Vi = initial speed = 10 km/h
Vf= final speed = 75 km/h
t = time = 7 sec
a = acceleration
First, express km/h in m/s
Vi = 2.78 m/s
Vf= 20.83 m/s
Apply
a = (vf-vi) / t
replace:
a = (20.83 - 2.78 ) / 7 = 2.58 m/s^2
Austin does his Power lifting every morning to stay in shape. He lifts a 90 kg barbell, 2.3 m above the ground.a) How much energy does it have when it was on the ground? Jb)How much energy does it have after being lifted 2.3 m? Jc) What kind of energy does it have after being lifted? d) How much work did Austin do to lift the barbell? Je) If he lifted it in 1.9s, what was his power? W
ANSWER
[tex]\begin{gathered} (a)0J \\ (b)2030J \\ (c)\text{ Potential energy} \\ (d)2030J \\ (e)1070W \end{gathered}[/tex]EXPLANATION
Parameters given:
Mass of barbell, m = 90 kg
Height above ground, h = 2.3 m
(a) We want to find the energy the barbell has on the ground. #
When it is on the ground, the barbell is stationary, which means its velocity is 0 m/s, hence, its kinetic energy is also 0 J, since kinetic energy is given as:
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ KE=\frac{1}{2}\cdot m\cdot0=0J \end{gathered}[/tex]Also, on the ground, it is at a height of 0 m, hence, its potential energy is 0 J:
[tex]\begin{gathered} PE=mgh \\ PE=m\cdot g\cdot0=0J \end{gathered}[/tex]where g = acceleration due to gravity
Therefore, on the ground, the energy the barbell had was 0 J.
(b) After it had been lifted 2.3 m, its height above the ground became 2.3 m.
Now, we can find the potential energy possessed by the barbell:
[tex]\begin{gathered} PE=90\cdot9.8\cdot2.3 \\ PE=2028.6J\approx2030J \end{gathered}[/tex]After it is lifted, it is once again stationary, hence, it has no kinetic energy.
Therefore, the energy the barbell has after it has been lifted 2.3 m is 2070J.
(c) As stated in (b) above, after being lifted, the barbell only possesses potential energy since it is at a height above the ground and it is not moving.
(d) The work done in lifting the barbell is equal to the force applied multiplied by the height moved by the barbell.
That is:
[tex]W=F\cdot d[/tex]The force applied is equal to the weight of the barbell:
[tex]\begin{gathered} F=W=mg \\ F=90\cdot9.8 \\ F=882N \end{gathered}[/tex]Therefore, the work done is:
[tex]\begin{gathered} W=882\cdot2.3 \\ W=2028.6J\approx2030J \end{gathered}[/tex](e) He lifted the barbell in 1.9 seconds. To find his power, we have to divide the work done by the time taken to do the work.
That is:
[tex]\begin{gathered} P=\frac{W}{t} \\ P=\frac{2030}{1.9} \\ P=1068.4W\approx1070W \end{gathered}[/tex]That was his power.
Your engineering team has created a 46.9 kg spider robot that moves along a strand of web for
Halloween. The spider begins at rest and moves straight down the strand increasing its speed
at a constant rate. It covers 3.04 m in a time of 6.08 s. What is the Tension, in Newtons, in
the strand of web?
The fabric of the spiderwebs is so thin that it adheres easily to the hook side of a Velcro strip. Push Velcro onto the object you want to attach your webs to, then remove the adhesive backing. Grab a few of the strands and put them onto the Velcro to keep the webs in place.
Explain about the web for Halloween?A black spider spread between two slices of buttered bread is said to offer a witch great power. Spiders were rumoured to help witches cast charms. Old customs claim that if you see a spider on Halloween, the ghost of a departed loved one is purportedly watching over you.
The spiderwebs stick in the hook side of a Velcro strip with ease because they are constructed of a very thin fabric. When you wish to attach your webs to something, push Velcro onto the surface and peel off the sticky backing. To ensure that the webs stay in place, grab a handful of the strands and push them onto the Velcro.
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A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. The scale of the figure's vertical axis is set by F = 20.0 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?
The work done by the force to move the object from the origin to x = 8 m is 160 J.
The mass of the block, m = 5 kg
The block moves in a straight line on a frictionless surface.
A force of F = 20 N is acting on the object.
