Answer:
F = 2.45 [N]
Explanation:
The force that the potato exerts on the Earth is equal to the product of the potato mass by gravitational acceleration.
[tex]F=m*g[/tex]
where:
m = mass = 0.25 [kg]
g = gravity acceleration = 9.81 [m/s²]
[tex]F=0.25*9.81\\F=2.45[N][/tex]
You are holding two balloons of the same shape and size. One is filled with helium, and the other is filled with ordinary air. On which balloon the buoyant force is greater?
a. The helium filled balloon experiences the greater buoyant force.
b. The air filled balloon experiences the greater buoyant force.
c. Both balloons experiences same buoyant force.
Answer:
Both balloons experiences same buoyant force.
Explanation:
Buoyant force is defined as the upward force that is exerted on an object which is wholly or partly immersed in a fluid. We can also define it as the upward force exerted by any fluid upon a body immersed in it. We may also refer to this buoyant force as the upthrust.
This buoyant force is the push of air on the balloon and it is independent of the contents of the balloons. Hence, both balloons experiences same buoyant force.
____ is factual information not subject to bias.
Interpretation
Analysis
Data
Opinion
Answer:
Data
Explanation:
Data is factual information not subject to bias.
This ultimately implies that, data connotes fact, thus, it is an information that is credible, accurate, a statement of truth, evidential and proven.
In Computer programming, a data dictionary can be defined as a centralized collection of information on a specific data such as attributes, names, fields and definitions that are being used in a computer database system.
In a data dictionary, data elements are combined into records, which are meaningful combinations of data elements that are included in data flows or retained in data stores.
This ultimately implies that, a data dictionary found in a computer database system typically contains the records about all the data elements (objects) such as data relationships with other elements, ownership, type, size, primary keys etc. This records are stored and communicated to other data when required or needed.
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substance by 1 °C. Specific heat capacity can be calculated using the following equation:
q = mc deltaT
In the equation q represents the amount of heat energy gained or lost in joules), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and AT is the temperature change of the substance in °C).
Goal: Calculate the specific heat capacities of copper, granite, lead, and ice.
Solve: When you mix two substances, the heat gained by one substance is equal to the heat lost by the other substance. Suppose you place 125 g of aluminum in a calorimeter with 1,000 g of water. The water changes temperature by 2 °C and the aluminum changes temperature by -74.95 °C.
A. Water has a known specific heat capacity of 4.184 J/g °C. Use the specific heat equation to find out how much heat energy the water gained (q).
B. Assume that the heat energy gained by the water is equal to the heat energy lost by the aluminum. Use the specific heat equation to solve for the specific heat of aluminum. Aluminum's accepted specific heat value is 0.900 J/g °C. Use this value to check your work.
Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
The heat gained by the water can be obtained solving the following equation:[tex]q_{g} = c_{w} * m * \Delta T (1)[/tex]
where cw = specific heat of water = 4.184 J/gºCm= mass of water = 1,000 gΔT = 2ºC Replacing these values in (1) we get:[tex]q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)[/tex]
B)
Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:[tex]q_{l} = c_{Al} * m_{Al} * \Delta T (3)[/tex]
⇒ [tex]-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)[/tex]
[tex]c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)[/tex]
which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.
what is the pressure on a swimmer 50 m below the surface of a lake
Answer:
P = 490500 [Pa]
Explanation:
The pressure at the bottom of a vessel and even of a lake or sea can be calculated by means of the following hydrostatic equation.
[tex]P=Ro*g*h[/tex]
where:
P = pressure [Pa] (units of pascal)
Ro = water density = 1000 [kg/m³]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 50 [m]
Now replacing:
[tex]P=1000*9.81*50\\P=490500[Pa][/tex]
in the periodic table, the properties repeat in what direction?
Answer:
Left to right and top to bottom
Explanation:
On the periodic table, the properties repeat from left to right and from top to bottom.
Periodic properties have a pattern from the top to the bottom or down a group or family.
Also, across the period from left to right, they also show a repeating pattern.
