A ball is hurled vertically upward, and it can go up to 50 meters. gravitational acceleration (g = 9.8 m/s²). The equation given as :
v^2−u^2=2(as).
where,
v is final velocity,
u is initial velocity,
a is acceleration due to gravity, 10m/s2,
s = h is height traveled.
hence, we can get u with info given above:
In this,
The time to reach maximum height be t, then applying 1st equation of motion we get,
v=u−g* t
⇒0=50−10t
⇒t=5 seconds
time 50seconds
Due to the fact that an ball being thrown upwards is decelerating against gravity, its velocity gradually diminishes until it eventually zeroes out at the object's highest point.
case, v = 0 at maximum height.
0- u^2 =2×10×50
-u^2 =1000
=31.6227766017
(speed) u = - (31.63) m/s
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The purpose of the auricle is to equalize air pressure on both sides of the eardrums.TrueFalse
The auricle, also known as pinna is the visible portion of outer ear.
The purpose of auricle is to collect sound waves and send them through ear canal.
The eustachian tube connecting middle ear and back of nose, equalizes air pressure on both sides of the eardrum.
Thus, the statement is false.
Question is in attachment!
Thanks!
Concept : Thermodynamics
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution :
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder =P1
Final pressure inside the cylinder =P2
Initial volume inside the cylinder =V1
Final volume inside the cylinder =V2
Ratio of specific heats, γ=CVCP=1.4
For an adiabatic process, we have:
P1V1γ=P2V2γ
The final volume is compressed to half of its initial volume.
∴V2=V1/2
P1V1γ=P2(V1/2)γ
P2/P1=V1γ/(V1/2)γ
=21.4=2.639
Hence, the pressure increases by a factor of 2.639.
Answer:
The cylinder is fully isolated from the rest of the environment.
There is no heat exchange between the system (cylinder) and its surroundings as a result of the design. As a result, the process is called adiabatic.
P1 represents the initial pressure inside the cylinder.
P2 is the final pressure within the cylinder.
V1 is the volume of the cylinder at its beginning.
The final volume of the cylinder is equal to V2.
The specific heat ratio, = Cp / Cv = 1.4,
We have the following for an adiabatic process:
P1V1γ = P2V2γ
After compression, the final volume is reduced to half of its original size.
Hence,
V2 = V1 / 2
P1V1γ = P2(V1 / 2)γ
P2 / P1 = V1γ / (V1 / 2)γ
= 21.4
We get,
= 2.639
Therefore, the pressure increases by a factor of 2.639
Explanation:
hope it helps you
A Porsche 914 accelerates uniformly from 0 to 60 miles perhour in 5 seconds. Assuming a frictionless air track and a glider mass of 550grams, find the mass M2 that one must hang from the glider in order to obtain an accelerationequal to that of the Porsche.
The mass M₂ that one must hang from the glider in order to obtain an acceleration equal to that of the Porsche is 0.664 kilograms.
The initial velocity U of the porche is 0miles/hours or 0m/s.
The final velocity V of the porche is 60miles/hours or 26.84 m/s.
Total time period is 5 seconds,
Using equation of motion,
V+U = at
Putting all the values,
26.84= a(5)
a = 5.364 m/s².
Now, the mass of the glider is 0.550 Kg.
Let us assume that the glider and the mass are connected by a string. The tension in the string is T.
The mass M2 is hanged from one end the glider is on the horizontal surface.
When the system is released,
Total force on glider,
(0.550)a = T
Total force on the mass M2,
M2(a) = M2g - T.
Putting the value of T,
M2a = M2g - (0.550)a
(0.550)a = M2(g-a)
M2 = (0.550)×5.364/(9.8-5.364)
M2 = 0.664 Kg.
The required mass to be hung is 0.664 Kg.
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A storage tank 20 m high is filled with pure water. (Assume the tank is open and exposed to the atmosphere at the top.)
(a) Find the gauge pressure at the bottom of the tank.
(b) Calculate the magnitude of the net force that acts on a square access hatch at the bottom of the tank that measures 0.6 m by 0.6 m.
PLEASE HELP!!!
The gauge pressure at the bottom of the tank is 297500 N/m².
