ANSWER
[tex]\begin{gathered} (a)\text{ Boat A} \\ \\ (b)\text{ }42\text{ km} \\ \\ (c)\text{ }0\text{ km/h} \end{gathered}[/tex]EXPLANATION
(a) To determine which boat wins, we have to find the total time the journey takes for each boat.
To do that, apply the formula for speed:
[tex]\begin{gathered} s=\frac{d}{t} \\ \\ t=\frac{d}{s} \end{gathered}[/tex]where d = distance traveled
t = time taken
For boat A, the time taken to go across and return is:
[tex]\begin{gathered} t=\frac{42}{42}+\frac{42}{42}=1+1 \\ \\ t=2\text{ hrs} \end{gathered}[/tex]For boat B, the time taken to go across and return is:
[tex]\begin{gathered} t=\frac{42}{21}+\frac{42}{63}=2+0.667 \\ \\ t=2.667\text{ hrs} \end{gathered}[/tex]We see that it takes boat B longer to complete the race. Hence, boat A will win the race.
(b) To determine how much boat A will win, we can look at the distances both boats cover. Notice that while getting across the lake, boat A travels twice as fast as boat B and travels at the same speed while returning.
This implies that in the time that boat A travels both laps of the race, boat B is just reaching the other end of the lake.
In other words, in the time that boat A travels 84 km, boat B travels 42 km.
Hence, boat A will beat boat B by a distance of:
[tex]\begin{gathered} 84-42\text{ km} \\ \\ 42\text{ km} \end{gathered}[/tex](c) In the race, the winning boat starts from one end of the lake, gets to the other end, and then, returns to the starting point.
This implies that the displacement of the boat is 0 km.
Velocity is given by dividing displacement by the time taken. Since the displacement is 0 km, the velocity of the winning boat, Boat A is 0 km/h
Study the following statements regarding energy and the Law of Energy Conservation: 1. The Law of Energy Conservation means it is saved for another time often for environmental reasons. 2. The Law of Energy Conservation means that there is the same amount of energy before the transfer took place as after 3. When energy is dissipated it means that it disappears. 4. When energy is dissipated it means some of the energy is converted into less useful forms. Which two of the above statements are true?
Answer:
2. The Law of Energy Conservation means that there is the same amount of energy before the transfer took place as after
4. When energy is dissipated it means some of the energy is converted into less useful forms.
Explanation:
Conservation means that energy is conserved through the transfer. It is equal at the start and at the end.
Energy cannot be created or destroyed, so it doesn't disappear, it changes (converted)
A worker pushes a crate with a horizontal force of 225 N a distance of 8.0 M. How much work did the worker do on the crate?11,200 J1,800 J3,920 J0 J
We area asked to determine the work done by pushing a crate a distance of 8 meters using a force of 225N. To do that we will use the following formula:
[tex]W=Fd[/tex]Where:
[tex]\begin{gathered} W=\text{ work} \\ F=\text{ horizontal force} \\ d=\text{ distance} \end{gathered}[/tex]Now, we plug in the values:
[tex]W=(225N)(8m)[/tex]Now, we solve the operations:
[tex]W=1800J[/tex]Therefore, the amount of work done is 1800 Joules.
Derive Kinetic Energy and Potential Energy.
Answer:
[tex]\begin{gathered} \Delta K=\frac{1}{2}m(v_f)^2-\frac{1}{2}m(v_i)^2 \\ P=m\cdot g\cdot h \end{gathered}[/tex]Explanation: We need to derive the formula for both Kinetic energy and potential energy, the derivation of these formulas is as follows:
(i) Kinetic energy:
[tex]\begin{gathered} \Delta K=\Delta W=F\cdot\Delta d\cdot\cos (\theta) \\ \theta=0 \\ \therefore\Rightarrow \\ \Delta K=F\cdot\Delta d\Rightarrow(1) \end{gathered}[/tex]By using the Kinematic equations of motions, equation (1) can be changed to the kinetic energy formula as follows:
[tex]\begin{gathered} (1)\Rightarrow\Delta K=F\cdot\Delta d=m\cdot a\cdot\Delta d\Rightarrow(2) \\ (v_f)^2=(v_i)^2+2a\Delta d \\ \therefore\Rightarrow \\ a\Delta d=\frac{(v_f)^2-(v_i)^2}{2} \\ \text{ Substituting above in the }(2)\text{ gives the formula:} \\ \Delta K=m\cdot\frac{(v_f)^2-(v_i)^2}{2} \\ \Delta K=\frac{1}{2}m(v_f)^2-\frac{1}{2}m(v_i)^2 \\ \text{ This is the kinetic energy formula} \end{gathered}[/tex](ii) Potential-energy:
Potential energy is basically the work needed to lift an object to a certain height, mathematically, the derivation would be as follows:
[tex]\begin{gathered} \Delta P=\Delta W=F\cdot\Delta d \\ \Delta d=\Delta h \\ \therefore\Rightarrow \\ \Delta P=m\cdot a\cdot\Delta h \\ a=g \\ \therefore\Rightarrow \\ \Delta P=m\cdot g\cdot\Delta h \\ \text{ Simple version:} \\ P=m\cdot g\cdot h \end{gathered}[/tex]A phonograph record has an initial angular speed of 33 rev/min . The record slows to 11 rev/min in 2.0 seconds. What is the records angular acceleration in rad/s2 during this time interval ?
