The equation to obtain the final speed of car is,
[tex]v^2=u^2+2as[/tex]Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(11.6m/s)^2+2a(0.287\text{ m)} \\ a=\frac{-134.56m^2s^{-2}}{2(0.287\text{ m)}} \\ \approx-234.4m/s^2 \end{gathered}[/tex]The negative sign of acceleration indicates that the car is deaccelerating.
The force required to stop the car is,
[tex]F=ma[/tex]Substitute the magnitude of known values,
[tex]\begin{gathered} F=(21.2kg)(234.4m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =4969.28\text{ N} \\ \approx4970\text{ N} \end{gathered}[/tex]Thus, the force required to stop the car is 4970 N.
Which picture below correctly identifies the effort length and lifting length of a lever?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
A lever is a simple machine made of a rigid beam and a fulcrum.
where
the Load is the object which we are lifting,Fulcrum is the point at which the lever is pivoted. and Effort is he force applied to make the object move
Step 1
check for the graph that:
the lifting length goes from the fulcrum to the load
and
the effort length goes from the fulcrum to the applied force,
therefore,
the answer is
d.D
I hope this helps you
my choice1A child pushes a toy truck up an inclined plane. The lifting force is...Select one:ut ofO a. the weight of the truck.O b. the length of the inclined plane.O c. the force the child uses to push the truck.uestionO d. equal to the total amount of work done.
When pushing a toy truck up an inclined plane, the force that makes the truck goes up is the force that the child uses to push the truck. (The force that pushes
Therefore the correct option is C.
50 POINTS
A velocity vs time graph is shown. What is the acceleration of the object?
Answer:
Explanation:
Given:
V₀ = 0 m/s
V = 20m/s
t = 5 s
___________
a - ?
The acceleration:
[tex]a = \frac{(V -V_{0} )}{t} \\\\[/tex]
[tex]a=\frac{(20 - 0)}{5} = 4 \frac{m}{s^{2} } \\[/tex]
If the mass m of the wrecking ball is 3920 kg , what is the tension TB in the cable that makes an angle of 40 ∘ with the vertical? What is the tension TA in the horizontal cable?
The tension TB in the cable makes an angle of 40 ∘ and the tension TA in the horizontal cable
TB=49380.9NTA=31741.4NThis is further explained below.
What is tension?Generally, To represent tension in a vertical direction, the term is:
[tex]T_B=\frac{m g}{\cos \theta}[/tex]
Substitute $3860kg for m, 9.8m/s^2 for g, and 40^0 for [tex]\theta[/tex].
[tex]T_B &=\frac{(3860 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)}{\cos 40^{\circ}} \\[/tex]
=49380.9N
Because the cosine of the tension in the cable, which is pushing up on the item, is equal to the weight force, which is pressing down on the ground, the ball is not moving and is thus in equilibrium.
The expression for the horizontal cable tension is,
[tex]T_A=T_B \sin \theta[/tex]
Substitute $49380.9N for T_B and $40^o for [tex]\theta[/tex]
[tex]T_A &=(49380.9 \mathrm{~N}) \sin 40^{\circ}[/tex]
=31741.4N
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A driver of a car going 90km/hr suddenly sees the lights of a barrier 40.0m ahead. It take the driver 0.75s before he applies the brakes (this is known as reaction time). Once he does begin to brake, he decelerates at a rate of 10m/s^2. Does he hit the barrier?
First, consider that the distance traveled by the car in 0.75s is:
[tex]x=v\cdot t[/tex]Convert 90km/h to m/s as follow:
[tex]\frac{90\operatorname{km}}{h}\cdot\frac{1h}{3600}\cdot\frac{1000m}{1\operatorname{km}}=\frac{25m}{s}[/tex]Then, the distance x is:
[tex]x=(\frac{25m}{s})(0.75s)=18.75m[/tex]Then, when the driver start to apply the brakes, the distance to the barrier is:
x' = 40.0 m - 18.75 m = 21.25 m
Next, calculate the distance that the car need to stop completely, by using the following formula:
[tex]v^2=v^2_o-2ad[/tex]where,
v: final velocity = 0m/s (the car stops)
vo: initial velocity = 25m/s
a: acceleration = 10m/s^2
d: distance = ?