The magnitude of the net force that acts on a square access hatch at the bottom of the tank is 107100 N.
What is the pressure at the bottom of the tank?The pressure at the bottom of the tank is the pressure due to the weight of the fluid above it and the atmospheric pressure.
The pressure due to the atmosphere = 101300 N/m²
The pressure due to the water above the tank is calculated with the formula below:
Pressure = height * density * g
height of tank = 20 m
the density of water = 1 * 10³ kg/m³
g = 9.81 m/s²
Pressure due to water = 20 * 1 * 10³ kg/m³ * 9.81
Pressure due to water = 192000 N/m²
Total pressure = 192000 + 101300 N/m²
Total pressure = 297500 N/m²
B. Force = pressure * area
area of the hatch = 0.6 * 0.6 = 0.36 m²
Force = 297500 N/m² * 0.36 m²
Force = 107100 N
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In laser eye surgery, the average laser pulse power is 74 kW, and contains 1 mJ of light energy. Each laser pulse lasts
nanosecond.
If an average laser pulse power is 74 kW, and contains 1 mJ of light energy, each laser pulse lasts for a time period of 0.074 ns
E = P t
E = Light energy
P = Pulse power
t = Time
E = 74 KW = 74 * 10³ W
P = 1 * [tex]10^{-3}[/tex] J
t = E / P
t = 74 * 10³ / 1 * [tex]10^{-3}[/tex]
t = 74 * [tex]10^{6}[/tex] s
t = 0.074 ns
Power is the rate of work done. It can also be called as rate at which energy is used. This can be used to all forms of work and all types of energy.
Therefore, each laser pulse lasts 0.074 nanosecond.
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If an object mass is 2 kg. And an applied force is acting on it at 15 N And the force of friction is 5 N and what is the objects acceleration
Given,
The mass of the object, m=2 kg
The applied force, F=15 N
The frictional force, f=5 N
The frictional force is a force that opposes the motion of an object. Hence it will be always be directed opposite to the direction of motion of the object.
The net force acting on the object is given by,
[tex]F_n=F-f[/tex]From Newton's second law of motion, the net force is given by,
[tex]F_n=ma[/tex]Where a is the acceleration of the object.
Therefore,
[tex]\begin{gathered} ma=F-f \\ \Rightarrow a=\frac{F-f}{m} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} a=\frac{15-5}{2} \\ =5\text{ m/s}^2 \end{gathered}[/tex]Therefore the acceleration of the object is 5 m/s²
In the diagram, 91 = -6.39*10^-9 C andq2 = +3.22*10^-9 C. What is the electricfield at point P? Include a + or - sign tomindicate the direction.q2q1P0.424 m-** 0.636 m(Remember, E points away from + charges,and toward - charges.)(Unit = N/C)Enter
The magnitud of an electric field is given as:
[tex]E=\frac{1}{4\pi\epsilon_0}\frac{\lvert q\rvert}{r^2}[/tex]For the charge 1 the magnitude is:
[tex]\begin{gathered} E_1=\frac{1}{4\pi\epsilon_0}\frac{\lvert-6.39\times10^{-9}\rvert}{(0.424)^2} \\ E_1=319.543 \end{gathered}[/tex]Now, since charge 1 is negative this means that this field points towards the charge, in this case to the left, then the electric field for charge one is:
[tex]\vec{E_1}=-319.543[/tex]For the charge 2 the magnitude is:
[tex]\begin{gathered} E_2=\frac{1}{4\pi\epsilon_0}\frac{\lvert3.22\times10^{-9}\rvert}{(0.636)^2} \\ E_2=71.565 \end{gathered}[/tex]Now, since charge 2 is positive this means that this field points away from the charge, in this case to the left, then the electric field for charge two is:
[tex]\vec{E_2}=-71.565[/tex]Now, the total field on point P is the sum of both electric fields, then the total electri field on this point is:
[tex]E=-391.108[/tex]An average froghopper insect has a mass of 12.7 mg and jumps to a maximum height of 278 mm when its takeoff angle is 57.0∘ above the horizontal. Find the takeoff speed of the froghopper.
