Answer:
The 59th element is Praseodymium it's symbol is Pr.
What test are included in the Physical Fitness Test?
PACER/Mile Run, Push Ups, Curl Ups, Trunk Lift, Back Saver Sit-and-Reach
PACER/Mile Run, Push Ups
O Push Ups, Curl Ups, Trunk Lift, Back Saver Sit-and-Reach
PACER/Mile Run, Long Jump, Push Up,
The test that is included in the Physical Fitness Test are: A. PACER/Mile Run, Push Ups, Curl Ups, Trunk Lift, Back Saver Sit-and-Reach.
What is physical fitness test?Physical fitness test can be defined as a measure of both the physical and mental soundness (wellness) of an individual to engage in sports, work and other day-to-day activities.
The examples of physical fitness test.In Health education, some examples of the test that is included in the Physical Fitness Test are:
PACER/Mile RunPush UpsCurl UpsTrunk LiftBack Saver Sit-and-Reach.Read more on physical fitness here: https://brainly.com/question/1809216
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V
Answer:
[tex]E=6.8Kv/m[/tex]
Explanation:
From the question we are told that
Distance b/w plate [tex]d=10cm=>0.1m[/tex]
P_1 Potential at 7.35 [tex]V=533v[/tex]
Generally the equation for electric field at a distance is mathematically given as
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{533}{7.85*10^-^2}[/tex]
[tex]E=6789.808917[/tex]
[tex]E=6.8*10^3[/tex]
[tex]E=6.8Kv/m[/tex]
A block is released from rest at the top of a hill of height h. If there is negligible friction between the block and the hill, the block arrives at the bottom of the hill with speed v. The block is released from rest at the top of another hill with a rough surface and height h. If one-half of the initial mechanical energy of the block-Earth system is lost due to friction as the block descends the hill, the block will reach the bottom of the hill with a speed of
Answer:
v₁ =√2gh, v₂ = v₁ /√2
Explanation:
Let's use the concepts of energy and work to analyze each case
hill without rubbing. Energy is conserved
starting point. Highest part
Em₀ = U = mg h
final point. Lower part
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v₁ =√2gh
rubbing hill
in this case the energy is not conserved because it is converted into work of the friction force, therefore the variation of the energy is the work of the friction
W = Em_{f} - Em₀
they indicate half of the initial mechanical energy is lost due to friction
W = ½ Em₀
we substitute
- ½ Em₀ = Em_{f} - Em₀
The negative sign is because the friction work always opposes the movement
Em_{f} = ½ Em₀
½ m v₂² = ½ m g h
v₂ = √½ √2gh
v₂ = v₁ /√2
When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period
Answer:
240 meters
Explanation:
The distance traveled by the vehicle can be calculated using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex] (1)
Where:
x: is the displacement
[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)
[tex]v_{0}[/tex]: is the initial speed = 60 m/s
a: is the acceleration = -7.5 m/s²
By solving equation (1) for x we have:
[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]
Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.
I hope it helps you!
