Answer:
58.8J
Explanation:
Given parameters;
Mass of ball = 4kg
Height above the floor = 1.5m
g = 9.8n/kg
Unknown:
Potential energy = ?
Solution:
The potential energy of a body is the energy due to the position of the body.
It is mathematically expressed as:
Potential energy = mass x acceleration due to gravity x height
Potential energy = 4 x 9.8 x 1.5 = 58.8J
According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car that accelerates from rest on a flat road. What force did the work that increased the car’s kinetic energy?
1. the force of the car engine
2. air resistance
3. the friction between the road and the tires
4. gravity
Answer:
1. The force of the car engine.
Explanation:
We shall see the effect and role played by each force, one by one, as follows:
1. The force of car engine:
The engine produces a force through combustion that is converted to mechanical work through the shaft. This work is then transmitted to the wheels of the car that cause the motion in the car and increase its kinetic energy.
2. Air Resistance:
It is the opposing force of air that tries to reduce the motion of the car and as a result, reduce its kinetic energy.
3. Frictional Force between road and tire:
It is also the opposing force of air that tries to reduce the motion of the car and as a result, reduce its kinetic energy.
4. Gravity:
Gravity pulls everything towards the center of Earth so it does not have much significant role in horizontal motion like this.
Hence, the force of the car engine did the work that increased the car's kinetic energy.
7. A motorcycle accelerates from rest at a rate of 4 m/s2 while traveling 60m. What is the motorcycle's velocity at
the end of this motion, to the nearest whole number?
A. 240 meters/second
B. 22 meters/second
c. 15 meters/second
D. O meters/second
Answer: C
Explanation: 60 divided by 4 =15
Velocity can be defined as the rate of change of distance with time
Given data
Acceleration = 4/ms^2
Distance = 60m
Initial Velocity U= 0
Final Velocity V= ?
The expression for velocity is given by
V^2= U^2+2as
Let us substitute our given data into the expression
V^2 = 0^2 + 2*4*60
V^2 = 480
Square both sides
V= √480
V= 21.9 meters/second
V= 22 meters/seconds Approx.
The correct answer is option B
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Which element has a complete valence electron shell?
selenium (Se)
oxygen (O)
fluorine (F)
argon (Ar)
Answer:
argon (Ar)
Explanation:
Argon is the element from the given choices with a complete valence electron shell.
The valence electron shell is the outermost shell of an atom.
Elements with complete outermost shell are found in the 8th group on the period table. In the 8th group, the elements are generally inert and unreactive. Elements with this configuration have 8 electrons in their outermost shell and 2 for helium. Some of the elements in this group are Helium and NeonAnswer:
argon (Ar)
Explanation:
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 44 m/s. The plane is in a 18 m/s wind blowing toward the southwest relative to earth.
a. letting x be east and y be north, find the components of \vec v_{\rm P/E} (the velocity of the plane relative to the earth.
b.Find the magnitude of \vec v_{\rm P/E}.
c.Find the direction of \vec v_{\rm P/E}.
Answer:
a) vx = -12.7 m/s vy = -56.7m/s
b) v= 58.1 m/s
c) θ = 77.4º S of W
Explanation:
a)
In order to get the components of the velocity of the plane relative to the earth, we need just to get the components of both velocities first:Since the nose of the plane is pointing south, if we take y to be north, and positive, this means that the velocity of the plane can be written as follows:[tex]v_{ps} = -44m/s (1)[/tex]
Since the wind is pointing SW, it's pointing exactly 45º regarding both directions, so we can find its components as follows (they are equal each other in magnitude)[tex]v_{we} = - 18m/s * cos (45) = -12.7 m (2)[/tex]
[tex]v_{ws} = - 18m/s * cos (45) = -12.7 m (3)[/tex]
The component of v along the x-axis is simply (2), as the plane has no component of velocity along this axis:[tex]v_{e} = v_{x} = -12.7 m/s (4)[/tex]
The component of v along the -y axis is just the sum of (1) and (3)[tex]v_{y} = -44 m/s + (-12.7m/s) = -56.7 m/s (5)[/tex]b)
We can find the magnitude of the velocity vector, just applying the Pythagorean Theorem to (4) and (5):[tex]v = \sqrt{(-12.7m/s)^{2} + (-56.7m/s)^{2}} = 58.1 m/s (6)[/tex]
c)
Taking the triangle defined by vx, vy and v, we can find the angle that v does with the negative x-axis, just using the definition of tangent, as follows:[tex]tg_{\theta} =\frac{v_{y} }{v_{x} } = \frac{(-56.7m/s)}{(-12.7m/s} = 4.46 (7)[/tex]
Taking tg⁻¹ from (7), we get:tg⁻¹ θ = tg⁻¹ (4.46) = 77.4º S of W. (8)
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.30 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 38.0 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
4.79m/s
Explanation:
According to law of conservation of momentum;
The sum of momentum of the bodies before collision is equal to the momentum after collision.