Takeoff speed of the froghopper is 2.786 m/s
By analyzing the vertical motion of the froghopper,
The initial vertical velocity of the froghopper as it jumps from the ground is given by
[tex]U_{y}[/tex] = U₀Sinθ
here, θ = 57°
Therefore Sin57° = 0.838
[tex]U_{y}[/tex]= 0.838U₀
Vertical motion is the motion that occurs when the object is thrown all the way up, i.e. , the initial velocity or force acts only in the vertical axis, therefore this motion has only vertical motion.
Maximum height reached by the froghopper is h = 278 mm
h = 0.278 m
As we know that vertical velocity at the point of maximum height will be zero,
[tex]V_{y} = 0[/tex]
Since the vertical motion is an accelerated motion with constant (de)acceleration, so by using Equation of motion,
[tex]V^{2} _{y} - U^{2} _{y} = 2gh[/tex]
[tex]U_y = \sqrt{V^{2} _{y} - 2gh }[/tex]
[tex]U_y = \sqrt{0 - 2(-9.81) (0.278) }[/tex]
[tex]U_{y}[/tex] = 2.335 m/s
[tex]U_{y}[/tex] = 0.838U₀
U₀ = [tex]\frac{U_y}{0.838}[/tex]
U₀ = [tex]\frac{2.335}{0.838}[/tex]
U₀ = 2.786 m/s
The takeoff speed of the froghopper is 2.786 m/s.
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Fernando, who has a mass of 43.0 kg, slides down
the banister at his grandparents' house. If the
banister makes an angle of 35.0° with the horizontal,
what is the normal force between Fernando and the
banister?
From the calculations, the normal force on the body is 345 N.
What is the normal force?Let us recall the theory of Newton that states that action and reaction are equal and opposite. We know that the force that is exerted on an object is equal to the reaction of that surface of the object on the source of the force.
We know that the normal force can be given by the formula;
N = mgcosθ
m = mass of the object
g = acceleration due to gravity
θ = angle involved
We would now have
N = 43.0 kg * 9.8 m/s^2 * cos 35
N = 345 N
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A object has a mass of 4 kg and is accelerating at 3 m/s2.The net force acting on the object is [] N.
In order to calculate the net force acting on the object, we can use the second law of Newton:
[tex]\sum ^{}_{}F=m\cdot a[/tex]So, for a mass of 4 kg and an acceleration of 3 m/s², we have:
[tex]\begin{gathered} F=4\cdot3 \\ F=12\text{ N} \end{gathered}[/tex]Therefore the net force is 12 Newtons.
A 180 N force acts at 190 degrees and a 140 N force acts at 260 degrees. Determine the magnitude and direction (include angle) of the resultant. Scale is 1cm =10N
NEED HELPPP ASAPPPPPP
The magnitude of the force is found to be 274.19N.
Force (F1)acting on θ1(190°) is 180N
Force (F2)acting on θ2(260°) is 140N
Therefore the angle between F1 and F2 is
θ2-θ1 = 180°-140° =40°
Now we calculate magnitude of force,
Magnitude of force ,F= F1+F2
F. F = [tex]\sqrt{ ( F1+F2). (F1+F2) }[/tex]
|F|² = [tex]\sqrt{F1 ²+F2 ²+F1 F2 COS θ}[/tex]
F= [tex]\sqrt{ 180²+140²+2×180×140× cos40°}[/tex]
F= [tex]\sqrt{32400+19600+23184}[/tex]
F= [tex]\sqrt{75184}[/tex]
F= 274.19N.
Thus, the magnitude of the force is found to be 274.19N.
The magnitude of the force is the entire quantity of forces acting on an object. When all forces are pulling in the same direction, the force becomes stronger. The strength of a force decreases as it is applied to an object from various angles.
Force has a magnitude and a direction, thus a vector quantity. The outcomes of two equal-sized forces acting in opposite directions one in the east and the other in the west are not the same.