3. What is the SI unit of force? What is this unit equivalent to in terms of fundamental units?
4. Why is force a vector quantity?
Answer:
force = mass * acceleration
therefore the SI unit is kg*m/s2 or newton's
it's a vector quantity because it has both direction(acceleration) and size (mass)
If pressurized air pressure is 350 kPa, atmospheric pressure is 100 kPa, initial atmospheric pressure is 100 kPa, initial acceleration of the water rocket is 0.5g, acceleration of the water rocket is 0.5g, mass of water is 0.5 kg and structural mass of water is 0.5 kg and structural mass is 0.5 kg. Calculate the diameter of mass is 0.5 kg. Calculate the diameter of the nozzle where water is leaving the the nozzle where water is leaving the bottle
Answer:
[tex]d=8.657mm[/tex]
Explanation:
From the question we are told that
Pressurized air pressure is [tex]P_{air}=350 kPa,[/tex]
Atmospheric pressure is [tex]P_a=100 kPa[/tex]
Initial acceleration of the water rocket is [tex]a_i=0.5g.[/tex]
Acceleration of the water rocket is [tex]a_r=0.5g[/tex]
Mass of water is [tex]M_w=0.5 kg[/tex]
Generally total mass is given mathematically given as
[tex]T_M=0.5+0.5=>1kg[/tex]
Generally the tension on the rocket is given mathematically given as
[tex]T=(P_{air}-P_a)A[/tex]
[tex]T=(350-100) \frac{\pi d^2}{4}[/tex]
T is also
[tex]T=\frac{3Mg}{2}[/tex]
Therefore
[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3Mg}{2}[/tex]
[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3*1*9.81}{2}[/tex]
[tex]d^2= \frac{3*1*9.81*4}{2(350-100) \pi}[/tex]
[tex]d=\sqrt{\frac{3*1*9.81*4}{2(350-100) \pi}}[/tex]
[tex]d=8.657mm[/tex]
therefore diameter of nozzle is mathematically given as
[tex]d=8.657mm[/tex]
In addition to the sources of error mentioned above, since the caliper jaws squeezed the flexible, rubber surface of the ball slightly, the measured diameters were slightly __________________ compared to what a non-contact method of measuring would provide. This represents an ________________ ________________ error in ________________. This additional source of error ________________ ________________ significant. When the caliper jaws closed, the zero mark on the sliding Vernier scale, ________________ line up with the zero mark on the measuring scale. This means the caliper ________________ calibrated correctly.
Answer:
MINORS, SYSTEMATIC, STATISTICAL, BOTOM LINE, ZERO MATCHES
Explanation:
In general the sources of error or uncertainty can be classified
* Statistics. Which are those that describe the statistical formulas, for example: average, standard deviation, absolute error, etc.
* Systematic. That they occur due to an inappropriate measurement or to an interaction between the system and the instrument that cannot be quantified, in general this error shifts the measurements towards an explicit side
* Random, so errors that sometimes occur in the measurement and sometimes not, for example temperature changes during the medical process
In this case, you are asked to complete the sentences with the appropriate word
the measured diameters were slightly ___ MINORS________ compared to what a non-contact method of measuring would provide
. This represents an ____SYSTEMATIC_______ error in ________________.
This additional source of error ________STATISTICAL________ significant.
When the caliper jaws closed, the zero mark on the sliding Vernier scale, BOTOM LINE AND THE __________ line up with the zero mark on the measuring scale.
This means the caliper ___ZERO MATCHES_____________ calibrated correctly.
Two steamrollers begin 105 mm apart and head toward each other, each at a constant speed of 1.20 m/s. At the same instant, a fly that travels at a constant speed of 2.50 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.
Required:
What distance does the fly travel?
Answer: 109.4 mm
Explanation: Distance is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time: d = vt
The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:
location of collision = [tex]\frac{0.105}{2}[/tex] = 0.0525 m
To reach that point, both steamrollers will have spent
[tex]v=\frac{\Delta x}{t}[/tex]
[tex]t=\frac{\Delta x}{v}[/tex]
[tex]t=\frac{0.0525}{1.2}[/tex]
t = 0.04375 s
The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:
d = 2.5*0.04375
d = 0.109375 m
Until it is crushed, the fly will have traveled 109.4 mm.
a) What magnitude point charge creates a 12596.37 N/C electric
held at a distance of 0.593 m?
