frictional force that must be overcome?
The magnitude of the frictional force that must be overcome is 42.04 N.
What is the magnitude of the frictional force?
The magnitude of the frictional force that must be overcome is calculated by applying Newton's second law of motion as follows;
Mathematically, the formula for the frictional force is given as;
F = μmg
where;
μ is the coefficient of frictionm is the mass of the flowerg is acceleration due to gravityThe magnitude of the frictional force that must be overcome is calculated as;
F = 0.39 x 11 kg x 9.8 m/s²
F = 42.04 N
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PLEASE HELP!!!!
An object is launched horizontally from a cliff. The cliff is 80 m high and the object has an initial launch velocity of 50 m/s.
What is the initial horizontal velocity?
What is the initial vertical velocity?
What is the final horizontal velocity?
How much time did it take for the object to hit the ground?
What is the final vertical velocity?
What is the final resultant speed?
How far from the base of the cliff will the projectile land?
Answer:
a. 50 m/s b. 0 m/s c. 50 m/s d. 4.04 s e. -39.6 m/s f. 63.78 m/s g. 202 m
Explanation:
a. What is the initial horizontal velocity?
Since the object is launched horizontally, it initial horizontal velocity is 50 m/s
b. What is the initial vertical velocity?
Since the object is launched horizontally, it has no initial vertical component. So, its initial vertical velocity is 0 m/s
c. What is the final horizontal velocity?
Its final horizontal velocity is 50 m/s since no force acts on it in the horizontal direction to change its value.
d. How much time did it take for the object to hit the ground?
We use the equation s = ut - 1/gt² since the object is falling under gravity where u = initial vertical velocity = 0 m/s, s = height of cliff = 80 m, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the object to hit the ground.
s = ut - 1/2gt²
80 m = 0 × t - 1/2 × -9.8 m/s² × t²
80 m = 4.9 m/s² × t²
t² = 80 m ÷ 4.9 m/s²
t² = 16.33 s²
t = √(16.33 s²)
t = 4.04 s
e. What is the final vertical velocity?
Using v = u + at where u = initial vertical velocity = 0 m/s, v = final vertical velocity, a = acceleration = -g = -9.8 m/s²and t = time it takes object to reach the ground = 4.04 s.
Substituting these values into the equation, we have
v = u + at
v = 0 m/s + (-9.8 m/s²) × 4.04 s
v = -39.6 m/s
f. What is the final resultant speed?
The final resultant speed v' is the resultant of the final horizontal velocity and the final vertical velocity. Let u' = final vertical velocity = 50 m/s.
v' = √(u'² + v²)
v' = √((50 m/s)² + (-39.6 m/s)²)
v' = √(2500 m²/s² + 1568.16 m²/s²)
v' = √(4068.16 m²/s²)
v' = 63.78 m/s
g. How far from the base of the cliff will the projectile land?
The distance from the base of the cliff, d where the projectile lands is
d = u't where u' = horizontal velocity = 50 m/s and t = time it takes object to land = 4.04 s
d = 50 m/s × 4.04 s
d = 202 m
Jesse drives 120km to a farm. His trip takes 2 1/2 hoursWhat is his speed?
Speed = distance / time
Speed = 120 km / 2 1/2 hours
Speed = 48 km per hour
A ball falls down 30 meters from the top of a building. If the ball weighed 1.2 kg, what is the gravitational potential energy lost by the ball? Estimate g as 9.81.
A steel ball moves from a position of +125 meters to a position of -75 meters. This motion takes 90.0 seconds. What is the velocity of the steel ball?
Answer:
2.22m/s to the left
Explanation:
Given parameters:
Initial position = +125m
Final position = -75m
Motion time = 90s
Unknown:
Velocity of the steel ball = ?
Solution:
The velocity of the steel ball is given as the displacement divided by the time;
Velocity = \(\frac{displacement}{time}\)
The net displacement of the ball = 125- (-75) = 200m to the left
Input the parameters and solve for the velocity;
Velocity = \(\frac{200}{90}\) = 2.22m/s to the left
When the balloon sticks to the wall (assuming it sticks to the wall). It is
because the balloon is negatively charged and the wall carries an extra
positive charge.
1.false
2.true