A. Igneous
Extrusive igneous rocks cool and solidify quicker than intrusive igneous rocks. They are formed by the cooling of molten magma on the earth's surface.
Sedimentary (sed-uh-MEN-tuh-ree) rocks, unlike the other two kinds of rocks, are formed on or near Earth's surface. They form from sediments eroded from other rocks. The sediments may come from igneous, metamorphic, or other sedimentary rocks.
Aiden takes 0.10s to throw a baseball, which leaves his hand with a velocity of 49m/s. The balls acceleration is
Your car is initially at rest when your hit that gas and the car begins to accelerate. The forward force on the car is 2830 N while the resisting forces total 130 N. The car has a mass of 1830 kg and the acceleration lasts for 2.8 s. What is the final speed of the car and how much ground does it cover during this acceleration?
The final speed of the car is 4.131 m/s and it covers 5.78 m during this acceleration .
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given that: The forward force on the car is 2830 N while the resisting forces total 130 N.
Net force acting on the car is = 2830 N - 130 N = 2700 N.
The car has a mass of 1830 kg.
The acceleration of the car is = force/mass = 2700 N / 1830 m/s^2
= 1.475 m/s^2.
The acceleration lasts for 2.8 s.
Hence, the final speed of the car is = 1.475 m/s^2 × 2.8 s = 4.131 m/s.
Hence, it covers during this acceleration = (acceleration × time²)/2
= ( 1.475 × 2.8²)/2 m.
= 5.78 m.
Learn more about acceleration here:
brainly.com/question/12550364
#SPJ3
If you drag a sled across the ground, the sled experiences what kind of friction?
A. Kinetic friction
B. rolling friction
C. static friction
D. cohesive friction
Answer: The correct answer is option A.
Explanation:
Kinetic friction is defined as the frictional force acting between the two objects in contact. It resists the motion of one object over the other.
Rolling friction is defined as the frictional force which acts on the rolling object.
Static friction is the frictional force that keeps an object at rest.
Cohesive friction is defined as the frictional force that hold the particles together.
We need to drag a sled across the ground, so the frictional force is acting between two objects. Here, one object is sled and other is the ground. So, in order to move the sled, it must overcome kinetic friction.
Hence, the correct option is A.
Which is a correct reason to physically connect objects with a bond wire when transferring flammable liquids? A. To create a pathway for the liquid to flow easily B. To eliminate a difference in the static charge potential C. To lessen the likelihood of a spill
To create a pathway for the liquid to flow easily is a correct reason to physically connect objects with a bond wire when transferring flammable liquids. Physically connect the two conductive objects together with a bond wire to eliminate a difference in static charge potential among them. The bond wire is attached between the containers in the process of flammable filling liquid actions except for a metallic path is found.
Answer
Correct Option is A
Explanation:
Correct Option is A that is to create a pathway for the liquid to flow easily
Option B
This option is not correct because objects are not conductive.
Option C
This option is also not correct because better options are there to lessen the spill
Hope this answer will help you
What should a graph of Force vs. Mass look like to show the direct proportionality between the two? What does the slope of the line represent?
slope= Force/Mass
force is measured in Newtons (N) and mass in kilograms (kg). This means that the units of the slope are N/kg. However a Newton is also equal to one kg*m/s2, so if we divide this by one kg we are left with a slope with units m/s2, which acceleration is measured in!
Rock a is dropped from a cliff and rock b is thrown upward from the same position on the cliff.
a. when each rock reaches the ground at the bottom of the cliff, which rock has a greater velocity?
In the A case, v0=0, in the B case v0 >0, so v is greater in the B case. In other words, v is related to the kinetic energy, since the rock B has a larger kinetic energy in the beginning, it also has a larger kinetic energy (larger velocity) in the end.
The correct answer is that rock a has a greater velocity when each rock reaches the ground at the bottom of the cliff.