Answer:
[tex]Q = 4.9216 * 10^{-7}C[/tex]
Explanation:
Given
[tex]E = 12596.37 N/C[/tex]
[tex]r = 0.593m[/tex]
Required
Determine the magnitude point charge (Q)
This question will be solved using [tex]the\ magnitude[/tex] of the electric field formula
[tex]E = \frac{kQ}{r^2}[/tex]
Where
[tex]k = 9 * 10^9\ Nm^2 / C^2[/tex]
Make Q the subject in [tex]E = \frac{kQ}{r^2}[/tex]
[tex]E * r^2 = kQ[/tex]
[tex]Q = \frac{E * r^2}{k}[/tex]
Substitute values for E, r and k
[tex]Q = \frac{12596.37 * 0.593^2}{9 * 10^9}[/tex]
[tex]Q = \frac{4429.50}{9 * 10^9}[/tex]
[tex]Q = \frac{492.16}{10^9}[/tex]
[tex]Q = 492.16 * 10^{-9}[/tex]
Express in standard form
[tex]Q = 4.9216 * 10^2 * 10^{-9}[/tex]
[tex]Q = 4.9216 * 10^{2-9}[/tex]
[tex]Q = 4.9216 * 10^{-7}C[/tex]
What Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
A trolley of mass 5.0 kg is moving at 1.0 ms to the right. A constant force of 25 N acts to the left for 0.75 seconds.
Calculate the change of kinetic energy of the trolley.
(Show Work)
Answer:
change in kinetic energy of the trolley is 53.91 J.
Explanation:
mass of the trolley, m = 5.0 kg
initial velocity of the trolley, u = 1.0 m/s
external force on the trolley, F = 25 N
time of force action, t = 0.75 s
The final velocity of the trolley at the end of 0.75 s is calculated as follows;
[tex]F = \frac{m(v-u)}{t} \\\\25 = \frac{5(v-1)}{0.75}\\\\5(v-1) = 18.75\\\\v-1 = \frac{18.75}{5} \\\\v-1 = 3.75\\\\v = 4.75 \ m/s \ in \ the \ direction \ of \ the \ applied \ force[/tex]
The change in kinetic energy of the trolley is calculated as;
Δ K.E = ¹/₂m(v² - u²)
Δ K.E = ¹/₂ x 5(4.75² - 1²)
Δ K.E = 53.91 J.
Therefore, change in kinetic energy of the trolley is 53.91 J.
A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD's information at a constant rate of 1.5 megabits per second. How many minutes does it take the player to read the entire CD? Express your answer using two significant figures
Answer:
69.62 minutes
Explanation:
From the information we have here,
1 byte is = 8 bits
So 1 megabyte = 8 megabits
Then
783.216 x 8 megabits = 6265.728 megabits
This player has its reading capacity at 1.5 megabits / second
So 1 minute = 60x1.5
= 90 megabits / min
Then we have the entire reading time of this CD player to be =
6265.728/90
= 69.62 minutes.
This answers the question
In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Columbiad is 900 ft long, but part of it is packed with poweder, so the spaceship accelerates over a distance of only 700 ft. Estimate the acceleration experienced by the occupants of the spaceship during launch. Give your answer in m/s2. (Verne realized that the "travelers would...encounter a violent recoil," but he probably didn't know that people generally lose consciousness if they experience accelerations greater than about 7g ~70 m/s2.)
Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)
Where:
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:
[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]
[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
) A 1000-nF capacitor with circular parallel plates with a radius of 1cm is accumulating charge at the rate of 52 mC/s at some point in time. What will be the induced magnetic field strength if you are positioned 20 cm radially outward from the center of the plates
Answer:
[tex]B=5.2*10^-^8T[/tex]
Explanation:
From the question we are told that
Capacitor [tex]c=1000nf[/tex]
Radius [tex]r=1cm =>0.001m[/tex]
Charge rate [tex]Q/t=52mC/s[/tex]
Distance [tex]d=20cm =0.2m[/tex]
Generally the rate of charge I is mathematically given as
[tex]I=\frac{dQ}{dt}[/tex]
[tex]I=52Cm/s[/tex]
[tex]I=52*10^-^3C[/tex]
Generally the the magnetic field intensity at distance d is mathematically given as
[tex]B=\frac{\mu I}{2\pi d}[/tex]
[tex]B=\frac{(4 * \pi *10^-^7)(52*10^-^3C)}{2\pi (0.2)}[/tex]
[tex]B=5.2*10^-^8T[/tex]
What is the difference between a wave and a medium?