Solve the previous equation for d and replace the values of the other parameters:
[tex]d=\frac{v^2_0-v^2}{2a}=\frac{(\frac{25m}{s})^2-(\frac{0m}{s})^2}{2(\frac{10m}{s})^{}}=31.25m[/tex]Then, the drive needs 31.25 m to stop. If you compare the previous result with the distance of the car related to the barrier when the driver applies the brakes
(x' = 18.75 m), you can notice that d is greater than x'.
Hence, the car does hit the barrier.
25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
If a student bikes at a pace of 5.0 m/s down a straight route. The student's bicycle weighs 120 kg in total. The pupil puts on the brakes and comes to a stop. The average braking force of the automobile would be 150 newtons if the stopping distance were 10 meters.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Work done by the braking force of the cycle = change in kinetic energy of the student
Force × distance = 1/2 × mass × velocity²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force of the cycle would be 150 Newtons.
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The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed. Is this true or false?
Yes, the given statement is true.
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed.
Need help 82x2682 please
ANSWER:
STEP-BY-STEP EXPLANATION:
We have the following multiplication:
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A planet's distance from _____ and its _____ both determine its overall gravity.
A) the sun; mass
B) the Kuiper belt; diameter
C) Mars; temperature
D) the Milky Way; perimeter
If the planet Mercury has a mass of planet 3.3×10²³ kg and a radius of 2400 km - calculate the magnitude of the gravitational field on its surface?
The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
Explanation:The mass of mercury, m = 3.3×10²³ kg
The radius, r = 2400 km
r = 2400 x 1000m
r = 2.4 x 10⁶m
Note that the magnitude of the gravitational field on the surface of the planet is the acceleration due to gravity on that planet
It is given by the formula:
[tex]g=\frac{Gm}{r^2}[/tex]Substitute the values of G, m, and r into the formula above
[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{6.67\times10^{-11}\times3.3\times10^{23}}{(2.4\times10^6)^2} \\ g=\frac{2.2\times10^{13}}{5.76\times10^{12}} \\ g=3.82m/s^2 \end{gathered}[/tex]The magnitude of the gravitational field on the surface of the planet mercury is 3.82 m/s²
Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet,
making contact over an area of 80.0cm2 with each foot. Both the ceramic and the
carpet are 2.00 cm thick and are 10.0ºC on their bottom sides. At what rate must
heat transfer occurs from each foot to keep the top of the ceramic and carpet at
33.0ºC?
The rate of heat transfer from each foot to keep the top of the ceramic and carpet at the desired temperature is 27.6 J/s.
What is thermal heat transfer?
Thermal heat transfer is defined as the movement of heat across the border of the system due to a difference in temperature between the system and its surroundings.
The rate at which heat transfer occurs from each foot to keep the top of the ceramic and carpet at the given temperature is calculated as follows;
Q/t = kA(T₂ - T₁)/d
where;
Q/t is the rate of heat transfer
k is heat transfer coefficientT₂ is the final temperatureT₁ is the initial temperatured is the distance between the wool and ceramics
The heat transfer coefficient of human skin = 0.3 W / m⁰C
Area = 80 cm² = 0.008 m²
thickness, d = 2 cm = 0.02 m
change in temperature, = 33⁰ C - 10 ⁰C = 23 ⁰C
Q/t = (0.3 x 0.008 x 23) / (0.002)
Q/t = 27.6 W = 27.6 J/s
Thus, the rate of heat transfer from each foot to keep the top of the ceramic and carpet at the desired temperature is 27.6 J/s.
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A PVC pipe has a length of 45.132 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? Include units in your answers.b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other? Include units in your answers.