In order to calculate the angular acceleration, we can use the following formula:
[tex]a=\frac{v_f-v_i}{t}[/tex]Where vf is the final angular speed, vi is the initial angular speed and t is the interval of time.
Since the speed is in rev/min, we need to convert to rad/s.
Knowing that 1 rev = 2π rad and 1 min = 60 s, we have:
[tex]\begin{gathered} 33\text{ rev/min}=33\cdot\frac{2\pi\text{ rad}}{60\text{ s}}=3.456\text{ rad/s} \\ 11\text{ rev/min}=11\cdot\frac{2\pi\text{ rad}}{60\text{ s}}=1.152\text{ rad/s} \end{gathered}[/tex]Now, using vf = 1.152, vi = 3.456 and t = 2, we have:
[tex]a=\frac{1.152-3.456}{2}=\frac{-2.304}{2}=-1.152\text{ rad/s2}[/tex]So the angular acceleration is -1.152 rad/s².
A pulse travels along a string that is fixed to a wall, as shown. The pulse reaches the wall and is reflected. Which of the following diagrams illustrates the orientation of the reflected pulse?
Using the concept of Simple Harmonic motion, we got that phase of the string wave changes by 180° and velocity gets reversed.
Simple Harmonic Motion or SHM is defined as motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards a mean position. The acceleration of particle executing simple harmonic motion is given by a(t) = -ω2 x(t). Here, ω is the angular velocity of particle.
A pulse of string wave when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of π with the incident wave but the wave's velocity after reflection reverses its direction.
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A stone is thrown straight upward and reaches a maximum height of 32.1 m above itslaunch point. What was the initial speed with which the stone was thrown upwards?Answer:m/s
The initial speed = 25.08 m/s
Explanation:The maximum height, H = 32.1 m
The initial speed, u = ?
The acceleration due to gravity, g = 9.8 m/s²
Write out the maximum height formula and solve for u
[tex]H=\frac{u^2}{2g}[/tex]Substitute H = 32.1 and g = 9.8
[tex]\begin{gathered} 32.1=\frac{u^2}{2(9.8)} \\ \\ u^2=32.1(2)(9.81) \\ \\ u^2=629.16 \\ \\ u=\sqrt{629.16} \\ \\ u=25.08\text{ m/s} \end{gathered}[/tex]The initial speed = 25.08 m/s
A grinding wheel 0.37 m in diameter rotates at 2400 rpm.(A) Calculate its angular velocity in rad/s.(B) What is the linear speed of a point on the edge of the grinding wheel? (C) What is the acceleration of a point on the edge of the grinding wheel?
A) Angular velocity = 251.36 rad/s
B) Linear speed = 46.50 m/s
C) Acceleration = 11687.8 m/s^2
Explanations:A) Angular velocity, w = 2400 rpm
Note that:
1 rpm = (2π)/60
Therefore:
2400 rpm = (2π/60) x 2400 rad/s
2400 rpm = 80π rad/s
2400 rpm = 80 x 3.142
2400 rpm = 251.36 rad/s
The angular velocity, w = 251.36 rad/s
B) The relationship between linear speed(v) and angular velocity(w) is:
v = wr
The diameter, d = 0.37m
Radius, r = d/2
r = 0.37/2
r = 0.185m
Substituting r = 0.185m and w = 251.36 rad/s into v = wr:
v = 251.36 x 0.185
v = 46.50 m/s
C) The acceleration a point on the edge of the grinding wheel is the centripetal acceleration, and is given by the formula:
[tex]\begin{gathered} a\text{ = }\frac{v^2}{r} \\ a\text{ = }\frac{46.50^2}{0.185} \\ a\text{ = }\frac{2162.25}{0.185} \\ a\text{ = }11687.8m/s^2 \end{gathered}[/tex]6) If a system has 48 J of work done on it and absorbs 22 J of heat, what is the value of ΔE for this change?