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Calculate the area of the plates of a 1 pF parallel plate capacitor in a vacuum if the separation of the plates is 0.1 mm. [ε0=8.85×10-12C2N-1m-2 ]
Answer:
For parallel plate capacitor ,the capacitance C=dAϵ0
Area, A=ϵ0Cd=8.85×10−121×10−3=1.13×108m2
Pls Check This Answer Is correct or not
actually i am not sure
Create a Graph : Include numbers, axes, and order pairs
a) Peak value = 20
b) Average value = 9.0
c) RMS value = 14.15
Explanation:The peak-to-peak value = 40
[tex]v_{p-p}=40[/tex][tex]\begin{gathered} v_{p-p}=2v_p \\ \\ 40=2v_p \\ \\ v_p=\frac{40}{2} \\ \\ v_p=20 \end{gathered}[/tex]c) The rms value
[tex]\begin{gathered} v_{rms}=\frac{v_p}{\sqrt{2}} \\ \\ v_{rms}=\frac{20}{\sqrt{2}} \\ \\ v_{rms}=14.14 \end{gathered}[/tex]b) Average value over alternation of the sine wave
[tex]\begin{gathered} v_{avg}=0.637v_p \\ \\ v_{avg}=0.637\times14.14 \\ \\ v_{avg}=9.0 \end{gathered}[/tex]A penny, starting from rest at Position 1, rolls down and then up a curved track towards Position 5. When it reaches Position 5, it rolls back down the track. Which description below is most consistent with what you can expect from the penny's motion?
A. The penny will roll as high as Position 3 since that position has the least potential energy.
B. The penny will roll past Position 1 since energy cannot be created or destroyed.
C. The penny will roll as high as Position 2 since some of the energy will be transformed to heat.
D. The penny will roll as high as Position 1 since energy cannot be created or destroyed
The penny will roll past Position 1 since energy cannot be created or destroyed, therefore the correct answer is option B.
What is mechanical energy?The sum of all the energy in motion (total kinetic energy) and all the energy that is stored in the system (total potential energy) is known as mechanical energy.
ME= PE + KE
As given in the problem if a penny, starting from rest at Position 1, rolls down and then up a curved track towards Position 5.
When it reaches Position 5, it rolls back down the track, then we have to find out Which description below is most consistent with what you can expect from the penny's motion,
Thus, the penny will roll past Position 1 since energy cannot be created or destroyed, therefore the correct answer is option B.
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How long must a 400 W electrical engine work in order to produce 300 kJ of work?
In order to calculate the time needed, let's use the formula below:
[tex]P=\frac{E}{t}[/tex]Where P is the power in Watts, E is the energy/work in Joules and t is the time in seconds.
First, let's convert the energy from kJ to J (1 kJ = 1000 J):
[tex]300\text{ kJ}=300000\text{ J}[/tex]So, for P = 400, we have:
[tex]\begin{gathered} 400=\frac{300000}{t} \\ t=\frac{300000}{400} \\ t=750\text{ s} \end{gathered}[/tex]Therefore the time needed is 750 seconds.
Hi Can you help out with this physics question? We may learn a new thing
Features that must be present for a stable well-designed racing car.
For, a racing car the speed is high and also to finish the race, the car should be easy to turn around.
Lower the centre of gravity more will be its stability.
A low centre of gravity is a must for a stable well-designed racing car.
An object is placed in front of a concave mirror between the center of curvature of the mirror and its focal point as shown in the diagram below three light rays are traced along with their corresponding reflected rays which statement below best describes the image formed
Given:
An object is placed in front of a concave mirror between the center of curvature of the mirror and its focal point.
To find:
The type of the image
Explanation:
The image is real if the rays after reflection from the mirror actually meet.
Here, the object is placed in front of a concave mirror between the center of curvature of the mirror and its focal point.
The rays after reflection actually meet. So, the image is real.
Hence, the image is real.
A 1.2kg ball rolls forward with an acceleration of 1.11 m/s. What is the net force on the ball
Answer:
1.332 N
Explanation:
Net Force = Mass x Acceleration
1.2 x 1.11 = 1.332 N
I'm so sorry if I'm wrong.
Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of 2.0 x 10^(-6) m^2Youngs modulus of steel is 2.1 x 10 ^11 Pa
Given data:
* The extension of the steel wire is 0.3 mm.
* The length of the wire is 4 m.