The work done by the force when the block moves from one place to another can be defined by the formula,
W = Fd where F is the force and d is the distance covered.
Now, we have F = 20 N and;
d = x = 8 m
Therefore, the work done by the object will be:
W = Fx
W = 20 × 8
W = 160 joules
The work done by the force to move the object to x = 8 m is 160 J.
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Modern roller coasters have vertical loops like the one shown in the figure. The radius of curvature is smaller at the top than on the sides so the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats.
1. What is the speed of the roller coaster, in meters per second, at the top of the loop if the radius of curvature there is 14 m and the downward acceleration of the car is 1.1g? Note that g here is the acceleration due to gravity.
2. The beginning of this roller coaster is at the top of a high hill. If it started from rest at the top of this hill, how high, in meters, above the top of the loop is this initial starting point? You may assume there is no friction anywhere on the track.
3. If it actually starts 7.5 m higher than your answer to the previous part (yet still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1800 kg.
1 ) The speed of the roller coaster = 12.28 m / s
2 ) Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction = 132 KJ
1 ) The speed of the roller coaster,
[tex]a_{c}[/tex] = v² / r
[tex]a_{c}[/tex] = Centripetal acceleration
v = Linear velocity
r = Radius
r = 14 m
[tex]a_{c}[/tex] = 1.1 g = 1.1 * 9.8
[tex]a_{c}[/tex] = 10.78 m / s²
v² = [tex]a_{c}[/tex] * r
v² = 10.78 * 14
v² = 150.9
v = 12.28 m / s
2 ) Initial starting point,
Considering hill as 1 and the loop as 2,
v1 = 0
h2 = 2 r = 2 * 14
h2 = 28 m
∑ [tex]F_{y}[/tex] = m [tex]a_{c}[/tex]
[tex]F_{N}[/tex] + [tex]F_{g}[/tex] = m [tex]a_{c}[/tex]
0 + m g = m v2² / r
v2² = g r
According to law of conservation of energy,
E1 = E2
m g h1 + 1 / 2 m v1² = m g h2 + 1 / 2 m v2²
m g h1 + 0 = 28 m g + 1 / 2 m g r
h1 = 28 + 1 / 2 ( 14 )
h1 = 35 m
Height of the hill above the top of the loop = h1 - h2
Height of the hill above the top of the loop = 35 - 28
Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction,
h1 = 35 + 7.5
h1 = 42.5 m
m = 1800 kg
v2² = g r
v2² = 9.8 * 14
v2² = 137.2 m / s
Since energy is lost due to friction,
KE1 + U1 = KE2 + U2 + W
0 + m g h1 = 1 / 2 m v2² + m g h2 + W
( 1800 * 9.8 * 42.5 ) = ( 0.5 * 1800 * 137.5 ) + ( 1800 * 9.8 * 28 ) + W
749700 = 123750 + 493920 + W
W = 749700 - 617670
W = 132030 J
W = 132 KJ
Therefore,
1 ) The speed of the roller coaster = 12.28 m / s
2 ) Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction = 132 KJ
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An object is launched at a speed of 20 m/s. If it rises to a height of 12 m, at what angle was it launched? ________° above the +x direction
Given:
The initial speed of the object is,
[tex]u=20\text{ m/s}[/tex]The maximum height is,
[tex]H=12\text{ m}[/tex]To find:
The angle of launch above the X-axis
Explanation:
The maximum height above the X-axis is,
[tex]H=\frac{u^2sin^2\theta}{2g}[/tex]Here, the angle above the X-axis is,
[tex]\theta[/tex]Substituting the values we get,
[tex]\begin{gathered} 12=\frac{(20)^2sin^2\theta}{2\times9.8} \\ sin^2\theta=\frac{12\times2\times9.8}{400} \\ sin\theta=0.7668\text{ \lparen taking positive value only\rparen} \\ \theta=50.1\degree \end{gathered}[/tex]Hence, the angle of launch is,
[tex]50.1\degree[/tex]true or false? a liter is a metric unit for area
A iter is defined as 1 cubic decimeter. It is the volume that fits inside a cube whose side measures 10 centimeters:
Therefore, the statement is false because a liter is a metric unit for volume.