m1u1 + m2u2 = (m1+m2)v
Given;
m1 = 0.3kg
u1 = 2.30m/s
m2 = 0.0225kg
u2 = 38m/s
Required
speed of the arrow after passing through the target v
Substituting the given data into the formula
0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v
0.69 + 0.855 = 0.3225v
1.545 = 0.3225v
v = 1.545/0.3225
v = 4.79m/s
Hence the speed of the arrow after passing through the target is 4.79m/s
How to find average speed in physics
Answer: you divide total distance by time. To get the time, divide total distance by speed. To get distance, multiply speed times the amount of time.
Explanation:
I hope this helps
A car pulls on to an onramp with an initial speed of 23.8 mph. The length of the onramp is 852 ft and the car needs to be moving at 45.7 mph at the end of the ramp to merge with traffic. What constant rate of acceleration (in ft/sec2) is required in order to accomplish this
Answer:
The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.
Explanation:
Let suppose that car accelerates uniformly in a rectilinear motion. Given that initial and final speeds and travelled distances are known, then the acceleration needed by the vehicle ([tex]a[/tex]), measured in feet per square second, is determined by the following kinematic formula:
[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot \Delta x }[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds, measured in feet per second.
[tex]\Delta x[/tex] - Travelled distance, measured in feet.
If we know that [tex]v_{o} = 34.907\,\frac{ft}{s}[/tex], [tex]v_{f} = 67.027\,\frac{ft}{s}[/tex] and [tex]\Delta x = 852\,ft[/tex], then acceleration needed to accomplish the task is:
[tex]a = 1.921\,\frac{ft}{s^{2}}[/tex]
The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0N, the second child exerts a force of 90.0 N, friction is 12.0 N, and the most of the third child plus wagon is 23.0 kga)what is the system of interest if the acceleration of the child in the wagon is to be calculated
Answer:
Explanation:
75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .
Friction force will act in the direction opposite to the direction of net force .
So friction force will act in the direction in which 75 N is acting .
Total force acting in the direction of 75 = 75 + 12 = 87 N
Net force acing on the third child = 90 - 87 = 3 N
Its direction will be that in the direction of 90 N .
46) Recoil is noticeable if we throw a heavy ball while standing on roller skates. If instead we go through the motions of throwing the ball but hold onto it, our net recoil will be
Answer:
Zero
Explanation:
Appearing to throw the ball but still holding on to it means the recoil velocity will be zero because the recoil velocity is defined as the backward velocity as a result of throwing an object or shooting a bullet. In this case the object was not thrown and thus there is no recoil velocity.
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers.
Answer:
D = -4/7 = - 0.57
C = 17/7 = 2.43
Explanation:
We have the following two equations:
[tex]3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)[/tex]
First, we isolate C from equation (2):
[tex]2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)[/tex]
using this value of C from equation (3) in equation (1):
[tex]3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}[/tex]
D = - 0.57
Put this value in equation (3), we get:
[tex]C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\[/tex]
C = 2.43
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim. Which of the following measuring tools can the student use to test the validity of the claim?
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
c. A motion detector that is oriented perpendicular to the direction that the car travels.
d. A mass balance to determine the mass of the car
Answer:
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
Explanation:
Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.
In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.
Therefore, to test the validity of the claim, the student should use the following measuring tools;
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.
b. A meterstick to measure the distance of the track that the car travels on.
Hence, with these two devices the student can effectively measure or determine the validity of the claim.
A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.40 cm from the wire and traveling at a speed of 6.20 * 104 m>s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron
Answer:
Explanation:
Magnetic field due to current at a distance of 4.4 cm
B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻² [ B = 10⁻⁷ x 2i / r = ]
= 2.36 x 10⁻⁵ T.
Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .
Force = 2.36 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴
= 23.41 x 10⁻²⁰ N .
This force will be perpendicular to the direction of current .
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
the masses as like terms)
Answer:
Force = 2240 Newton.
Explanation:
Given the following data;
Mass A= 65kg
Mass B = 215kg
Acceleration = 8m/s²
To find the force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
First of all, we would have to find the total mass.
Total mass = Mass A + Mass B
Total mass = 65 + 215
Total mass = 280kg
Substituting into the equation, we have
[tex] Force = 280 * 8 [/tex]
Force = 2240 Newton.