* The area of cross section of wire is,
[tex]A=2\times10^{-6}m^2[/tex]* The young modulus of the steel is,
[tex]Y=2.1\times10^{11}\text{ Pa}[/tex]Solution:
The young modulus of the steel in terms of the force and extension is,
[tex]Y=\frac{F\times l}{A\times dl}[/tex]where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,
Substituting the known values,
[tex]\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}[/tex]Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.
3. A uniform light beam is pivoted halfway along its length. At one end it supports a load of 5 kN while the other end is tethered to a fixed point by a rope inclined at 45° to the horizontal. If the beam is in equilibrium, what is the tension in the rope?
Answer:
Approximately [tex]7.1\; {\rm kN}[/tex] ([tex]5 \sqrt{2}\; {\rm kN}[/tex].)
Explanation:
Let [tex]F_{1}[/tex] and [tex]F_{2}[/tex] denote the two forces that act on this beam. Let [tex]s_{1}[/tex], [tex]s_{2}[/tex], [tex]\theta_{1}[/tex], and [tex]\theta_{2}[/tex] denote the distance from pivot and angle relative to the beam of the two forces, respectively. The magnitude of the torques that the two forces exert on this beam will be [tex]F_{1}\, s_{1}\, \sin(\theta_{1})[/tex] and [tex]F_{2}\, s_{2}\, \sin(\theta_{2})[/tex], respectively.
The two forces in this question act on the beam from opposite sides of the pivot. Hence, for the beam to be in equilibrium, the torque from the two forces need to be equal in magnitude. In other words:
[tex]F_{1}\, s_{1}\, \sin(\theta_{1}) = F_{2}\, s_{2}\, \sin(\theta_{2})[/tex].
Let [tex]F_{1}[/tex] denote the [tex]5\; {\rm kN}[/tex] force that the load exerts on this beam; [tex]\theta = 90^{\circ}[/tex] since this load is placed directly on the beam. The normal force from the load will be perpendicular to the beam.
Let [tex]F_{2}[/tex] denote the force that the rope exerts on this beam; [tex]\theta = 45^{\circ}[/tex].
Note that [tex]s_{1} = s_{2}[/tex] since the pivot is exactly halfway between the two forces.
Rearrange the equation [tex]F_{1}\, s_{1}\, \sin(\theta_{1}) = F_{2}\, s_{2}\, \sin(\theta_{2})[/tex] to find the unknown [tex]F_{2}[/tex]:
[tex]\begin{aligned}F_{2} &= \frac{F_{1}\, s_{1}\, \sin(\theta_{1})}{s_{2}\, \sin(\theta_{2})} \\ &= \frac{F_{1}\, \sin(\theta_{1})}{\sin(\theta_{2})} && (\text{since $s_{1} = s_{2}$}) \\ &= \frac{5\; {\rm kN}\, \sin(90^{\circ})}{\sin(45^{\circ})} \\ &= \frac{5\; {\rm kN}}{(1 / \sqrt{2})} \\ &= 5 \sqrt{2}\; {\rm kN} \\ &\approx 7.1\; {\rm kN} \end{aligned}[/tex].
The tension in the rope will be equal in magnitude to the force exerted on the beam: approximately [tex]7.1\; {\rm kN}[/tex] ([tex]5 \sqrt{2}\; {\rm kN}[/tex].)
7.0 mm diameter copper wire carries a current of 7.0 μA. What is the current density?
The current density of a 7.0 mm diameter copper wire that carries a current of 7.0 μA is 18.19 × 10⁻³ A/m².
The quantity of electric modern-day journeying in step with the unit move-phase region is known as current density and is expressed in amperes per rectangular meter. The more present-day a conductor, the higher could be the modern-day density. current density is the quantity of electrical modern flowing in step with the unit move-sectional region of a material.
subsequently, the SI unit of current density needs to be Ampere/meter2. Ohm's law relates the current flowing via a conductor to the voltage V and resistance R; this is, V = IR. An alternative assertion of Ohm's law is I = V/R.