To understand why rock a has a greater velocity upon impact with the ground, let's consider the physics of free-falling objects under the influence of gravity. Both rocks are subject to the acceleration due to gravity, which is approximately[tex]\(9.81 \, \text{m/s}^2\)[/tex] downward.
For rock a, which is simply dropped from rest, its initial velocity is zero. The only force acting on it is gravity, pulling it downward. Its velocity at any time \(t\) before it hits the ground can be calculated using the equation for uniformly accelerated motion:
[tex]\[ v_{a} = u_{a} + gt \][/tex]
where[tex]\(v_{a}\)[/tex] is the final velocity, [tex]\(u_{a}\)[/tex] is the initial velocity (which is zero for rock a), [tex]\(g\)[/tex] is the acceleration due to gravity, and [tex]\(t\)[/tex] is the time of fall. Since [tex]\(u_{a} = 0\)[/tex], the equation simplifies to:
[tex]\[ v_{a} = gt \][/tex]
For rock b, which is thrown upward, its initial velocity[tex]\(u_{b}\)[/tex] is upward and has some positive value. As it rises, it slows down due to the acceleration due to gravity acting in the opposite direction of its motion. At its highest point, its velocity will be zero, and then it will start to fall back down, accelerating due to gravity until it hits the ground. The velocity of rock b just before impact can be calculated using the same equation for uniformly accelerated motion, but taking into account that it starts with an initial upward velocity and then reverses direction:
[tex]\[ v_{b} = u_{b} - gt_{up} + gt_{down} \][/tex]
where [tex]\(t_{up}\)[/tex]is the time it takes for rock b to reach its highest point and \[tex](t_{down}\)[/tex]is the time it takes to fall back to the ground. Since the acceleration due to gravity is the same for both rocks, and assuming that air resistance is negligible, the time of ascent [tex]\(t_{up}\)[/tex] for rock b is equal to the time of descent [tex]\(t_{down}\), so \(t_{up} = t_{down} = t/2\).[/tex] The equation for rock b's velocity upon impact becomes:
[tex]\[ v_{b} = u_{b} - g(t/2) + g(t/2) \] \[ v_{b} = u_{b} \][/tex]
Since rock a was dropped from rest, its velocity upon impact is due solely to the acceleration due to gravity over the entire time \(t\) of its fall, while rock b's velocity upon impact is its initial launch velocity minus the effect of gravity over half the time (as it rises) and plus the effect of gravity over the other half of the time (as it falls).
Because rock a starts from rest and only accelerates downward, it will have a greater velocity upon impact than rock b, which has its initial upward velocity reduced by the same gravitational acceleration over the time of its ascent and then increased by the same gravitational acceleration over the time of its descent. Thus, rock a's velocity at the moment of impact will be greater than rock b's velocity at the moment of impact, assuming they both hit the ground at the same time and air resistance is negligible.
A 500 kg cannon exerts a force of 600 N on a 10 kg cannon ball. Find (a) the acceleration of the cannon and (b) the acceleration of the cannon ball.
A 500 kg cannon exerts 600 N force on 10 kg cannon ball
Now here we can say that as per Newton's III law every action has equal and opposite reaction force
So here when cannon exert a force of 600 N on the ball then equal and opposite magnitude force will act on cannon itself
Part a)
Okay now we will find the acceleration of cannon using the same force
so as per Newton's formula
[tex]F = ma[/tex]
[tex]600 = 500 * a[/tex]
[tex]a = 1.2 m/s^2[/tex]
Part b)
Okay now we will find the acceleration of cannon ball using the given force
so as per Newton's formula
[tex]F = ma[/tex]
[tex]600 = 10 * a[/tex]
[tex]a = 60 m/s^2[/tex]
HELP!! A car driving down a hill contains many types of energy. DESCRIBE at least 3 types of energy the car has.