Thermal equilibrium implies that:A:the state of restB:absolute zero temperatureC:the maximum temperatureD:equilibrium temperature
Explanation
Heat is the flow of energy from a high temperature to a low temperature. When these temperatures balance out, heat stops flowing
so, after some time both regions reach thermal equilibrium and no more energy is transfered.so we can conclude that
thermal equilibrium inplies that there is equilibrium temperature
D.equilibrium temperature
I hope this helps you
if the dolphin is moving horizontally when it goes through the hoop how high above the water is the center of the hoop
We are given that a dolphin moves describing a projectile motion. This can be represented in the following graph of position vs time:
Since the dolphin moves horizontally as he goes through the hoop this means that the hoop is at the maximum height of the motion. The maximum height of a projectile motion is given by:
[tex]h_{\max }=\frac{v^2\sin ^2\theta}{2g}[/tex]Where:
[tex]\begin{gathered} h_{\max }=\text{ max}imum\text{ height} \\ v=velocity_{} \\ \theta=\text{ initial angle} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we plug in the values:
[tex]h_{\text{max}}=\frac{(10\frac{m}{s})^2(\sin (41))^2}{2(9.8\frac{m}{s^2})}[/tex]Solving the operations:
[tex]h_{\max }=2.2m[/tex]Therefore, the hoop is at 2.2 meters above the water.
A basketball player jumps for a rebound and reaches a maximum height of 1.5 m. with what speed did he jump off the floor? How long was he in the air?
The final speed of the player can be given as,
[tex]v^2=u^2-2gh[/tex]At the maximum height, the final speed of player is zero.
Plug in the known values,
[tex]\begin{gathered} (0m/s)^2=u^2-2(9.8m/s^2)(1.5\text{ m)} \\ u^2=29.4m^2s^{-2} \\ u=5.42\text{ m/s} \end{gathered}[/tex]Thus, the speed with which it jump off the floor is 5.42 m/s.
The time for which the player was in player is,
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Plug in the known values,
[tex]undefined[/tex]What do you diagram to analyze orbital motion ?
The diagram to analyze the orbital motion can be shown as,
Here, a is the acceleration of moon and v is the speed.
The above diagram indicates the orbital motion of the moon around the earth. The moon is more towards the earth than the sun due to larger gravity of earth and at the same time the moon has its velocity that tends moon to move. Therefore, the moon has balanced gravitational and centripetal force to keep in an uniform orbital motion.
You exert a force of 5.3 N on a book to slide it across a table. If you do 2.5 J of work in the process, how far did the book move?
We will have the following:
[tex]2.5J=5.3N\cdot x\Rightarrow x=\frac{2.5J}{5.3N}[/tex][tex]\Rightarrow x=\frac{25}{53}m\Rightarrow x\approx0.47m[/tex]So, the book moved 25/53 meters, that is approximately 0.47 meters.
A person is at the top of a slide with 200 J of potential energy. Explain what happens to the energy as they slide down.
Answer:
the person changes from potential energy , to kinetic energy back to potential
Identify as many different ways as you can for giving energy to a basketball? (Select all that apply)
To answer this question we need to remember each kind of energy:
• Potential energy is the energy held by an object because of its position relative to other objects.
,• Kinetic energy is the energy held by an object due to is motion.
,• Internal energy is the energy due to the movement of the molecules of the object.
With this in mind we conclude that the following are ways of giving energy to a basketball:
• You can give a basketball kinetic energy by pushing on it with your hand, as in throwing or dribbling.
• You can give a basketball kinetic energy by spinning it on your finger.
• You can give a basketball potential energy by lifting it upward with your hand, as when shooting a free throw.
,• You can give a basketball internal energy by heating it.
If a copper wire with a 0.300 mm diameter is to have a resistance of 0.500 Ω at 20.0ºC, how long should it be? The resistivity of copper is 1.68 * 10^-8 Ω*m
Given:
• Diameter = 0.300 mm
,• Resistance = 0.500 Ω
,• Temperature = 20.0ºC
,• Resistivity = 1.68 * 10⁻⁸ Ω*m
Let's find the length of the copper wire.