ANSWER
ΔE = 70 J
EXPLANATION
Given:
• The work done on the system, W = 48 J
,• The heat absorbed by the system, Q = 22 J
Find:
• The change of energy of the system, ΔE
We will use the convention where:
• Heat transferred to the system: positive (otherwise negative)
,• Work done on the system: positive (otherwise negative)
By the first law of thermodynamics,
[tex]\Delta E=Q+W[/tex]Replace with the known values and solve,
[tex]\Delta E=22J+48J=70J[/tex]Hence, the change in the system's internal energy is 70 J.
Finding the coefficient of friction,I need how I would find the friction coefficient
the coefficient is 0.267
ExplanationFree-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object
Step 1
so, to graph the forces we use the Cartessian plane
let the x-axis paralle to the inclided plane
therefore,the weigth is a force acting directly to the center of the earth , ( bottomof page), so we have to find the x and y components of the weitgh
hence
the Free body diagram would be
Step 2
how to find the friction coefficient
the force of friction (ff) is given by
[tex]\begin{gathered} ff=\mu *N \\ where\text{ }\mu\text{ is the coefficient } \\ Nis\text{ the normal force} \end{gathered}[/tex]so, to find the coefficient we need the Normal force
if we do the sum of forces in y -axis equal zero
[tex]\begin{gathered} N-wcos(15)=0 \\ N=wcos(15) \end{gathered}[/tex]now, x -axis
[tex]\begin{gathered} ff-wsin(15)=0 \\ ff=wsin(15) \end{gathered}[/tex]finally, replace and solve for the coefficient
[tex]\begin{gathered} ff=\mu N \\ w*sin(15)=\mu *wcos(15) \\ divide\text{ both sides by w cos\lparen15\rparen} \\ \frac{wsin(15)}{wcos(15)}=\frac{\mu *w*cos(15)}{wcos(15)} \\ \frac{sin(15)}{cos(15)}=\mu \\ tan(15)=\mu \\ 0.267=\mu \end{gathered}[/tex]therefore, the coefficient is 0.267
I hope this helps you
4. A 45.8 kg block is placed on an inclined plane that is 44.2 degrees from the horizontal. What isthe acceleration of the block? Ignore friction.Challenge Question: 5. The same block in problem #4 is placed on the same inclined plane from problem #4. However, now there is friction. If the coefficient of kinetic friction is 0.25, what is the Friction force?(I already answered question 4, i just need help with question 5)
Explanation
Step 1
free body diagram
Step 2
sum of the forces on x-axis
a)x
[tex]\begin{gathered} F=\text{mg}\cos \text{ }\theta=ma \\ \text{replace} \\ F=\text{ 45.8}\cdot9.81\cdot\cos \text{ 44.2=ma} \\ 322.10=45.8a \\ a=\frac{322.10}{45.8} \\ a=703\text{ }\frac{\text{m}}{s^2} \end{gathered}[/tex]
What is the extension force (in lbf) of a 14 inch diameter cylinder with a 10 inch rod and pressure of 700psi?
The given problem can be exemplified in the following diagram:
To determine the extension force we will use the definition of pressure:
[tex]P=\frac{F}{A}[/tex]Where "F" is the force, and "A" is the area. To determine the area we will use the following equation:
[tex]A=\frac{\pi D^2}{4}[/tex]Where "D" is the diameter. The force acts over the area of the 10 inches diameter alone, therefore we don't need to have into account the area of the rod. Replacing the value of the diameter we get:
[tex]A=\frac{\pi(14in)^2}{4}[/tex]Solving the operation:
[tex]A=153.94in^2[/tex]Replacing in the formula for the pressure we get:
[tex]P=\frac{F}{153.94in^2}[/tex]Since we are required to determine the force, we will multiply both sides by the area:
[tex]153.94in^2P=F[/tex]Replacing the given value of the pressure we get:
[tex](153.94in^2)(700\frac{lbf_{}}{in^2})=F[/tex]Solving the operations we get:
[tex]107756.62lb_f[/tex]Therefore, the extension force is 107756.62 lbf.