current I = 0.7 μA
r = 3.5mm
J = I/A = I/πr³
J = 0.7 × 10⁻⁶ / 3.14 × (3.5 × 10⁻³)²
J = 18.19 × 10⁻³ A/m²
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Question 2 of 25Which of the following is an effect of increasing the wavelength of anelectromagnetic wave?O A. Energy decreases.O B. Speed decreases.O C. Speed increases.O D. Energy increases.SUBMIT
Thesppeed of electromagnetic waves remains the same an is eqal to the speed of light. The wavelength is rearlated tothe energy bty the following ormula:
[tex]\lambda=\frac{c}{f}[/tex]Wher:
[tex]\begin{gathered} \lambda=\text{ wavelength} \\ c=\text{ speed of light} \\ f=\text{ frequency} \end{gathered}[/tex]Therefore, if the wavelength increases the frequency must decrease.-
Since the energy is proportiona to the frequency if the dfrequency decreases the energy decreases. Therefore, the right answer is A.
Answer:
energy decreases
Explanation:
hope this helps!
POSSIBLE POINTS: 2The graph below shows the relationship between the force acting on an object and the acceleration of the object. What is the acceleration of the objectwhen a 3 - newton force acts on it?____________What is the object's mass? ____________
From the graph we notice that when the force is 3 N the acceleration is 0.6 meters per second per second.
Now, to find the mass we use Newton's second law:
[tex]F=ma[/tex]for the point we found earlier we have:
[tex]\begin{gathered} 3=0.6m \\ m=\frac{3}{0.6} \\ m=5 \end{gathered}[/tex]Therefore the mass of the object is 5 kg
This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me
a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,
Frequency = 9/5 = 1.8 oscillations per second
Frequency = 1.8 Hz
Period = 1/frequency = 1/1.8
Period = 0.056 s
b) When we differenctiate displacement with respect to time, the result is velocity.
Recall, period = 1/f = 5/9 cycles
1/4 cycle behind = 1/4 x 5/9 = 5/36
It is delayed with 5/36 sec with respect to displacement.
5/36 sec = 0.139 sec
Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =
5/36 = 0.139 sec delayed with respect to velocity
= 5/18 = 0.2777 secs delayed with respect to displacement
Thus, the number of seconds out of phase with the displacements is 0.278 seconds
c) The formula for calculating the period of an ideal pendulum anywhere is
T = 2π√length/local gravity). We would calculate the local gravity.
From the information given,
length = 0.2
T = P = 5/9
Thus,
5/9 = 2π√0.2/local gravity)
(5/9)/2π = √0.2/local gravity
Square both sides. It becomes
[(5/9)/2π]^2 = 0.2/local gravity
local gravity = 0.2/[(5/9)/2π]^2
local gravity = 25.56 m/s^2
Thus,
acceleration due to gravity = 25.56 m/s^2
Recall, earth's gravity = 9.8 m/s^2
number of g forces = 25.56/9.8
number of g forces = 2.61
Jack drops off 10m high red bridge into swimming waters below. Assuming acceleration to be 9.91 m/s^2
A. What Jack speed as he hits the water?
Jack speed as he hits the water is 14.13m/s
What is speed?
The speed of an object, also known as v in kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time, making it a scalar quantity. The instantaneous speed is the upper limit of the average speed as the duration of the time interval approaches zero. The average speed of an item in a period of time is equal to the distance travelled by the object divided by the duration of the period. Velocity and speed are not the same thing.
The only force on Jack is gravity, which points straight down, and has an acceleration of 9.91m/s². Whether he's falling off a bridge or simply standing still on the surface of the earth, the acceleration due to gravity is the same.
If you're looking for the speed of Jack as he hits the water, you just need to solve for v in the equation[tex]$v^2 = v_0^2 + 2ad$[/tex].
Here,
v is Jack's speed as he hits the water,
[tex]$v_0$[/tex] is Jack's initial speed (which is zero since he's not moving when he jumps),
a is Jack's acceleration (which is 9.91m/s²), and
d is Jack's displacement (which is the height of the bridge, which you said is 10m).
This gives you an answer of [tex]$v = \sqrt{2ad} = \sqrt{2(9.91)(10)} = 14.13 \ \mathrm{m/s}$[/tex]
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How far will a car travel going at a speed of 18 m/s in 42 minutes?
O 756 meters.
O2,520 meters.
O 45,360 meters.
O 140 meters.