To find the length of the wire, apply the formula:
[tex]\begin{gathered} R=\frac{\rho L}{A} \\ \\ R=\frac{\rho L}{\pi r^2} \end{gathered}[/tex]Where:
R is the resistance = 0.500 Ω
p is the resistivity = 1.68 * 10⁻⁸ Ω*m
L is the length
r is the radius = diameter/2 = 0.300mm/2 = 0.15 mm
Convert the radius to meters, where:
1 m = 1000 mm
0.15mm = 0.15 x 10⁻³ m
Now, input values into the formula and solve fo L:
[tex]0.500=\frac{1.68\times10^{-8}\times L}{\pi\times(0.15\times10^{-3})^2}[/tex]Solving further, rewrite the equation for the length L:
[tex]\begin{gathered} L=\frac{0.500\pi\times(0.15\times10^{-3})^2}{1.68\times10^{-8}} \\ \\ L=\frac{3.534\times10^{-8}}{1.68\times10^{-8}} \\ \\ L=2.1\text{ m} \end{gathered}[/tex]Therefore, the length of the copper wire is 2.1 meters.
ANSWER:
2.1 m
The velocity-time table represents the motion of a rightward-
moving motorcycle.
Magnitude =
Time (s)
0.0
0.5
1.0
1.5
2.0
What is the magnitude (i.e., value) and direction of the
acceleration?
Direction =
Velocity (m/s)
24.0, right
22.0, right
20.0, right
18.0, right
16.0, right
m/s/s
(No -
sign.)
(Tap field to change.)
NE
The direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form. In essence, velocity is a vector quantity. It is the speed at which distance is changing.
What is velocity?Galileo Galilei, an Italian physicist, is credited with being the first to calculate speed by dividing it by the required amount of time and the distance traveled.
Speed, according to Galileo, is the distance traveled in a set amount of time.
The speed at which an object is moving is referred to as its velocity.
Examples of fast motion include an automobile driving north on a highway or a rocket blasting off.
The equation v = u + at, where v signifies the ultimate speed, can be used to express an object's final velocity, which is equal to its starting velocity plus acceleration times the distance traveled.
Therefore, the direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form. In essence, velocity is a vector quantity. It is the speed at which distance is changing.
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What physical property is described in the following statement?: The asphaltis smooth.O A. TextureO B. ShapeO C. HardnessO D. SizeSUBMIT
A. Texture
Texture is the feel or appearence of a surface.
The average radius of Earth is 6,371 km. If the average thickness of oceanic crust is 7.5 km and the average thickness of continental crust is 35 km, what fraction of Earth's radius is each type of crust? Please show how to solve .
If the average thickness of the oceanic crust is 7.5 km and the average thickness of the continental crust is 35 km, then the fraction of the Earth's radius is each type of crust would be 0.00117720922 and 0.005493 respectively.
What is the percentage of a number?It is the relative value that represents the hundredth part of any number for example 2% of any number represents, 2 multiplied by the 1/100th of that number.
As given in the problem average thickness of the oceanic crust is 7.5 kilometers and the average thickness of the continental crust is 35 km,
The fraction of the earth's radius as oceanic crust = 7.5 / 6371
= 0.00117720922
The fraction of the earth's radius as continental crust = 35 / 6371
= 0.005493
Thus, the fraction of the Earth's radius is each type of crust would be 0.00117720922 and 0.005493 respectively.
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The train above is traveling at a constant velocity because the forces acting on it are in equilibrium. Therefore, the missing force must have a magnitude (blank) of newtons to the (blank).
The missing force has a magnitude of 800 N to the right
Explanation:The forces acting on the train are in equilibrium.
This means that the sum of all the forces acting in the right direction equals the sum of all the forces acting in the left direction
Let the missing force be represented by F
1700 = 900 + F
F = 1700 - 900
F = 800 N
Therefore, the missing force has a magnitude of 800 N to the right
but 9 volt battery is connected to a 4 ohm resistor and the 5 Ohm resistor as shown in the diagram. How much current flows through the 4 ohm resistor ?
Given data
The resistance of the first resistor is R1 = 4 ohm
The resistance of the second resistor is R2 = 5 ohm
The potential difference of the battery is V = 9 V
The resistors are connected in series. The expression for the equivalent resistance is given as:
[tex]\begin{gathered} R=R_1+R_2_{} \\ R=4\text{ }\Omega+5\text{ }\Omega \\ R=9\Omega \end{gathered}[/tex]The expression for the current in the 4-ohm resistor is given as:
[tex]\begin{gathered} I=\frac{V}{R} \\ I=\frac{9\text{ V}}{9\text{ }\Omega} \\ I=1\text{ A} \end{gathered}[/tex]Thus, the magnitude of the current flows through the 4-ohm resistor is 1 A.