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas.
Required:
a. How much heat flows into or out of the gas?
b. What is the change in internal energy of the gas?
c. Does its temperature rise or fall?
Answer:
[tex]Q=0[/tex][tex]U=2820[/tex]Energy increasesExplanation:
From the question we are told that
Work done [tex]W=2820[/tex]
a)Generally the heat flow for an adiabatic process is 0 (zero)
[tex]Q= U + W =>0[/tex]
[tex]Q=0[/tex]
b)Generally Change in internal energy of gas is mathematically given by
Since [tex]W=-2820J[/tex]
Therefore
[tex]U=2820[/tex]
Giving
[tex]Q= 2820 -2820[/tex]
[tex]Q=O[/tex]
c)With increases in internal energy brings increase in temperature
Therefore
Energy increases
An engineer claims to have measured the characteristics of a heat engine that takes in 150 J of thermal energy and produces 50 J of useful work. What is the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs?
Answer:
1.4999
Explanation:
Efficiency can be calculated using below expresion
Efficiency = W/Q.............eqn(1)
Where W= work = 50 J
Q= thermal energy= 150 J
But
W/Q= (Th-Tc)/Th ...........Eqn(2)
Th= temperature of the hot
Tc= temperature of the cold
Where Th/ Tc= ratio of the temperature hot and cold reservoirs?
If we simplify eqn(2) we have
W/Q = 1-Tc/Th.........eqn(3)
If we make the ratio subject of the formula we have
Tc/Th = 1-(W/Q)
Th/Tc = 1/(1-W/Q )
Then substitute the values
= 1/(1-50/150) = 1.4999
Hence, the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs is 1.4999
Consider two points in an electric field. The potential at point 1, V1, is 24V. The potential at point 2, V2, is 154V. A proton is moved from point 1 to point 2.
(a) Write an equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e.
(b) Find the numerical value of the change of the electric potential energy in electron volts (eV).
(c) Express v2, the speed of the electron at point 2, in terms of AU, and the mass of the electron me.
(d) Find the numerical value of v2 in m/s
Answer:
[tex]\triangle U=-e (V_2-V_1)[/tex]
[tex]\triangle U=130eV[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
Explanation:
From the question we are told that
The potential at point 1, [tex]V_1 = 24V[/tex]
The potential at point 2, [tex]V_2 = 154V[/tex]
a)Generally work done by proton is given as
[tex]w=-\triangle U[/tex]
[tex]e\triangle V=-\triangle U[/tex]
[tex]\triangle U=-e (V_2-V_1)[/tex]
Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as
[tex]\triangle U=-e (V_2-V_1)[/tex]
b)Generally the electric potential energy in electron volts (eV). is mathematically given as
[tex]\triangle U=-e (154-24)V[/tex]
[tex]|\triangle U| =|-e (130)V|[/tex]
[tex]\triangle U=130eV[/tex]
c) Generally according to the law of conservation of energy
[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]
[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]
[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
he potential energy between two atoms in a particular molecule has the form U(s) = 2.6/x^8 - 4.3/x^4 where the units of x are length and the numbers 2.G and 4.3 have appropriate units so that U(x) has units of energy. What b the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)?
Answer:
x = 1.04866
Explanation:
Force can be defined from power energy by the expressions
F = [tex]- \frac{ dU}{ dx}[/tex]
in this case we are the expression of the potential energy
U = [tex]\frac{2.6}{x^{8} } - \frac{4.3}{ x^{4} }[/tex]
let's find the derivative
dU / dx = 2.6 ( [tex]\frac{-8}{x^{9} }[/tex]) - 4.3 ([tex]\frac{-4}{ x^{5} }[/tex])
dU / dx = [tex]- \frac{20.8}{ x^{9} } + \frac{17.2 }{ x^{5} }[/tex]
we substitute
F = + \frac{20.8}{ x^{9} } - \frac{17.2 }{ x^{5} }
at the equilibrium point the force is zero, so
[tex]\frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }[/tex]
20.8 / 17.2 = x⁴
x⁴ = 1.2093
x = [tex]\sqrt[4]{ 1.2093}[/tex]
x = 1.04866
Consider a turnbuckle that has been tightened until the tension in wire AD is 350 N. Draw the FBD that is required to determine the internal forces at point J. (You must provide an answer before moving on to the next part.) The FBD that is required to determine the internal forces at point J is
Answer:
yes
Explanation:
yes
According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.