Answer:
Mediums in which the speed of sound is different generally have differing acoustic impedances, so that, when a sound wave strikes an interface between
Explanation:The propagation of a wave through a medium will depend on the properties of the medium. For example, waves of different frequencies may travel
As electric current moves through a wire, heat generated by resistance is conducted through a layer of insulation and then convected to the surrounding air. The steady-state temperature of the wire can be computed as
Answer:
Hello your question is incomplete attached below is the complete question
answer : ri ( thickness of wire ) = 14.167 m
Explanation:
attached below is a detailed solution
using the given data to determine the thickness of wire
A torsional pendulum is formed by attaching a wire to the center of a meter stick with a mass of 5.00 kg. If the resulting period is 4.00 min, what is the torsion constant for the wire
Answer:
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
Explanation:
The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:
[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)
Where:
[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
The angular frequency and the moment of inertia are represented by the following formulas:
[tex]\omega = \frac{2\pi}{T}[/tex] (2)
[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)
Where:
[tex]T[/tex] - Period, measured in seconds.
[tex]m[/tex] - Mass of the stick, measured in kilograms.
[tex]L[/tex] - Length of the stick, measured in meters.
By (2) and (3), (1) is now expanded:
[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]
[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]
[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]
If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:
[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]
[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbed to the top of the building
it would help if work was shown :,)
Answer:
579600J
Explanation:
Given parameters:
Height of the building = 828m
Weight of the man = 700N
Unknown:
Work done by the man = ?
Solution:
The work done by the man is the same as the potential energy expended.
Work done:
Work done = Weight x height = 700 x 828
Work done = 579600J
An ambitious physics major decides to check out the Uncertainty Principle for macroscopic systems. She goes to the top of the UD tower and drops a marble of mass m to the ground, trying to hit one of the cracks between bricks on the mall. To aim her marble, she teeters precariously directly over the desired crack and uses a very sophisticated apparatus of the highest possible precision, which she has borrowed from the General Physics Lab. Alas, try as she might, she cannot hit the crack.
Required:
Prove that the marble will inevitably miss the crack.
Answer:
The order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]
Explanation:
To miss the crack at a given distance is apparently not the same as the uncertainty that occurred in the distance while falling from the tower. However, it is believed that the uncertainties in both cases appear to be the same.
So, let's work it out together
According to Heisenberg's uncertainty principle:
[tex]\Delta s. \Delta p =\dfrac{h}{2} =\dfrac{h}{4 \pi}[/tex]
Also; if we recall from the equation of motion that:
[tex]v = u + at ---(1) \\ \\ v^2 - u^2 = 2as --- (2) \\ \\ s = ut + \dfrac{1}{2}at^2 --- (3)[/tex]
So, if u = 0 and a = g
Then;
[tex]v = gt --- (1) \\ \\ v^2 = 2gs - - - ( 2) \\ \\ s = \dfrac{1}{2}gt^2 --- (3)[/tex]
From (2)
Making (s) the subject, we have:
[tex]s = \dfrac{v^2}{2g}[/tex]
[tex]s = \dfrac{p^2}{2gm^2}[/tex]
By differentiation;
[tex]ds = d (\dfrac{p^2}{2gm^2})[/tex]
[tex]ds = \dfrac{2pdp}{2gm^2}[/tex]
[tex]\Delta \ s = \dfrac{p \Delta p}{gm^2 }[/tex]
where;
[tex]\Delta p = \dfrac{h}{4 \pi \Delta \ s}[/tex] from uncertainty principle
This implies that:
[tex]\Delta s = \dfrac{p(\dfrac{h}{4 \pi \Delta s }) }{gm^2}[/tex]
[tex]\Delta s = p(\dfrac{h}{4 \pi gm^2 }) \times \dfrac{1}{ \Delta s}}[/tex]
[tex](\Delta s)^2 = \dfrac{hmv} {4 \pi gm^2 }[/tex]
here;
v = 2gH
So;
[tex](\Delta s)^2 = \dfrac {h \sqrt{2gH} }{4 \pi gm }[/tex]
[tex]\mathbf{(\Delta s)^2 = \sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }[/tex]
Thus, the order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun.