A 228-turn, 24.506-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 27 degrees away from vertical increases from 0.807 T to 4.68 T in 13.843 s. Determine the emf induced in the coil.
Given:
• Number of turns, N = 228
,• Diameter, d = 24.506 cm
,• θ = 27 degrees
,• Initial Magnetic field, B1 = 0.807 T
,• Final, B2 = 4.68 T
,• Time , t = 13.843 s
Let's find the induced emf in the coil.
To find the induced EMF, apply Faraday's law:
[tex]\begin{gathered} E=N\frac{d}{dt}(B*A) \\ \\ E=N*Acos\theta\frac{d}{dt}(B) \\ \\ E=N*(\pi r^2)cos\theta(\frac{B_2-B_1}{t}) \end{gathered}[/tex]Where:
A is the area in meters.
Rewrite the diameter from cm to meters.
Where:
100 cm = 1 meters
24.056 cm = 0.24506 m
Now, the radius will be:
radius = diameter/2 = 0.24506/2 = 0.12253 m
Now, plug in the values and solve for E:
[tex]\begin{gathered} E=228*(\pi *(0.12253)^2)cos(27)*(\frac{4.68-0.807}{13.843}) \\ \\ E=228*0.0471666*cos(27)*0.27978 \\ \\ E=2.6\text{ volts} \end{gathered}[/tex]Therefore, the EMF induced in the coil is 2.6 volts.
ANSWER:
2.6 v
A spring with spring constant 40 N/m is compressed .1m past it natural length. A mass of .5kg is attached to the spring. A. What is the elastic potential energy stored in the spring?B. The spring is released. What is the speed of the masses as it reaches the natural length of the spring?
Given data
*The given spring constant is k = 40 N/m
*The given compressed length is x = 0.1 m
*The given mass is m = 0.5 kg
(a)
The formula for the elastic potential energy stored in the spring is given as
[tex]U_p=\frac{1}{2}kx^2[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} U_p=\frac{1}{2}(40)(0.1)^2 \\ =0.2\text{ J} \end{gathered}[/tex]Hence, the elastic potential energy stored in the spring is 0.2 J
(b)
The formula for the speed of the masses is given by the conservation of energy as
[tex]\begin{gathered} U_p=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=x\sqrt[]{\frac{k}{m}} \end{gathered}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} v=(0.1)\sqrt[]{\frac{40}{0.5}} \\ =0.89\text{ m/s} \end{gathered}[/tex]Hence, the speed of the masses as it reaches the length of the spring is v = 0.89 m/s
What is the displacement of the particle in the time interval 7 seconds to 8 seconds?OA. O metersОВ.1.5 metersOC. 3 metersOD. 7 meters
Given,
A velocity-time graph.
The area under the curve of a velocity-time graph of an object gives us the displacement of the object.
From the graph, we can see that the area under the curve from 7 seconds to 8 seconds is a triangle.
The height of the triangle is h=6 m/s.
And the base of the triangle is b=1 s.
The area of a triangle is given by,
[tex]A=\frac{1}{2}bh[/tex]On substituting the known values,
[tex]\begin{gathered} A=0.5\times6\times1 \\ =3\text{ m} \end{gathered}[/tex]Therefore the displacement of the particle in the time interval 7 s to 8 s is 3 meters.
Thus, the correct answer is option C.
jerome pitches a baseball of mass 0.300 kg. the ball arrives at home plate with a speed of 40.0 m/s and is batted straight back to jerome with a return speed of 52.0 m/s. what is the magnitude of change in the ball's momentum?
The change in the momentum of the ball is 3.6 Kgm/s.
What is momentum?The term momentum has to do with the product in the velocity of a body and mass of the body. We have to recall at this point that rate of change of momentum is directly related to the impressed force and this is in accordance with the Newton second law of motion.