What is statement?A statement, question, or phrase that is presented as a problem to be solved and has a dual or disguised meaning is called a riddle. Enigmas, which are difficulties typically presented in metaphoric or allegorical language that call for inventiveness and careful thought to solve, and conundra, which are problems that rely on puns either in the question or the answer, are two different forms of riddles.
Across many nations and even entire continents, many riddles take on a similar format. Riddles might be borrowed close to home as well as long distances. A man and his son were rock climbing on a particularly dangerous mountain when they slipped and fell. the man was killed, but the son lived and was rushed to a hospital.
Therefore, According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.
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According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units
This question is incomplete, the complete question is;
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g
where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.
Do you think this equation is valid in any system of units
Answer:
YES, the equation is a general equation that is valid in any system of units
Explanation:
Given the data in the question;
h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]
so
[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)
[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)
∴ [ L ] = (0.04 to 0.09) [L]
So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.
Therefore, YES, the equation is a general equation that is valid in any system of units
A roller coaster car is released from rest as shown in the image below. If
friction is neglected, the car will oscillate back and forth across the "dip" in
the roller coaster. What is the approximate velocity of the roller coaster car
each time it reaches the bottom of the roller coaster in the image? (Recall
that g = 9.8 m/s2.)
TAS
81 m
O A. 40 m/s
B. 25 m/s
C. 30 m/s
D. 15 m/s
Answer:
40m/s
Explanation:
a=g
u=0
s=81
v²=u²+2as
v²=2(9.81)(81)
v=√1587.6=39.8446985181≈40m/s
The velocity of the roller coaster car each time it reaches the bottom is 40 ms⁻¹. The correct option is (A).
The rate at which the position of an object changes with respect to time is described by the physical quantity known as velocity. It has both magnitude and direction because it is a vector quantity.
Given:
Initial velocity, u = 0 m/s
Acceleration, a = -9.8 ms⁻²
Distance, d = 81 m
From the third equation of motion:
v² = u² - 2as
v² = 0 - 2×(-9.8)×81
v = 40 ms⁻¹
Hence, the velocity of the roller coaster car is 40 ms⁻¹. The correct option is (A).
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Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Answer:
e.
Explanation:
Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according Newton's 2nd Law, as we can see below:[tex]F_{g} = m*g = m*a (1)[/tex]⇒a = g = 9.8m/s² (pointing downward)Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:[tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]
Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:[tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]
Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.What health consequences is most likely to result from alcohol school?
Answer:
difficulty concentrating
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver's perception/reaction time before and after drinking. (Assume practical stopping distance.)
Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
A 50.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 20.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.00 ms, what is the magnitude of the average acceleration of the ball during this time interval
Answer:
The magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the ball, u = 29.5 m/s
final velocity of the ball, v = -20.0 m/s (negative because it rebounds)
time of contact of the ball and the wall, t = 4 ms = 4 x 10⁻³ s
The force exerted on the brick wall by the ball is given as;
[tex]F = ma\\\\ma = \frac{m(v-u)}{t} \\\\a = \frac{v-u}{t} \\\\a = \frac{(-20) - 29.5}{4.0 \ \times \ 10^{-3}} \\\\a = \frac{-49.5}{4.0 \ \times \ 10^{-3}} \\\\a = -1.238 \times 10^4 \ m/s^2\\\\|a| = 1.238 \times 10^4 \ m/s^2[/tex]
Therefore, the magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².
What is the period of an objects motion?
Which characteristic involves cleavage and fracture?
the way a mineral breaks apart
the color of a mineral’s powder
the light that is reflected from a mineral’s surface
the number and angle of crystal faces of a mineral
Answer:
the way a mineral breaks apart
Explanation:
The way a mineral breaks apart involves fracture and cleavage.
These characteristics are very important in mineral identification especially during physical observations. Minerals have different cleavage properties and fracture planes.
Cleavage of a mineral is the ability of a mineral to split along preferred weakness planes. These planes are usually ingrained within the mineral during its formation. Fracture is the plane along through which minerals are able to break.Answer:
A:the way a mineral breaks apart
Explanation:
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a ramp of 64.8 degrees at a speed of 25.4 m/s. What would be the largest number of buses he can clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 10.0 m long
Answer: he can only make it over 5 buses
Explanation:
Given the data in the question;
we know that range is expressed as;
R = (V₀²sin2∅₀)/g
V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),
so we substitute
R = ((25.4)²sin2(64.8))/9.8
R = 50.7 m
now, them number of buses will be;
n = R / bus length
given that bus length is 10.0 m
we substitute
n = 50.7 m / 10.0
n = 5.07 ≈ 5
Therefore, he can only make it over 5 buses
A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.
Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.
Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s