1. Kinetic
2. Chemical
3. Potential
whats the standard unit for measuring mass
Four students are comparing vectors and scalars. The chart contains each students statements about vectors and scalars
In summary, vectors are quantities with both magnitude and direction, and their addition and multiplication by scalars follow specific rules. Scalars are quantities with only magnitude and are added by simple arithmetic. The correct option for any statement about vectors and scalars would be the one that accurately reflects these properties
To provide a solution to the question, we need to understand the key differences between vectors and scalars as described by the students in the chart. Here are the typical characteristics that distinguish vectors from scalars:
1. Magnitude and Direction: Vectors have both magnitude and direction, whereas scalars have only magnitude. This is the fundamental difference between the two. For example, velocity is a vector quantity because it describes both the speed (magnitude) and the direction of motion, while speed alone is a scalar quantity.
2. Representation: Vectors are often represented by arrows, where the length of the arrow corresponds to the magnitude and the arrowhead indicates the direction. Scalars are typically represented by a number with units, if applicable.
3. Addition: When adding vectors, both their magnitudes and directions must be considered, which often involves geometric construction or trigonometric calculations. In contrast, scalar addition is straightforward and involves simply adding the numerical values.
4. Multiplication by a Scalar: Multiplying a vector by a scalar changes the magnitude of the vector but not its direction. The result is a vector that is parallel to the original vector.
5. Dimensionality: Vectors can be one-dimensional (straight line), two-dimensional (plane), or three-dimensional (space). Scalars are zero-dimensional, as they are just single values without direction.
6. Physical Quantities: Examples of vector quantities include displacement, velocity, acceleration, and force. Examples of scalar quantities include mass, temperature, energy, and time.
Based on these characteristics, the students' statements can be analyzed to determine if they are correct or incorrect. For instance, if a student says that a vector has only magnitude, this statement is incorrect because vectors must have both magnitude and direction.
To solve the question accurately, we would need to see the specific statements made by each student in the chart. However, without the actual statements, we can only provide a general explanation of the differences between vectors and scalars.
In summary, vectors are quantities with both magnitude and direction, and their addition and multiplication by scalars follow specific rules. Scalars are quantities with only magnitude and are added by simple arithmetic. The correct option for any statement about vectors and scalars would be the one that accurately reflects these properties."
Dalton's model of an atom is best described as _____. A. A solar system B. A cloud C. A scoop of chocolate chip ice cream D. A solid sphere
Solid sphere is the correct answer.
Answer:
D) A Solid Sphere
Explanation:
As per Dalton's atomic description he said that atom is not divisible and that is the smallest part of atom.
It has no net charge and it can not be divided further. So till Dalton's theory of atom electron and nucleons were not discovered.
So we can consider atom as a model of Solid sphere because similar to a solid sphere the atom is considered to be unique and not destructible.
So here correct answer will be
D) A Solid Sphere
The decomposition of water into hydrogen gas h2 and oxygen gas o2 can be moldled by the balanced chemical equation
If you had a positive 8 score on the sit-an-reach test you
Answer:you are very flexible
Explanation:
That’s a very high score.
Which is defined as a disturbance in matter that carries energy from one place to another?
an acoustic wave carries energy
A mechanical wave----
1.
When driving in rain the road becomes the most slippery:
Just after the rain quits
Right after it starts to rain
After it has rained for 20 minutes
When the wind is blowing at least 25 mph
When driving in rain, the road becomes the most slippery right after it starts to rain.
As the first rainwater hits the road, it mixes with the oil, rubber tire particles, and other gunk on the pavement, forming a very slippery coating on the surface.
After it has rained for a while, a lot of that gunk gets washed away, and most of the slipperiness is just water ... which is pretty slippery on its own, but not as slippery as it is when it has oil and rubber particles in it too.
The current in the circuit shown is 2.0 A.
What is the value of R3?