Calculate the power of a 2057 Ω night light plugged into a 120 V source.
Lets write the data down:
R = 2057 Ω
V = 120 V
The power of the night light is given by:
[tex]P=\frac{V^2}{R}[/tex]Plug in the data in the formula:
[tex]\begin{gathered} P=\frac{120^2}{2057} \\ P=\frac{14,400}{2057} \\ P\approx7.0W \end{gathered}[/tex]The power of the night light is 7.0 W.
4. Thunderstorms are caused by warm and cold air colliding and making a large cloud which is positively charged near the top of the cloud, but negatively charged near the bottom of the cloud. (a) What charge is the surface of the ground? Explain how you know. (b) The thunderclouds are 15,000 meters high off the ground and if there’s a tall tree, 70 meters tall, how much voltage does a single lightning bolt containing 6.25 x 1019 electrons have as it strikes the tree?
let's go,
Generally,
Thunderstorms are caused by warm and cold air colliding and making a large cloud that is positively charged near the top of the cloud but negatively charged near the bottom of the cloud. the charge on the surface of the ground will be positive.
To calculate voltage --- By measuring the length of a lightning strike, multiplying it by the energy per length required.
[tex]\begin{gathered} V=(15000-70)\times(6.25\times10^{19}) \\ \\ V=93,312.5\times10^{19} \\ \\ V=9.33125\times10^{23} \end{gathered}[/tex]which property justifies the statement below? x(y-3)=xy-3xis it associative, transitive, distributive, or communitive
Given the equation:
x(y - 3) = xy - 3x
Let's identify the property which justifies the statement.
Here, the property which was applied is the distributive property.
The law of distributive property states that multiplying the sum of two or more addends by a given value will amount to the same result as multiplying the addends individually by the number and adding the products.
In this equation, we can see x is distributed into the values in the parentheses.
Therefore, the property that justifies this statement is the distributive property.
ANSWER:
Distributive property
Two charges are separated by 6.74 cm. Object A has a charge of 5.0 μ C , while object B has a charge of 7.0 μ C . What is the force on Object A?
ANSWER
69.34 N
EXPLANATION
Given:
• The charge of object A, qA = 5 x 10⁻⁶ C
,• The charge of object B, qB = 7 x 10⁻⁶ C
,• The distance between the objects, r = 6.74 cm = 0.0674 m
Find:
• The force on Object A, FA
The magnitude of the force between two charged particles, q₁ and q₂, separated by a distance r is given by Coulomb's Law,
[tex]F=k\frac{q_1q_2}{r^2}[/tex]Where k is Coulomb's constant and has a value of approximately 9x10⁹ Nm²/C²,
[tex]F=9\times10^9\frac{Nm^2}{C^2}\cdot\frac{5\times10^{-6}C\cdot7\times10^{-6}C}{0.0674^2m^2}\approx69.34N[/tex]Hence, the force on object A is 69.34 N.
Determine the volume in cubic inches (in. ^3) of a 2 liter bottle
Given
Volume of the bottle, V=2 l
To find
The volume in cubic inch
Explanation
We know,
[tex]\begin{gathered} 1\text{ l=61.024 cubic inch} \\ \Rightarrow2\text{ l=2}\times61.024\text{ cubic inch} \\ =122.048\text{ cubic inch} \end{gathered}[/tex]Conclusion
The required volume is 122.048 cubic inch
A fire is an example of ________________ reaction, because heat is _____________ the system.
A. exothermic; leaving
B. exothermic; entering
C. endothermic; leaving
D. endothermic; entering
Because heat is being introduced into the system, a fire is just an example of an exothermic reaction.
The right response is B.
What is exothermic example?Any process was considered to be exothermic if it produces heat while also undergoing a net decrease in standard enthalpy change. Examples include any type of combustion, iron rusting, and water freezing. Exothermic processes are those that discharge warmth and energy into the surrounding environment.
What process are exothermic?If heat is released from the system into the environment, the reaction or change is exothermic. The temperature of the environment rises because it is absorbing heat from the system.
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A flashlight bulb uses 190.1W of power when the current in the bulb is 0.64 A. What is the voltage?