ANSWERS
a. f₁ = 380 Hz; f₂ = 760 Hz; f₃ = 1140 Hz
b. f₁ = 190 Hz; f₃ = 570 Hz; f₅ = 950 Hz
EXPLANATION
a. For a pipe of length L open at both ends, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{2L} \\ \\ f_2=2f_1=\frac{v}{L} \\ \\ f_3=3f_1=\frac{3v}{2L}\end{cases}[/tex]Assuming that the speed of the wave is the speed of sound: 343 m/s and knowing that the length of the pipe is L = 45.132 cm = 0.45132 m we can find the frequencies of the first three harmonics:
[tex]\begin{cases}f_1=\frac{343m/s}{2\cdot0.45132m}\approx380Hz \\ \\ f_2=2f_1=2\cdot380Hz\approx760Hz \\ \\ f_3=3f_1=3\cdot380Hz\approx1140Hz\end{cases}[/tex]b. For a pipe of length L closed at one end and open at the other, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_2=DNE \\ \\ f_3=3f_1=\frac{3v}{4L}\end{cases}[/tex]In a closed pipe, there can only be odd harmonics (1, 3, 5...). Therefore, the second harmonic does not exist and the "third harmonic" would be the 5th,
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_3=3f_1=\frac{3v}{4L} \\ \\ f_5=5f_1=\frac{5v}{4L}\end{cases}[/tex]Again, the length of the pipe is 45.132 cm = 0.45132 m, so the first three harmonics are:
[tex]\begin{cases}f_1=\frac{343m/s}{4\cdot0.45132m}\approx190Hz \\ \\ f_3=3f_1=3\cdot190Hz=570Hz \\ \\ f_5=5f_1=5\cdot190Hz=950Hz\end{cases}[/tex]an ice has a volume of 8975 ft^3. what is the mass in kilograms of the iceberg? the density of ice 0.917 g/cm^3
The density is given by:
[tex]\rho=\frac{m}{V}[/tex]where V is the volume and m is the mass.
To determine the mass we have to solve the equation for m:
[tex]m=\rho V[/tex]Now, before we can calculate the mass we have to convert the volume given to cubic meter, this comes from the fact that the density is given in g/cm^3 units. We have to remember that a ft is equal to 30.48 cm, then we have:
[tex]8975ft^3(\frac{30.48\text{ cm}}{1\text{ ft}})(\frac{30.48\text{ cm}}{1\text{ ft}})(\frac{30.48\text{ cm}}{1\text{ ft}})=2.54\times10^8[/tex]Hence the volume of the iceberg is:
[tex]2.54\times10^8cm^3[/tex]Now that we have the volume in the correct units we plug its value and the density in the equation for the mass above:
[tex]\begin{gathered} m=2.54\times10^8(0.917) \\ m=2.32\times10^8 \end{gathered}[/tex]Hence the mass of the iceber is 2.32x10^8 g.
Therefore the mass of the iceberg in kilograms is:
[tex]2.32\times10^5\text{ kg}[/tex]Why is it helpful to break vectors into their horizontal and vertical components before adding vectors?
It is helpful to break vectors into their horizontal and vertical components before adding vectors because it helps to simplify calculations.
A vector at an angle will have both horizontal and vertical component in it. So resolving the vector into its horizontal and vertical components makes it easier and clearer to solve both the components separately.
Solving the components is done using trigonometry ratios and Pythagoras theorem. Some of the trigonometric ratios used are:
sin θ = Vertical component / Actual vector
cos θ = Horizontal component / Actual vector
tan θ = Vertical component / Horizontal component
Therefore, it is helpful to break vectors into their horizontal and vertical components before adding vectors because it helps to simplify calculations.
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A penny is dropped from a building and it takes 7.00 seconds to hit the ground.
Based on the given question, the final velocity of the penny is 68.6m/s
Calculations and ParametersBased on the definition of velocity, we can see that it has to do with the speed at which an object moves with at a particular direction.