A) 10 Ω
B) 15 Ω
C) 20 Ω
D) 55 Ω
The value of R3 is A) 10 Ω
Answer:
Resistor R₃ = 10 Ω
Explanation:
It is given that,
Current flowing in the circuit, I = 2 A
Resistors R₁ = 2 Ω
Resistor R₂ = 3 Ω
Voltage, V = 30 V
We have to find the value of R₃. All three resistors are connected in series. In series combination, the current flowing through the all resistors is same. Firstly, calculating equivalent resistance of three resistors as :
[tex]R_{eq}=R_1+R_2+R_3[/tex]
[tex]R_{eq}=2\ \Omega+3\ \Omega+R_3[/tex]
[tex]R_{eq}=5\ \Omega+R_3[/tex]..............(1)
Using Ohm's law : [tex]V=IR_{eq}[/tex]
[tex]30\ V=2\ A\times (5 + R_{3})[/tex]
On solving above equation : [tex]R_3=10\ \Omega[/tex]
Hence, the correct option is (A) " 10 Ω "
If you were standing on the moon and you had a radio signal transmitter, you could send the radio signal to Earth in 1.28 seconds. If radio waves travel at 3.00 X 1018 m/s, what is the distance from the moon to the earth in kilometers?
Answer:
Distance between moon and earth = [tex]3.84*10^{5}km[/tex]
Explanation:
Distance traveled by radio waves = Velocity of radio waves * Time taken by radio waves
Velocity of radio waves = [tex]3*10^{8}m/s[/tex]
Time taken by radio waves = 1.28 seconds
So distance traveled by radio wave = [tex]3*10^{8}*1.28=3.84*10^{8}m[/tex]
Distance between moon and earth = [tex]3.84*10^{8}m[/tex]=[tex]3.84*10^{5}km[/tex]
Identify two factors that determine the intensity of sound
Amplitude and distance.
amber applies a net force of 35 N to pull her younger brother (32 kg) in a wagon (3 kg). at what rate does the wagon accerlerate
A. 1.09 m/s2
B. 0.91 m/s2
C. 1.00 m/s2
D. It doesn’t accelerate because force and mass are equal
Answer:
C. 1.00 m/s^2
Explanation:
We can solve the problem by using Newton's second law:
[tex]F=ma[/tex]
where
F is the net force acting on an object
m is the mass of the object
a is its acceleration
In this case, the net force applied by Amber is F=35 N, while the total mass of the brother+the wagon is
[tex]m=32 kg+3 kg=35 kg[/tex]
Therefore, we can re-arrange the previous equation to find the acceleration:
[tex]a=\frac{F}{m}=\frac{35 N}{35 kg}=1.00 m/s^2[/tex]
A car's position in relation to time is plotted on the graph. What can be said about the car during segment B?
A) The car travels for 80 seconds during segment B.
B) The car has come to a stop and has zero velocity.
C) The car is traveling faster during segment B than in segment C.
D) The car is traveling with a constant velocity due to the flat line of the graph.
we know that
the speed is equal to
[tex]speed=\frac{distance}{time}[/tex]
The slope of the line on the graph is equal to the speed of the car
so
during the segment B the slope of the line is equal to zero
that means
the speed of the car is zero
therefore
the answer is the option B
The car has come to a stop and has zero velocity
Option (B) is correct .i.e. the car has come to a stop and has zero velocity.
Further Explanation:
According to the theory of graphs, ‘the straight line parallel to x axis’ means the quantity given on y axis is not changing or is constant with respect to the change in the quantity on x axis. This is why because the slope of that particular straight line is zero.
Concept:
Mathematically, the slope is defined as the tangent (trigonometry) of the angle made by the segment with the x axis or it is also defined as the ratio of change in the quantity on y-axis to change in the quantity on x-axis.
[tex]\boxed{\text{slope}=\tan(\theta)}[/tex]
Here, [tex]\theta[/tex] is the angle made by the segment of graph with x-axis
The slope can also be expressed as:
[tex]\boxed{\text{slope}=\dfrac{\text{change in quantity given on y-axis}}{\text{change in quantity given on x-axis}}}[/tex]
Here, the quantity given on y axis is car’s position and that given on x axis is time taken by the car to travel.