Certain properties increase from left to right and decreases from top to bottom. E.g. electronegativity. Also, some properties decreases from left to right and increases from top to bottom e.g. atomic radius.Which is in the Moneran kingdom? A. plants B. bacteria C. animals D. mushrooms
Answer: b
Explanation:
Lisa skips 5 m North to the playground. She realizes she forgot to bring water so she turns around and goes 3 m South to the convenience store
Answer:
-2 South
Always subtract where they ended first is what my teacher said
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
What statement is not an example of Newton’s first law of motion
Answer:
c
Explanation:
im smart....................... i think
This law is about inertia, and the law displayed in A is Newton's third law of equal and opposite reactions, so option A is correct.
What is Newton’s first law of motion?The basis of classical mechanics is laid out in three assertions known as Newton's laws of motion, which were first articulated by English physicist and mathematician Isaac Newton. These laws describe the relationships between forces acting on a body and its motion.
Unless a force acts on a body that is at rest or moving in a straight line at a constant speed, Newton's first law asserts that it will continue to be at rest or move in that direction.
This law is about inertia (an object wanting to stay in its state of motion) and the law displayed in A is Newton's third law of equal and opposite reactions, therefore, it is not an example of Newton’s first law of motion.
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A motorist is traveling 40ms-¹ and applies brakes and slow down at a rate of 2ms-² the available distance for the the motorist to stop is 400m will the motorist be able to stop?
Answer:
[tex] \underline{ \boxed{ yes}}\\[/tex]
Explanation:
[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]
A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is the corresponding raft speed?
Answer:
The answer is below
Explanation:
Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.
Momentum = mass * velocity
The principle of conservation of momentum states that momentum cannot be created or destroyed but can be transferred. Therefore the momentum before and after an action is equal.
Initial momentum = Final momentum
Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.
m = 71 kg, M = 500 kg, v = 6 m/s
Initial both the raft and diver are at rest, hence u and U is zero, hence:
mu + MU = mv + MV
71(0) + 500(0) = 71(6) + 500(V)
0 = 426 + 500(V)
500(V) = -426
V = -426/500
V = -0.852 m/s
At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2
Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
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A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed 9.2 m/s It collides in a perfectly elastic collision with a second block of mass M that is at rest. After the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s The block of mass M travels to the right at a speed V of 6.7 m/s what is M
Answer:
[tex]m_2=6.3\:\mathrm{kg}[/tex]
Explanation:
In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:
[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]
Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].
Plugging in all given values, we have:
[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot9.2^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot m_2\cdot 6.7^2,\\m_2=\fbox{$6.3\:\mathrm{kg}$}[/tex].
Assuming no friction, how does the initial gravitational potential energy of
the marble on a downward slope compare to the final kinetic energy?
a) they are the same
b) the initial gravitational potential energy is greater than the final kinetic energy
c) the initial gravitational potential energy is less then the final kinetic energy
Answer:
a) They are the same.
Explanation:
Assuming no friction, there should be no energy transfer and thus the Law of Conservation of Energy says:
[tex]PE=KE,\\mgh=\frac{1}{2}mv^2[/tex]
These types of problems also disregard any air resistance the surface of the object may cause. Therefore, no energy is transferred and from the Law of Conservation of Energy, [tex]100\%[/tex] of energy is preserved.
A 5 cm diameter orfice discharges fluid from a tank with a head of 5 meters. The discharge rate, Q, is measured at 0.015 m^3/s. The actual velocity at the vena contracta is 9 m/s. The coefficient of discharge is nearest to:________
Answer:
[tex]\mathbf{ C_d = 0.86}[/tex]
Explanation:
Given that:
diameter of the orifice D = 5 cm = 5 × 10⁻² m
discharge rate Q = 0015 m³/s
Actual velocity V = 9 m/s
By using the formula
[tex]Q = C_d \times A \times \sqrt{2gh}[/tex]
where;
[tex]v = \sqrt{2gh}[/tex] = 9 m/s
[tex]Q = C_d \times A \times v}[/tex]
where;
[tex]A = \dfrac{\pi D^2}{4}[/tex]
[tex]A = \dfrac{\pi (5 \times 10^{-2} ) ^2}{4}[/tex]
[tex]A = 0.00196 \ m^2[/tex]
Again,
[tex]Q = C_d \times A \times v}[/tex]
[tex]0.015 = C_d \times 0.00196 \times 9[/tex]
[tex]C_d = \dfrac{0.015 }{ 0.00196 \times 9}[/tex]
[tex]\mathbf{ C_d = 0.86}[/tex]
Hence, the coefficient of discharge [tex]\mathbf{ C_d = 0.86}[/tex]
Two objects are interacting but stay stationary. Which best describes what is happening to he action and reaction forces
Answer: B
Explanation: The forces are equal and opposite each other.