If we observe the 1st law of motion:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time taken
We need to find the velocity as it was not given.
We already know that
u = 0 (it was at rest),
t = 7s
v = unknown
Let us put in the given values based on the equation and solve for the answer.
v = u + gt
v = 0 + ( 9.8m / s ² × 7.00s )
v = 9.8m / s ² × 7.00s
v = 68.6m/s
Therefore, the final velocity of the penny is 68.6m/s
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A penny is dropped from a building, and it takes 7 seconds to hit the ground, so the final velocity will be 68.6 m/s.
What is Velocity?The ratio of displacement to time is referred to as velocity of the object. It has SI unit meter per second or m/s and has a dimension formula LT⁻¹.
We already know that
Initial speed, u = 0
Time, t = 7 seconds
Use the equation of motion to find the final velocity,
v = u + gt
v = 0 + ( 9.8 m / s ² × 7.00s )
v = 9.8 m / s ² × 7.00s
v = 68.6 m/s
Hence, the penny hit the ground with a final velocity of 68.6 m/s.
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Given v = 520 sin (30t - 5π/4), what is the phase angle?Question 4 options:-225 degrees90 degrees-135 degrees-90 degrees
Given data:
The voltage can be expressed as,
v = 520 sin (30t - 5π/4) ...... (1)
Now, the general equation of the sine wave can be given as,
[tex]v=V_m\text{ sin}(\omega t+\phi)\ldots\ldots\text{ }(2)[/tex]Here,
[tex]\phi[/tex]is the phase angle,
[tex]V_m[/tex]is the maximum voltage, and
[tex]\omega[/tex]is the angular frequency.
Compare equations (1) and (2), we get:
[tex]\begin{gathered} \phi=-\frac{5\pi\text{ rad}}{4}(\frac{180\degree}{\pi}) \\ =-225\degree \end{gathered}[/tex]Thus, the phase angle is
[tex]-225\degree[/tex]and the first option (-225 degrees) is correct.
Analyze the benefits and consequences of sleep create (continue) their own fitness programs incorporating sleep as a fundamental pillar of fitness
Explanation:
The Intimate Relationship Between Fitness and Sleep
*“If you don’t sleep, you undermine your body,” says W. Christopher Winter, MD, the president of Charlottesville Neurology and Sleep Medicine and the author of "The Sleep Solution: Why Your Sleep Is Broken and How to Fix It."
When it comes to working out, you know that what you do in the gym is important. But what you do outside the gym — what you eat, what you drink, and especially how you sleep, is just as crucial. In fact, you must sleep in order for exercise to actually work.
“We exercise for a purpose: for cardiovascular health, to increase lean muscle mass, to improve endurance, and more. All of these 'goals' require sleep,” says W. Christopher Winter, MD, the president of Charlottesville Neurology and Sleep Medicine and the author of The Sleep Solution: Why Your Sleep Is Broken and How to Fix It.
In other words, without sleep, exercise does not deliver those benefits, Dr. Winter explains. “If you don’t sleep, you undermine your body.”
Determine the order of magnitude of the following small number by entering the appropriate exponent n below: 4,870×1021kg≈10nkg .(PLEASE LOOK AT THE SCREENSHOT FOR CORRECT NUMBERING)
To find the order of magnitude we move the decimal point until we have only one interger before it:
[tex]4870\times10^{21}=4.870\times10^{24}[/tex]Now that we have in this form we conclude that the order of magnitude of the number is 24
How much power is used by a contact lens heating unit that draws 0.05 A of current from a 197 V line?
Given,
The current drawn by the contact lens heating unit, I=0.05 A
The supply voltage, V=197 V
The electric power is given by the product of the current drawn and the supply voltage.
Thus the power used by the given device is given by,
[tex]P=VI[/tex]On substituting the known values,
[tex]\begin{gathered} P=197\times0.05 \\ =9.85\text{ W} \end{gathered}[/tex]Thus the power used by the contact lens heating unit is 9.85 W
A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?
Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:
[tex]\begin{gathered} d=vt \\ =(3.1\frac{m}{s})(314s) \\ =973.4m \end{gathered}[/tex]That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:
[tex]\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=\frac{d}{2\pi} \\ =\frac{973.4m}{2(3.14...)} \\ =154.9...m \end{gathered}[/tex]The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:
[tex]a_c=\frac{v^2}{r}[/tex]Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:
[tex]a_c=\frac{(3.1\frac{m}{s})^2}{(154.9m)}=0.062\frac{m}{s^2}[/tex]Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.
A box is standing on a conveyor belt that is not in motion. At one point the belt starts moving with some acceleration. At that point the box starts moving too (without slipping). Which force is responsible for the acceleration of the box. a. The air resistance force. b. The force of the pull. c. The force of friction. d. The normal force.
Given that a box is standing on a conveyor belt that is not in motion.
When the belt starts moving with some acceleration, the box starts moving too without slipping.
Let's determine the force that is responsible for the acceleration of the box.
Here, since the box starts moving without slipping when the belt starts moving, there will be static friction between the box and the belt since the belt was fixed.
Now, the force which is responsible for the acceleration of the box will be the force of gravity and the normal force.
Applying the Newton's second law, if the there is only force of gravity and the normal force acting on the box, there will be zero horizontal acceleration.
In order for the box to accelerate without slipping, the force responsible will be the static frictional force.
ANSWER:
c. The force of friction.
Car A is traveling with a constant velocity of 18 [m/s]. Car B speeds up from 0 [m/s] to 10 [m/s] in 4 seconds. Which car has a greater acceleration?Car ACar BCar A and Car B have the same accelerationNeither car is accelerating
Given:
Car A is traveling with a constant velocity of,
[tex]18\text{ m/s}[/tex]The initial speed of car B is,
[tex]v_i=0\text{ m/s}[/tex]After t=4 s, carB's speed is,
[tex]v_f=10\text{ m/s}[/tex]To find:
Which car has a greater acceleration
Explanation:
The acceleration of Car A is Zero as there is no change in velocity with time.
The acceleration of car B is,
[tex]\begin{gathered} a_B=\frac{v_f-v_i}{t} \\ =\frac{10-0}{4} \\ =2.5\text{ m/s}^2 \end{gathered}[/tex]So, Car B has greater acceleration.
Hence, Car B has greater acceleration.
If John Glenn weighed 640 N on Earth's surface, a) how much would he haveweighed if his Mercury spacecraft had (hypothetically) remained at twice thedistance from the center of Earth? b) Why is it said that an astronaut is nevertruly "weightless?"
Given:
The weight of John Glenn, w=640 N
To find:
a) The weight if the distance was twice that of the initial value.
b) Why is an astronaut never weightless.
Explanation:
a)
Let the distance between the spacecraft and the earth be r.
If it becomes twice, then the distance is 2r.
The initial gravitational force on John Glenn is,
[tex]F=w=\frac{GMm}{r^2}[/tex]Where G is the gravitational constant, M is the mass of the earth and m is the mass of John Glenn.
The force when the distance is twice,
[tex]\begin{gathered} w_n=\frac{GMm}{(2r)^2} \\ =\frac{GMm}{4r^2} \\ =\frac{w}{4} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} w_n=\frac{640}{4} \\ =160\text{ N} \end{gathered}[/tex]b)
Even when the astronaut is in space they still have the mass and so does the earth. Thus there will always be a gravitational force of attraction between the earth and the astronaut. The astronaut does not feel the weight because there will be nothing in space that pushes them back. That is why an astronaut is never truly weightless.
Final answer:
a) Thus the weight of John Glenn will be 160 N
The pressure is greater at the bottom of the bucket filled with water ,why?