The slope of position-time graph gives velocity. So, by the definition of slope, we have
[tex]{\text{velocity}=\dfrac{\text{change in position}}{{\text{change in time}}}[/tex]
Here, as shown in the graph, the two ends of the segment B have same values of position on y axis. So, change in position will be zero.
Substitute [tex]0[/tex] for [tex]\text{change in position}[/tex] in above expression.
[tex]\text{Velocity}=0\text{ m/s}[/tex]
This shows that the speed of the car in the segment B of the graph will be zero as there is no change in the position of the car with respect to time.
Thus, option (B) is correct .i.e. the car has come to a stop and has zero velocity.
Learn More:
1. A highway patrolman traveling at the speed limit is passed by a car https://brainly.com/question/2456051
2. Head-on collision of the two bodies https://brainly.com/question/2097915
Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Position, car, relation to time, segment B, 80 sec, zero velocity, travelling, faster, graph, position vs time, constant velocity, slope, change in position, change in time.
The acceleration due to gravity on the moon is 1.6 m/s2, about a sixth that of Earth’s. Which accurately describes the weight of an object on the moon? An object on the moon is 1/6 times lighter than on Earth. An object on the moon is 1/6 times heavier than on Earth. An object on the moon is six times lighter than on Earth. An object on the moon is six times heavier than on Earth.
The answer is C.
An object on the moon is six times lighter than on Earth.
The statement accurately describes the weight of an object on the moon is An object on the moon is 1/6 times lighter than on Earth.
What is weight?The weight is the product of mass of the object and acceleration due to gravity on the specific planet.
When the acceleration due to gravity is less, the weight measured will be less. The acceleration due to gravity on the moon is 1.6 m/s2, about a sixth that of Earth’s.
Thus, an object on the moon is 1/6 times lighter than on Earth.
Learn more about Weight.
https://brainly.com/question/10069252
#SPJ2
Are external computer flash drive can hold 1 gigabyte of data. How many bytes is this?
The prefix "giga..." means 1 billion. We talk about that flash drive as if its capacity is exactly 1 billion bytes.
But with digital things, a "giga..." Is actually 2^30 = 1,073,741,824 .
While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the top of the science building. Determine the distance the bag has traveled after falling for 1.5 seconds assuming it has reach free fall and given the gravitational acceleration of 9.8 m/sec2.
In the experiment of free fall bob released a bag of mass 1 lb
so here we can say that initial speed of the bag is Zero
time taken by the bag to free fall is given as
t = 1.5 s
also the acceleration of free fall is given as
a = 9.8 m/s^2
now we will use kinematics equation here for finding the distance of free fall
[tex]d = v_i * t + \frac{1}{2} at^2[/tex]
[tex]d = 0 + \frac{1}{2}*9.8* 1.5^2[/tex]
[tex]d = 4.9 * 2.25[/tex]
[tex]d = 11.025 m[/tex]
so the bag will fall down by total distance of 11.025 m from its initial released position.
The 1.0 lb bag of feathers has traveled approximately 11.025 meters after falling for 1.5 seconds, assuming it is in free fall and subject to Earth's gravitational acceleration of 9.8 m/s².
Explanation:To determine the distance the 1.0 lb bag of feathers has traveled after falling for 1.5 seconds, assuming it has reached free fall and is only affected by gravitational acceleration, we can use the formula for the distance covered by an object in free fall:
d = ½ * g * t²,
where g is the acceleration due to gravity (9.8 m/s²) and t is the time in seconds.
Plugging in the values:
d = ½ * 9.8 m/s² * (1.5 s)²
d = ½ * 9.8 * 2.25
d = ½ * 22.05
d = 11.025 meters
Therefore, the bag of feathers has traveled approximately 11.025 meters after falling for 1.5 seconds.