Estimate gravity on the surface on this star.
I know the solution of this question it is as the picture shows but I only need to add *10^3 to the lower part of the division lower part to get the correct answer. But I don't know why I should add it can anyone explain?
Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
Answer:
[tex]g=4.38*10^{7} m/s^2[/tex]
Explanation:
To solve this, we need to know the mass of the sun, and the radius of the moon
[tex]M_{s} = 1.989*10^{30}kg \\R_{m} = 1737400m[/tex]
Now we can plug our values into our equation:
[tex]g=G*\frac{M_{E} }{r^{2} }[/tex]
This gives us:
[tex]g=6.67*10^{-11}*\frac{1.989*10^{30}}{1737400^{2}}[/tex]
This equals:
[tex]g=4.38*10^{7} m/s^2[/tex]
Suppose you were digging a well into saturated sediments. Why is the sediment’s permeability an important factor in deciding where to put your well?
Answer:
The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Explanation:
When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 3.89 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
Explanation:
from coulumbs law, The force that is acting over both charge can be computed as
F=( kq1q2)/r^2..............eqn(1)
Where
F=electrostatic force= 3.89 x 10-21 N
k= column constant= 9 x 10^9 Nm^2/C^2
q1 and q2= magnitude of the charges
r= distance between the charges= 1.08 x 10-3 m.
Since both charges are experiencing the same force, eqn(1) can be written as
F=( kq^2)/r^2.
We can make q subject of the formula
q= √(Fr^2)/k
= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9
q= 71.043×10^-20 C
Hence, the charge is 71.043×10^-20 C
From quantization law, the number of electron can be computed as
N=q/e
Where
N= number of electron
q= charges
=1.6×10^-19C
N=71.043×10^-20/1.6×10^-19
=4.44
Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
A heavy book is launched horizontally out a window from the first floor, a height, h, above the ground, with initial velocity, v0, and it hits the ground a horizontal distance X1 away from the window. Another book is similarly launched (same initial velocity) from the second floor window, a height 2h above the ground. Where does the second book land relative to the first book
Answer:
x₂ / x₁ = √2
Explanation:
To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0
0 = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2y_o}{ g} }[/tex]
t = \sqrt{ \frac{2 \ 2h}{ g} }
with this time we calculate the horizontal distance traveled
x = v₀ t
x₂ = v₀ [tex]\sqrt{ \frac{4h}{g} }[/tex]
now let's calculate the time it takes him to get to the floor when he leaves from the first floor
t =\sqrt{ \frac{2y_o}{ g} }
the horizontal distance traveled is
x₁ = v₀ [tex]\sqrt{ \frac{2h}{g} }[/tex]
therefore the difference in distance between the two runs is
Δx = x₂-x₁
Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }
Δx = v₀ \sqrt{ \frac{2h}{g} } √2
Δx =√2 x₁
the relationship between the two distances is
x₂ / x₁ = √2
Suppose that material scientists are testing a new alloy that has remarkably large coefficients of thermal expansion. A sample of the alloy is formed into a thin, uniform, solid disk with diameter 2.00 cm when at room temperature. If the temperature of the disk is increased by an amount ΔT the diameter of the disk increases by an amount ΔD.
a. True
b. False
Answer:
True
Explanation:
When the temperature of solid is increased its size increases. It is due to the increase in internal energy of the solid and then there is expansion in the solid. The temperature gives heat to the solid object and it expands in size. The expansion cannot be linear it can be increased in size by the volume.