55. Write the shortened notation for the ion symbol next to each element given below. A.) Calcium loses 2 negative electronsB.) Chlorine gains 1 negative electronsC.) Hydrogen gains 1 negative electron
We are asked to write shortened notation for the ion symbol next to each element given below.
Convention:
First of all, write the shortened notation of the element.
The superscript (at the top) is the number of charges on the ion and also a + sign for a positive ion and - sign for a negative ion.
When an atom loses an electron then the charge is positive (+).
When an atom gains an electron then the charge is negative (-)
A.) Calcium loses 2 negative electrons
Since it loses 2 electrons it will have 2+ superscript.
Calcium has an atomic number of 20 (the number of protons)
[tex]_{20}Ca^{2+}[/tex]B.) Chlorine gains 1 negative electron
Since it gains 1 electron it will have - superscript (we don't write 1 when the charge is 1)
Chlorine has an atomic number of 17 (the number of protons)
[tex]_{17}Cl^-[/tex]C.) Hydrogen gains 1 negative electron
Since it gains 1 electron it will have - superscript
Hydrogen has an atomic number of 1 (the number of protons)
[tex]_1H^-[/tex]what is the velocity of the boat from point x to point y
The distance between the points X and Y is,
[tex]\begin{gathered} d=9\text{ km} \\ =9000\text{ m} \end{gathered}[/tex]The time taken to move the distance is,
[tex]\begin{gathered} t=12\text{ min} \\ =12\times60\text{ s} \end{gathered}[/tex]The velocity is given by,
[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{9000\text{ m}}{12\times60\text{ s}} \\ =12.5\text{ m/s} \end{gathered}[/tex]Hence the velocity is 12.5 m/s.
An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?AluminumNickela = 23.10-6C+ B = 69.10-6 0-1a = 13.10-6C1 B = 39.10-6-1(Unit = m)Enter
The difference in length between the two rods = 0.0025m
Explanations:Linear expansivity of a material is given by the formula:
[tex]\alpha\text{ = }\frac{l_2-l_1}{l_1(\theta_2-\theta_1)}[/tex]For the Aluminium rod:
[tex]\begin{gathered} l_{A1}\text{ = 5.0m} \\ \theta_{A1}=20^0C \\ \theta_{A2}=70^0C \\ \alpha_A\text{ = }23\times10^{-6}C^{-1} \\ \alpha_A\text{ = }\frac{l_{A2}-l_{A1}}{l_{A1}(\theta_{A2}-\theta_{A1})} \\ \text{ }23\times10^{-6}\text{ = }\frac{l_{A2}-5}{5(70-20)} \\ 5\times50\times\text{ }23\times10^{-6}=\text{ }l_{A2}-5 \\ l_{A2}=\text{ (}5750\text{ }\times10^{-6})\text{ + 5} \\ l_{A2}=\text{ 0.00575+5} \\ l_{A2}=\text{ 5.00575m} \end{gathered}[/tex]For the Nickel rod:
[tex]\begin{gathered} l_{N1}\text{ = 5.0m} \\ \theta_{N1}=20^0C \\ \theta_{N2}=70^0C \\ \alpha_N=\text{ 13}\times10^{-6}C^{-1} \\ \alpha_N\text{ = }\frac{l_{N2}-l_{N1}}{l_{N1}(\theta_{N2}-\theta_{N1})} \\ \text{ 1}3\times10^{-6}\text{ = }\frac{l_{N2}-5}{5(70-20)} \\ 5\times50\times\text{ 1}3\times10^{-6}=\text{ }l_{A2}-5 \\ l_{N2}=\text{ (32}50\text{ }\times10^{-6})\text{ + 5} \\ l_{N2}=\text{ }0.00325+5 \\ l_{N2}=\text{ 5.00325m} \end{gathered}[/tex]The difference in length between the two rods will be given as:
[tex]\begin{gathered} l_{A2}-l_{N2}=\text{ 5.00575-5.00325} \\ l_{A2}-l_{N2}=0.0025m \end{gathered}[/tex]The difference in length between the two rods = 0.0025m
having been given a newly discovered mineral, carry out a method in the physics lab on how to determine its specific gravity
ANSWER
Weigh a piece of the material, drop it in water and weigh the amount of water displaced. Specific gravity is the quotient between the mass of the mineral and the mass of water.