Which of these are minerals? Select all that apply
A) mica
B) quartz
C) steam
D) copper
From that particular list:
Mica (A), Quartz (B), and Copper (D) are minerals.
Steam (C) isn't.
When the amount of water in a river increases so much that the river overflows its channel, a flood occurs.True or false
That's true, and there are also other mechanisms that can also create floods.
a ball is thrown upwards at an unknown speed. in a time of
.5 second the ball has a height of 6 meters.
what is the initial speed of the ball?
what is the balls speed at a height of 6 meters?
what would be the balls maximum height? (hint: you dont know the time to reach maximum height)
Answer:
a) Initial speed of the ball = 14.45 m/s
b) At height 6 m speed of ball = 9.55 m/s
c) Maximum height reached = 10.65 m
Explanation:
a) We have equation of motion [tex]s=ut+\frac{1}{2} at^2[/tex], where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.
s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8[tex]m/s^2[/tex]
Substituting
[tex]6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s[/tex]
Initial speed of the ball = 14.45 m/s
b) We have equation of motion [tex]v^2=u^2+2as[/tex], where v is the final velocity
s = 6 m, u = 14.45 m/s, a = -9.8[tex]m/s^2[/tex]
Substituting
[tex]v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s[/tex]
So at height 6 m speed of ball = 9.55 m/s
c) We have equation of motion [tex]v^2=u^2+2as[/tex], where v is the final velocity
u = 14.45 m/s, v =0 , a = -9.8[tex]m/s^2[/tex]
Substituting
[tex]0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m[/tex]
Maximum height reached = 10.65 m
Briefly describe the history of the metric system as it applies to the meter and how the definition of a meter has changed over time.
Answer and explanation;
In 1670 Gabriel Mouton, Vicar of St. Paul’s Church and an astronomer proposed the swing length of a pendulum with a frequency of one beat per second as the unit of length.
In 1791 the Commission of the French Academy of Sciences proposed the name meter to the unit of length. It would equal one tens-millionth of the distance from the North Pole to the equator along the meridian through Paris.It is realistically represented by the distance between two marks on an iron bar kept in Paris.
In 1889 the 1st General Conference on Weights and Measures define the meter as the distance between two lines on a standard bar that made of an alloy of 90%platinum with 10%iridium.
In 1960 the meter was redefined as 1650763.73 wavelengths of orange-red light, in a vacuum, produced by burning the element krypton (Kr-86).
In 1984 the Geneva Conference on Weights and Measures has defined the meter as the distance light travels, in a vacuum, in 1299792458⁄ seconds with time measured by a cesium-133 atomic clock which emits pulses of radiation at very rapid, regular intervals.
The metric system was first put into practice in 1799, during the French Revolution, when the existing system of measures became impractical for trade and was supplanted by a decimal system based on the kilogram and the meter.
What is metric system?During the French Revolution, the existing system of measures became impractical for trade and was replaced by a decimal system based on the kilogram and the meter, and the metric system was born.
In 1793, the meter was defined as one ten-millionth of the distance from the equator to the North Pole along a great circle, implying that the Earth's circumference is approximately 40000 km.
The meter was redefined in 1799 in terms of a prototype meter bar.
The meter was introduced as a new unit of length, defined as one ten-millionth of the shortest distance between the North Pole and the Equator passing through Paris, assuming an Earth flattening of 1/334.
Thus, this is the history of the metric system as it applies to the meter.
For more details regarding metric system, visit:
https://brainly.com/question/25966695
#SPJ2
What are the two possible solutions to 5x^2−15=0?
I have attached the solution as image.
Final Answer:
The two possible solutions of [tex]5x^{2} - 5 = 0[/tex] are [tex]\( x = \pm \sqrt{3} \)[/tex].