When two objects are interacting but stay stationary, then the forces are equal and opposite each other.
What do you mean by Force?Force may be defined as the process of pushing and pulling an object with an actual mass that stimulates its velocity to be changed. It is a type of vector quantity because it has both magnitude and direction.
It is the simple and fundamental concept of physics that when two or more objects are interacting with one another but do stimulate any change in their position, the forces among them are definitely equal and opposite to one another. It is the most plausible explanation of Newton's third law of motion.
Therefore, when two objects are interacting but stay stationary, then the forces are equal and opposite each other.
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Your question seems incomplete. The most probable complete question is as follows:
The forces are equal and opposite each other. The forces are not equal and opposite to each other. The forces are equal but not opposite to each other.The forces are not equal but opposite to each other.10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends soccer ball towards the goal.
G
U
E
S
S
Formula
11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?
G
U
E
S
S
Formula
please I need help I don't understand it and I had to deliver it yesterday helpp:(
Answer:
10. 65 N
11. 35 N
Explanation:
10. Determination the force that sends the soccer ball towards the goal.
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Initial velocity (u) of the ball = 0 m/s
Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s
Acceleration (a) of the ball =?
a = (v – u) /t
a = (32.5 – 0) / 0.21
a = 32.5 / 0.21
a = 154.76 m/s²
Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:
Mass (m) of the ball = 0.42 kg
Acceleration (a) of the ball = 154.76 m/s²
Force (F) =?
F = ma
F = 0.42 × 154.76
F = 65 N
Thus, the force that sends soccer ball towards the goal is 65 N
11. Determination of the force exerted by the rocket on the satellite.
We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:
Initial velocity (u) of satellite = 0 m/s
Final velocity (v) of satellite = 0.63 m/s
Time (t) = 1296 s
Acceleration (a) of the satellite =?
a = (v – u) /t
a = (0.63 – 0) / 1296
a = 0.63 / 1296
a = 4.861×10¯⁴ m/s²
Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:
Mass (m) of the satellite = 72000 Kg
Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²
Force (F) =?
F = ma
F = 72000 × 4.861×10¯⁴
F = 35 N
Thus, the force exerted by the rocket on the satellite is 35 N
Calculate the extension of a 20cm spring that has a spring contrast of 45000N/m and 1500N of elastic potential energy.
Answer:
See the answer below
Explanation:
The elastic potential energy can be calculated by means of the following equation:
[tex]E_{el}=\frac{1}{2} *k*x^{2}[/tex]
where:
Eel = elastic energy = 1500 [J]
k = spring constant = 45000 [N/m]
x = extension [m]
[tex]1500=\frac{1}{2} *45000*x^{2} \\3000=45000*x^{2} \\x=\sqrt{0.06666}\\x=0.258[m] = 25.8 [cm][/tex]
Which is the best explanation for why Toms technique works ?
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
In the picture shown below A represents a characteristic of only geocentric model, B represents a characteristic common to both geocentric and heliocentric models, C represents a characteristic of only heliocentric model, and D represents a characteristic which the geocentric and heliocentric models do not have.
Under which label will the characteristic, "The sun and planets revolve around a central moon in the solar system" fall?
A
B
C
D
What is the correct organization of living things, from smallest to largest?
Cells - Tissues - Organs - Organ Systems - Organism
Organs - Tissues - Cells - Organ Systems - Organism
Cells - Organs - Tissues - Organism - Organ Systems
Cells - Organism - Tissues - Organ Systems - Organs
A racecar makes 24 revolutions around a circular track of radius 2 meters in
162 seconds. Find the racecar's frequency
Answer:
[tex]0.15\: \mathrm{Hz}[/tex]
Explanation:
The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.
Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.
Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).
The radius of the track is irrelevant in this problem.