I REALLY NEED HELP!!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.How much impulse does Max experience as a result of the collision? (list unknown variable and known variables, write an equation, plug in numbers, and get answer with unit.)
Answer:
400 Ns
Explanation:
Impulse = Change in momentum
i.e I = ΔP
So that,
Impulse experienced by Max = Change in Max's momentum
Change in Max's momentum = m(v - u)
Where m is the mass, v is the velocity after collision, and u is the velocity before collision.
m = 100.0 kg, v = 2.0 m/s, u = 6.0 m/s
Change in Max's momentum = 100 x (2 -6)
= -400 kg m/s
The negative sign shows that the change in momentum was against his direction of motion.
Impulse experienced by Max = 400 Ns.
Thus,
Max experienced an impulse of 400 Ns as a result of the collision.
Mary is trying to pull Julie on a sled across a flat snowy field. Mary pulls on the rope attached to the sled. Her pulling force is directed horizontally. Julie weighs 109 pounds. The sled weights 12 pounds. If the coefficient of static friction between the sled runners and the snow is 0.42, how much force must Mary pull with (in lbs) to start moving the sled
Answer: F = 498.04 lbs
Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.
Force of friction is given by the formula
[tex]F_{f}=\mu.F_{N}[/tex]
where
μ is coefficient of friction
[tex]F_{N}[/tex] is normal force
Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.
In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So
[tex]F_{N}=m.g[/tex]
Coefficient of friction is how much friction exists between two surfaces.
Rearraging friction force is
[tex]F_{f}=\mu.m.g[/tex]
Mass for this system is the sum of Julie and the sled, therefore
m = 109 + 12
m = 121 lb
Calculating Friction Force:
[tex]F_{f}=0.42.121.9.8[/tex]
[tex]F_{f}=[/tex] 498.04 lbs
LBS is a unit of measurement referred as pound by weight.
In conclusion, force Mary needs to start moving the sled is 498.04 lbs
effects of heat on matter
Answer:
it can melt orcan put them past their boiling point
Explanation:
car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 kg how much force must the breaks exert to stop the car in time?
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.
Anyone can help me out with this question ? Just number 2,
Answer:
- 21⁰C .
Explanation:
Speed of jet = 2.05 x 10³ km /h
= 2050 x 1000 / (60 x 60 ) m /s
= 569.44 m / s
Mach no represents times of speed of sound , the speed of jet
1.79 x speed of sound = 569.44
speed of sound = 318.12 m /s
speed of sound at 20⁰C = 343 m /s
Difference = 343 - 318.12 = 24.88⁰C
We know that 1 ⁰C change in temperature changes speed of sound
by .61 m /s
So a change in speed of 24.88 will be produced by a change in temperature of
24.88 / .61
= 41⁰C
temperature = 20 - 41 = - 21⁰C .
Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 39 cm. (a) How much work (in J) is needed to stretch the spring from 30 cm to 35 cm
Answer:
Workdone = 0.89 Joules
Explanation:
Given the following data;
Workdone = 6J
Extension = 39 - 26 = 13cm to meters = 13/100 = 0.13m
The workdone to stretch a string is given by the formula;
Workdone = ½ke²
Where;
k is the constant of elasticity.
e is the extension of the string.
We would solve for string constant, k;
6 = ½*k*0.13²
6 = ½*k*0.0169
Cross-multiplying, we have;
12 = 0.0169k
k = 12/0.0169
k = 710.06
a. To find the workdone when e = 30, 35.
Extension = 35 - 30 = 5 to meters = 5/100 = 0.05m
Workdone = ½*710.06*0.05²
Workdone = 355.03*0.0025
Workdone = 0.89 Joules
Therefore, the amount of work (in J) needed to stretch the spring from 30 cm to 35 cm is 0.89.