Explanation:
To find the solutions to the quadratic equation [tex]\(5x^2 - 15 = 0\)[/tex], we first add 15 to both sides to isolate the term involving [tex]\(x^2\)[/tex], yielding [tex]\(5x^2 = 15\)[/tex]. Then, we divide both sides by 5 to solve for [tex]\(x^2\)[/tex], giving us [tex]\(x^2 = 3\)[/tex]. Taking the square root of both sides, we obtain [tex]\(x = \pm \sqrt{3}\)[/tex]. Therefore, the two possible solutions are [tex]\(x = \sqrt{3}\)[/tex] and [tex]\(x = -\sqrt{3}\)[/tex]. These solutions represent the values of x for which the equation is true.
A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are running at full speed, with the gazelle 96.8 m ahead, how long before the cheetah catches its prey? Answer in units of s.
The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape? (After 7.5 s the speed of cheetah is less than that of the gazelle.) Answer in units of m.
(1)
Cheetah speed: [tex]v_c = 97.8 km/h=27.2 m/s[/tex]
Its position at time t is given by
[tex]S_c (t)= v_c t[/tex] (1)
Gazelle speed: [tex]v_g = 78.2 km/h=21.7 m/s[/tex]
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
[tex]S_g(t)=S_0 +v_g t[/tex] (2)
The cheetah reaches the gazelle when [tex]S_c=S_g[/tex]. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
[tex]v_c t=S_0 + v_g t[/tex]
[tex](v_c -v_g t)=S_0[/tex]
[tex]t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s[/tex]
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah: [tex]S_c = v_c t =(27.2 m/s)(7.5 s)=204 m[/tex]
Gazelle: [tex]S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m[/tex]
So, the gazelle should be ahead of the cheetah of at least
[tex]d=S_c -S_g =204 m-162.8 m=41.2 m[/tex]
The cheetah catches the prey [tex]\fbox{17.78 s}[/tex] before the gazelle.
The minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is [tex]\fbox{40.83 m}[/tex].
Further Explanation:
The speed is the rate of change of the distance and the relative speed is the speed of the one object in respect of another object. If two body moves in same direction then the relative speed in respect of the one body is given by subtract of both speed and it they are moving in opposite direction the relative speed in respect of the one body is given by sum of both speeds.
Given:
The maximum speed of the cheetah is [tex]97.8 km/h[/tex].
The maximum speed of the gazelle is [tex]78.2 km/h[/tex].
The distance between the starting points of cheetah and the gazelle is [tex]96.8 m[/tex].
Concept:
The expression for the position of the cheetah at some time t is:
[tex]{S_1}={v_c}t[/tex] …… (1)
Here, [tex]{S_1}[/tex]the position of the cheetah at some time [tex]t[/tex], [tex]{v_c}[/tex] is the speed of the cheetah and [tex]{S_1}[/tex] is the distance covered at time [tex]t[/tex].
The expression for the position of the gazelle at some time [tex]t[/tex] is:
[tex]{S_2}={S_o}+{v_g}t[/tex] …… (2)
Here, [tex]{S_2}[/tex] the position of the gazelle at some time [tex]t[/tex], [tex]{v_g}[/tex] is the speed of the gazelle, [tex]{S_o}[/tex] is the distance between the starting points of cheetah and the gazelle.
At the time when the cheetah reached the gazelle then [tex]{S_1}={S_2}[/tex].
Equate equation (1) and (2).
[tex]\begin{aligned}{v_c}t&={S_o}+{v_g}t\hfill\\{v_c}t-{v_g}t&={S_o}\hfill\\t\left( {{v_c} - {v_g}}\right)&={S_o}\hfill\\t&=\frac{{{S_o}}}{{\left( {{v_c} - {v_g}} \right)}}\hfill\\\end{aligned}[/tex]
Substitute [tex]97.8 km/h[/tex] for [tex]{v_c}[/tex], [tex]78.2 km/h[/tex] for [tex]{v_g}[/tex] and [tex]96.8 m[/tex] for [tex]{S_o}[/tex] in the above equation.