A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and released, calculate;the spring constant,the angular velocity, amplitude of oscillation, maximum velocity, maximum acceleration of the mass,period, frequency
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
In designing buildings to be erected in an area prone to earthquakes, what relationship should the designer try to achieve between the natural frequency of the building and the typical earthquake frequencies?
A) The natural frequency of the building should be exactly the same as typical earthquake frequencies.
B) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly lower
C) The natural frequency of the building should be very different frem typical earthquake frequencies
D) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly higher.
Answer:
C) The natural frequency of the building should be very different from typical earthquake frequencies
Explanation:
We shall apply the concept of resonance in this problem .
When a body is applied an external harmonic force ( forced vibration) such that natural frequency of body is equal to frequency of external force or periodicity of external force , the body vibrates under resonance ie its amplitude of vibration becomes very high .
In the present case if natural frequency of building becomes equal to the earthquake's frequency ( external force ) , the building will start vibrating with maximum amplitude , resulting into quick collapse of the whole building . So to avoid this situation , natural frequency of building should be very different from typical earthquake frequencies .
If a car is traveling at an average speed of 20 m/s, how long will it take to travel 500 meters?
A. 0.04 seconds
B. 25 seconds
C. 520 seconds
D. 10,000 seconds
Answer:
B. 25 seconds
Explanation:
500÷20=25
Where do the rain forest plants get their nutrients?
from plants that fall to the forest floor
from animals that die on the forest floor
from plants and animals that die and fall to forest floor
from sunlight
Answer:
From plants and animals that die and fall to forests floor
Explanation:
Hope it helps
Mark brainliest
When a glass rod is rubbed with silk, the silk becomes negatively charged. Which of the following explanations best describes that is happening?
a. Electrons move from silk to glass
b. Electrons move from glass to silk
c. Protons move from silk to glass
d. Protons move from glass to silk
Answer: b. Electrons move from glass to silk
Explanation:
When the glass rod is rub over against the silk, the glass loses the electrons due to the physical contact with the silk and the friction so produced. The surface of the glass becomes positively charged and the surface of the silk becomes negatively charged. As the surface of the glass loses electrons which are accepted by the silk surface.
Show that the speed of a moving particle over a time interval is constant if and only if its velocity and acceleration vectors are perpendicular over the time interval.
Answer:
[tex]|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C[/tex]
[tex]2v(t)\cdot \frac{dv(t)}{dt}=0[/tex]
[tex]v(t)\cdot a(t)=0[/tex]
Explanation:
Let's start with the definition of a constant velocity.
If the velocity magnitude, in three dimensions, is a constant value (C) we have a constant velocity, which means.
[tex]|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C[/tex]
Now, we know that the dot product between v(t) and v(t) is the |v|².
[tex]v(t)\cdot v(t)=|v|^{2}(t)[/tex]
If we take the derivative whit respect to time in both sides of this equation we will have:
[tex]\frac{d}{dt}(v(t)\cdot v(t))=\frac{d}{dt}|v|^{2}(t)[/tex]
We apply the product rule on the left side and the right side will zero because the derivative of a constant is 0.
[tex]\frac{dv(t)}{dt}\cdot v(t)+v(t)\cdot \frac{dv(t)}{dt}=0[/tex]
[tex]2v(t)\cdot \frac{dv(t)}{dt}=0[/tex]
We know that dv(t)/dt = a(t) (using the acceleration definiton)
Therefore, we conclude:
[tex]v(t)\cdot \frac{dv(t)}{dt}=0[/tex]
[tex]v(t)\cdot a(t)=0[/tex]
If the dot product is 0, it means that v(t) and a(t) are orthogonal.
I hope it helps you!
A 62 kg student, starting from rest, slide down an 10.6 m high water slide. How fast is he going at the bottom of the slide? Use g = 10 m/s2
Answer:
14.6m/s
Explanation:
Given parameters:
Mass of the student = 62kg
Initial velocity = 0m/s
Height of slide = 10.6m
g = 10m/s²
Unknown:
Speed at the bottom of the slide = ?
Solution:
The speed at the bottom of the slide is the final velocity;
v ² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + 2x 10 x 10.6
v² = 212
v = 14.6m/s