Now we have to look at the few pieces of information that we can be able to glean from the question;
Mass of the object = 0.300 kg
Initial speed of the object = 40.0 m/s
Final speed of the object = 52.0 m/s
Given that the change in the velocity of the object is given by;
m( v - u)
m = Mass of the object
v = Final speed of the object
u = Initial speed of the object
change in the velocity of the object = 0.300(52 - 40)
= 3.6 Kgm/s
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what is the mass on grams of 0.56 moles of NaCl
Answer:
1 mole of Na = 23 g
1 mole of Cl = 35 g
1 mole of NaCl = 58 g
.56 * 58 g = 32.5 g
Consider a fluid of density 3.43 g⋅cm−3 flowing through a pipe of varying cross-section. The diameter of the pipe in one section is 9.1 cm, while the diameter in a second section is 12.6 cm. When the diameter of the pipe is 9.1 cm, the flow speed of the fluid is 339 cm⋅s−1 and the pressure is 2.93 × 105 Pa.A)Calculate the flow speed (in m⋅s−1) of the fluid when the diameter of the pipe is 12.6 cm. B)Calculate the pressure (in × 105 Pa) when the pipe has a diameter of 12.6 cm
Given that the pipe has varying cross-sections.
The diameter of one section is d1 = 9.1 cm and the diameter of second section is d2 = 12.6 cm.
Also, the fluid has the density,
[tex]\rho=3.43gcm^{-3}[/tex]The area of the cross-section for the first section is
[tex]\begin{gathered} A_1=\frac{\pi(d1)^2}{4} \\ =\frac{\pi(9.1)^2}{4}cm^2 \end{gathered}[/tex]The area of the cross-section for the second section is
[tex]\begin{gathered} A_2=\frac{\pi(d2)^2}{4} \\ =\frac{\pi(12.6)^2}{4}cm^2 \end{gathered}[/tex]The flow speed for the first section is v1 = 339 cm s^-1
The flow speed for the second section will be v2.
(a) The flow speed for the second section can be calculated as
[tex]\begin{gathered} A_1v1=A_2_{}v2 \\ v2=\frac{A_1v1}{A_2} \\ =\frac{\pi(9.1)^2\times339\times4}{4\times\pi\times(12.6)^2} \\ =\text{ 176.82 cm/s} \\ =1.7682\text{ m/s} \end{gathered}[/tex](b) The pressure for first section is p1 = 2.93 x 10^5 Pa
The pressure for the second section will be p2.
The pressure for the second section can be calculated by the formula,
[tex]\begin{gathered} p2=p1+\frac{1}{2}\rho\mleft\lbrace(v1)^2-(v2\mright)^2\} \\ =2.93\times10^5+\frac{1}{2}\times3.43\mleft\lbrace(339)^2-(176.82)^2\mright\rbrace \\ =4.36\text{ }\times10^5\text{ Pa} \end{gathered}[/tex]7.0 J of work is done to draw a bowstring back. The bow launches an arrow with a mass of 0.09 kg straight upward.
(a) What is the arrow's kinetic energy as it leaves the bow? (Round your answer to one decimal place.)
J
(b) What is the arrow's speed? (Round your answer to one decimal place.)
m/s
(c) What maximum height does the arrow reach? (Round your answer to one decimal place.)
m
The final answer is
(a) The KE = 3.88 * 10
(b) The speed of the arrow is 2.93 * 10
(c) Height which the arrow attains is 4.38 * 10
When the arrow is drawn from a bowstring back then there will be some work done. The pulling of the bowstring will have a distance moved and the energy used to do this.
Given,
Work done = 7 J
Mass of the arrow = 0.09 kg
The computations for a and b are the following:
To calculate the Kinetic energy,Work done = force x distance
7.0 J = force x 0.09
force = 7/0.09=77.77 = 7.77 x 10
force = m. a = 0.09 x a
77.77 / 0.09 = a = 864.1 = 8.64*10²
KE = 0.5 x m x v²
= 0.5 x 0.09 x 864.1
= 3.88 * 10
The speed of the arrow can be calculated asspeed 38.8/(0.5 x 0.09) = v² = 862 sq-root
Hence, v = 2.93*10¹ m/s is the answer
Height that the arrow reaches29.3² = 2gh
= 29.3²/(2 x 9.8) = 858.49 / (2 x 9.8) = 43.8 m
h = 43.8 m or 4.38 * 10¹ is the answer
Therefore, we can conclude that the kinetic energy of the arrow that leaves bowstring is 3.88*10 with the speed of 29.3 m/s and it reaches the height of 43.8 m.
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A fireman standing on a 14 m high ladderoperates a water hose with a round nozzle ofdiameter 2.65 inch. The lower end of the hose(14 m below the nozzle) is connected to thepump outlet of diameter 3.49 inch. The gaugepressure of the water at the pump isCalculate the speed of the water jet emerg-ing from the nozzle. Assume that water is anincompressible liquid of density 1000 kg/m3and negligible viscosity. The acceleration ofgravity is 9.8 m/s?Answer in units of m/s.