Answer:
d = rt
Explanation:
distance = ?
rate = 18 m/s
time = 42 x 60 = 2520 s
distance = 18 x 2520 = 45360 m
Review the statements below. Select the one that best describes the difference between astronomy and cosmology.
A) Astronomy is a much broader study of the universe as a whole, while cosmology is a more focused study on specific objects or celestial bodies contained within the universe.
B) Astronomy is the study of objects in the universe—how they are formed, their composition, and how they change over time. Cosmology is the study of the entire universe—what objects are contained in the universe, and what is happening to the universe over time, including how it may or may not end.
C) Astronomy is a branch of science that focuses on the beginning, middle, and end of the universe. Cosmology is the study of how objects move, interact, and change.
D) Astronomy is the study of space in general through the use of telescopes and other tools, while cosmology is the study of specific objects in space.
The statement that correctly differentiates between astronomy and cosmology is as follows: Astronomy is a much broader study of the universe as a whole, while cosmology is a more focused study on specific objects or celestial bodies contained within the universe (option A).
What is astronomy?Astronomy is the study of the physical universe beyond the Earth's atmosphere, including the process of mapping locations and properties of the matter and radiation in the universe.
On the other hand, cosmology is the study of the physical universe, its structure, dynamics, origin and evolution, and fate.
Astronomy studies objects and phenomena beyond Earth, whereas cosmology is a branch of astronomy that studies the origin of the universe and how it has evolved.
Therefore, it can be said that astronomy studies the universe from a much broader sense while cosmology is much more specific study.
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FPressure is a function of force and area on which the force is exerted: P =AWhat is the effect of an increase in temperature of a sealed container of gaswith no change in volume?hA. The change in temperature decreases the force of the collisions ofthe container and gas particles, so the pressure decreases.B. The change in temperature increases the force of the collisions ofthe container and gas particles, so the pressure increases.C. The change in temperature decreases the area over whichcollisions of the container and gas particles occur, so the pressureincreases.D. The change in temperature increases the area over whichcollisions of the container and gas particles occur, so the pressuredecreases.
B: The change in temperature increases the force of the collisions of
the container and gas particles, so the pressure increases.
P = F/A
Force is directly proportional to Force, so , if force increases also Pressure.
what is the formula to use when measuring specific heat capacity by method of mixtures?
Answer: Cs = [m1c1(T - T1) + mcCc(T - Ts)]/Ms(Ts - T)
Explanation:
The concept of heat exchange which is expressed as
heat lost by a hot body = heat gained by a cold body
The specific heat of a solid substance as determined by the method of mixture using the heat exchange concept would be
Heat lost by substance = heat gained by liquid + heat gained by calorimeter
Thus,
MsCs(Ts - T) = m1c1(T - T1) + mcCc(T - Tc)
Dividing both sides by Ms(Ts - T)
The formula for calculating the specific heat capacity by method of mixtures is
Cs = [m1c1(T - T1) + mcCc(T - Ts)]/Ms(Ts - T)
Where
m1 = mass of liquid
c1 = specific heat of liquid
ms = mass of substance
Cs =specific heat of substance
mc = mass of calorimeter
Cc = specific heat of material used in making the calorimeter
T = Final temperature of the mixture
T1 = initial temperature of the liquid
Ts = initial temperature of the substance
What is the acceleration of a car,
moving along a straight line, that
increases its velocity from 0 to 15
m/s in 10 s?
The acceleration of a car, moving along a straight line, that increases its velocity from 0 to 15 is A=1.5m/s.
Acceleration = Velocity Change / Time Interval
So,
A=15/10
A=1.5m/s .
A slowing car slows down. It also accelerates because the speed changes. Imagine a car driving down the street. As speed increases, the car has positive acceleration. When a car turns a corner at a constant speed it changes direction and thus accelerates. The faster the rotation the greater the acceleration.
Therefore, any change in velocity in magnitude and direction causes acceleration. The main causes of poor acceleration are air and fuel issues and sensor issues. However mechanical problems can also cause poor performance. The car is moving forward and accelerating in the positive direction so the acceleration is in the same direction as the car is moving. An object can also have positive acceleration if it slows down while moving in a negative direction.
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