[tex]\begin{aligned}t&=\frac{{96.8{\text{ m}}}}{{\left( {97.8 - 78.2} \right){\text{km/h}}\left({\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)}}\\&=\frac{{96.8{\text{ m}}}}{{{\text{19}}{\text{.6 m/s}}\left( {\frac{5}{{18}}}\right)}}\\&=17.78{\text{s}}\\\end{aligned}[/tex]
Therefore, the cheetah catches the prey [tex]17.78 s[/tex] before the gazelle.
To calculate the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape we should subtract the distance covered by the cheetah and gazelle in [tex]7.5 s[/tex].
The expression for the distance covered is:
[tex]S = vt[/tex]
Here, [tex]S[/tex] is the distance covered, [tex]v[/tex] is the speed and [tex]t[/tex] is the travel time.
Substitute [tex]97.8 km/h[/tex] for [tex]v[/tex] and [tex]7.5 s[/tex] for [tex]t[/tex] in the above equation.
[tex]\begin{aligned}S&=\left( {97.8{\text{ km/h}}} \right)\left( {\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)\left( {7.5{\text{ s}}} \right)\\&=\left( {97.8{\text{ m/s}}} \right)\left( {\frac{5}{{18}}} \right)\left( {7.5{\text{ s}}} \right)\\&=203.75{\text{m}}\\\end{aligned}[/tex]
Therefore, the distance covered by the cheetah in [tex]7.5 s[/tex] is [tex]203.75 m[/tex].
Substitute [tex]78.2 km/h[/tex] for [tex]v[/tex] and [tex]7.5 s[/tex] for [tex]t[/tex] in the above equation.
[tex]\begin{aligned}S&=\left( {78.2{\text{ km/h}}} \right)\left( {\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)\left( {7.5{\text{ s}}} \right)\\&=\left( {78.2{\text{ m/s}}} \right)\left( {\frac{5}{{18}}} \right)\left( {7.5{\text{ s}}} \right)\\&=162.92{\text{ m}}\\\end{aligned}[/tex]
Therefore, the distance covered by gazelle in [tex]7.5 s[/tex] is [tex]162.92 m[/tex].
The minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is:
[tex]\begin{aligned}d&=\left( {203.75 - 162.92} \right){\text{ m}}\\&=40.83{\text{ m}}\\\end{aligned}[/tex]
Therefore, the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is [tex]\fbox{40.83 m}[/tex].
Learn more:
1. Conservation of momentum https://brainly.com/question/7031524.
2. Motion under friction https://brainly.com/question/11023695.
3. Motion under force https://brainly.com/question/4033012.
Answer Details:
Grade: High school
Subject: Physics
Chapter: Kinematics
Keywords:
Cheetah, 97.8 km/h, gazelle, 78.2 km/h, full speed, 96.8 m. ahead, catches, prey, 7.5 s, minimum, ahead, escape, answer, unit, s, m, 40.83 m.
What are two examples of common units for each of the above measurements
Density: g/mL, kg/cubic meter
Volume: L, teaspoon
Mass: g, MeV/sq. C
The common units for length, weight, and time in the American system are foot, pound, and second, respectively. In the metric system, the common units for length, weight, and time are meter, kilogram, and second, respectively.
Explanation:In the American system of measurement, the common units for length, weight, and time are foot, pound, and second, respectively. For example, the ton (2240 lb) is a larger unit of weight, the yard (3 ft) is a smaller unit of length, and the minute (60 s) is a smaller unit of time.
In the metric system, the common units for length, weight, and time are meter, kilogram, and second, respectively. For example, the kilometer (1000 m) is a larger unit of length, the gram (0.001 kg) is a smaller unit of weight, and the millisecond (0.001 s) is a smaller unit of time.
Learn more about Units for Measurements here:https://brainly.com/question/18982750
#SPJ3