Given data,
The height, H = 14 m
The diameter, D = 2.65 inch
The gauge pressure, P = 317.84 kPa
We need to calculate the speed of the water jet emerging from the nozzle.
Using Bernoulli's equation,
[tex]\begin{gathered} \frac{1}{2}\rho(v^2_n-v^2_p)=P_{gauge\text{ }}-\rho gh \\ (v^2_n-v^2_p)=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{A_n}{A_p})^2v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{r_n}{r_p_{}})^4v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \end{gathered}[/tex]Further solved as,
[tex]\begin{gathered} v_n=\sqrt[]{\frac{(\frac{2}{\rho})P_{gauge}-2gh}{1-(\frac{r_n}{r_p})^4}} \\ v_n=\sqrt[]{\frac{(\frac{2}{1000})\times317.84\times10^3-2\times9.8\times14}{1-(\frac{1.325_{}}{1.745_{}})^4}} \\ v_n=\sqrt[]{\frac{635-274.4}{0.667}} \\ v_n=\sqrt[]{540.62} \end{gathered}[/tex]Thus, the speed of the water jet is
[tex]v=23.25\text{ m/s}[/tex]Three bulbs of resistance 100. Ω, 200, Ω and 300 Ω are connected in parallel to a 120. V DC power supply. Draw the diagram and find thea) current in each bulb b) current drawn from the power supplyc) total power drawn power supply d) the net resistance of all bulbs
Let's use the formula for electric current.
[tex]I=\frac{V}{R}[/tex]Where V is the power supply 120 V, and R is the resistance. Let's find the current in each bulb.
[tex]\begin{gathered} I=\frac{120V}{100\Omega}=1.20A \\ I=\frac{120V}{200\Omega}=0.6A \\ I=\frac{120V}{300\Omega}=0.4A \end{gathered}[/tex](a) The current in each bulb is 1.20A, 0.6A, and 0.4A, respectively.(b) (c) The diagram of the circuit isTo find the net resistance, we use the following formula.
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{100\Omega}+\frac{1}{200\Omega}+\frac{1}{300\Omega} \\ \frac{1}{R}=\frac{6+3+2}{600\Omega} \\ \frac{1}{R}=\frac{11}{600\Omega} \\ R=\frac{600}{11}\Omega \\ R\approx54.55\Omega \end{gathered}[/tex]Therefore, the net resistance of all bulbs is 54.55 ohms.which of the following are independent of the mass of an object falling freely near earth's surface: (may have more than 1 answer) 1) acceleration of the object 2) gravitational force acting on the object 3) gravitational force acting on the object 4) magnitude of the gravitational field
As the object is falling freely, the acceleration of the object will be equal to the acceleration due to gravity.
It is given as,
[tex]g=\frac{GM}{R^2}[/tex]Here, G is the univarshal gravitational constant and M is the mass of the Earth.
means acceleration of the object is constant and independent of the mass of the object.
so option 1 is correct.
now the gravitational force on that object is,
[tex]F=\frac{GMm}{R^2}[/tex]here this is dependent on the mass of the object(m).
NOw the gravitational field means the force per unit mass and is given by,
[tex]E=\frac{GM}{R^2}[/tex]Here we can se that this gravitational field is also independent of the mass of the object.
So, option 1 and 4 are correct.
A car travels 400 km in the first 4.5 hours of a trip. It stops for an hour and then travels final 300 km in 2.5 hours. Find the average speed of the car.
Given data:
Distance traveled by car in t_1=4.5 hr is s_1=400 km.
Distance traveled by car in t_2=1 hr is s_2=0 km (as the car was stopped).
Distance traveled by car in t_3=2.5 hr is s_3=300 km.
The average speed is given as,
[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \end{gathered}[/tex]Substitute all known values,
[tex]\begin{gathered} v_{avg}=\frac{(400\text{ km})+(0\text{ km})+(300\text{ km})}{(4.5\text{ hr})+(1\text{ hr})+(2.5\text{ hr})} \\ =87.5\text{ km/h} \end{gathered}[/tex]Therefore, the average speed of the car is 87.5 km/h.