EXPLANATION
The specific gravity is the ratio of the mineral's density to the density of water at 23°C,
[tex]specific\text{ }gravity=\frac{density\text{ }of\text{ }mineral}{density\text{ }of\text{ }water}[/tex]Density, denoted with the greek letter ρ, is the ratio between the mass and volume of the substance,
[tex]\rho=\frac{m}{V}[/tex]If the mineral is a solid piece, we can weigh it - so we know the mass, and can find its volume by dropping it into water. Then, if we weigh the amount of water displaced, we have the mass of water as well - by Archimedes principle, the volume of water will be the volume of the object,
[tex]specific\text{ }gravity=\frac{\frac{m_{mineral}}{V}}{\frac{m_{water}}{V}}=\frac{m_{mineral}}{m_{water}}[/tex]An intrepid hiker reaches a large crevasse in his hiking route. He sees a nice landing ledge 60.0 cm below his position but it is across a 2.3 m gap. He spends 1.2 s accelerating horizontally at 5.92 m/s2 [right] in an attempt to launch himself to the safe landing on the far side of the gap. Does he make it?
The hiker made it to a safe landing on the other side of the gap after travelling horizontally at 2.49 m.
What is the time motion from the vertical height?
The time taken for the hiker to fall from the given height is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity = 0t is the time of motiong is acceleration due to gravityh is the height of fallh = ¹/₂gt²
t = √(2h/g)
t = √[(2 x 0.6) / (9.8)]
t = 0.35 seconds
The horizontal velocity of the hiker during the period of acceleration is calculated as follows;
Vₓ = at
Vₓ = (5.92 m/s²) x (1.2 s)
Vₓ = 7.104 m/s
The horizontal distance travelled during the time period of 0.35 seconds;
X = Vₓt
X = 7.104 x 0.35
X = 2.49 m
Thus, the hiker made it to a safe landing on the other side of the gap which is 2.3 m wide and smaller to his horizontal displacement of 2.49 m.
Learn more about horizontal distance here: https://brainly.com/question/24784992
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The weight of a proton is 1.64×10−26 N. The charge on a proton is +1.60×10−19 C. If a proton is placed in a uniform electric field so that the electric force on the proton just balances its weight, what is the magnitude and direction of the field?
Given:
The weight of the proton is: W = 1.6 × 10^(-26) N.
The charge on a proton is: q = 1.60 × 10^(-19) C
To find:
The magnitude and the direction of the electric field.
Explanation:
The weight of the proton is the force that the proton experiences due to its mass and acceleration. The electric force balances the weight of the proton. Thus we have,
F = W
Here, F is the electric force a proton experiences when it is placed in an electric field and W is the weight of the proton,
The force experienced by a photon when it is placed in an electric field is given as,
[tex]F=Eq[/tex]Here, E is the electric field.
Rearranging the above equation and substituting the values, we get:
[tex]\begin{gathered} E=\frac{F}{q} \\ \\ E=\frac{1.64\times10^{-26}\text{ N}}{1.60\times10^{-19}\text{ C}} \\ \\ E=1.025\times10^{-7}\text{ N/C} \end{gathered}[/tex]Thus, the magnitude of the electric field is 1.025 × 10^(-7) N/C.
The charge on the proton is positive and when it is placed in the electric field, the electric force on the proton is balanced by the weight of the proton. Thus, The direction of the electric force is opposite to the direction of the weight of the proton which is radially outward.
Final answer:
The magnitude of the electric field is 1.025 × 10^(-7) N/C and it has a radially outward direction that is opposite to the wight of the proton.