Given:
[tex]\begin{gathered} P=190.1\text{ W} \\ I=0.64\text{ A} \end{gathered}[/tex]The power is given as,
[tex]P=VI[/tex]Here, P is the power, V is voltage and I is the current.
Rearranging equation in order to get voltage,
[tex]V=\frac{P}{I}[/tex]Putting all known values,
[tex]\begin{gathered} V=\frac{190.1\text{ W}}{0.64\text{ A}} \\ =297.03\text{ V} \end{gathered}[/tex]Therefore, the voltage is 297.03 V.
A uniform rod of mass m=1.0 kg and length L=2 meters is free to rotate about its center as shown in the figure. A constant torque is applied by a constant force of magnitude F=5.0 N to one end of the rod, as shown. To be clear, the force is always perpendicular to the rod, which rotates about the axis indicated by the red line.a) If the rod is initially at rest, how long does it take to reach an angular velocity omega=30 radians/s. b) Assuming the torque stops when the angular velocity is 30 radians/s, what is the total rotational energy of the rod? c) The rod suddenly turns into a uniform sphere of radius R=1m and mass M=1.0 kg rotating about its center. Assuming angular momentum is conserved, what is the angular velocity of the sphere omegasphere?
a)
If the length is 2 meters and the rod rotates about its center, so 2 meters is the diameter, and the radius of rotation is 1 meter.
Then, if the force is 5 N, let's calculate the torque:
[tex]\begin{gathered} \tau=F\cdot r \\ \tau=5\cdot1=5\text{ Nm} \end{gathered}[/tex]Then, calculating the rotational inertia and the angular acceleration, we have:
[tex]\begin{gathered} I=\frac{1}{2}mr^2 \\ I=\frac{1}{2}\cdot1\cdot1^2=0.5 \\ \\ \alpha=\frac{\tau}{I}=\frac{5}{0.5}=10 \end{gathered}[/tex]The angular acceleration is 10 rad/s², so to reach an angular velocity of 30 rad/s, it takes 3 seconds.
b)
The rotational energy can be calculated with the formula below:
[tex]\begin{gathered} E_k=\frac{1}{2}I\cdot\omega^2 \\ E_k=\frac{1}{2}\cdot0.5\cdot30^2 \\ E_k=225\text{ J} \end{gathered}[/tex]c)
The angular momentum is given by:
[tex]\begin{gathered} L=I\cdot\omega \\ L=0.5\cdot30 \\ L=15 \end{gathered}[/tex]Then, since the rod turns into a sphere, the new rotational inertia is:
[tex]\begin{gathered} I=\frac{2}{5}mr^2 \\ I=\frac{2}{5}\cdot1\cdot1^2 \\ I=\frac{2}{5}=0.4 \end{gathered}[/tex]So the new angular velocity is:
[tex]\begin{gathered} L=I\cdot\omega \\ 15=0.4\cdot\omega \\ \omega=\frac{15}{0.4}=37.5\text{ rad/s} \end{gathered}[/tex]You are told that the value of a car of the given model, based on its age, is less than $6000. Give a range of values that could represent the car.
The following equation represents the value (y) of a car based on its age (x).
[tex]y=-800x+12000[/tex]We are told that the value of a car of the given model, based on its age, is less than $6000.
So, we can write
[tex]6000<-800x+12000[/tex]Let us solve the above inequality for x.
[tex]\begin{gathered} 6000<-800x+12000 \\ 6000-12000<-800x \\ -6000<-800x \\ \frac{-6000}{-800}<\frac{-800x}{-800} \\ 7.5>x \\ x<7.5 \end{gathered}[/tex]This means that the age of the car is less than 7.5 years.
Therefore, the possible range of values that could represent the age of the car is 0 to 7.5 years.
[tex]0A white crowned sparrow flying horizontally with a speed of 1.80m/s folds its wings and begins to drop in free fall. (a) How far does the sparrow fall after traveling a horizontal distance of 0.500m? (b) If the sparrows initial speed is increased, does the distance of fall increase, decrease or stay the same?
First, find the time that it takes for the sparrow to travel 0.500m at a speed of 1.80 m/s. Since speed is the ratio between distance and the time taken to travel that distance, then:
[tex]v=\frac{d}{t}[/tex]Isolate t from the equation:
[tex]\Rightarrow t=\frac{d}{v}[/tex]Replace d=0.500m and v=1.80 m/s to find the time:
[tex]t=\frac{0.500m}{1.80\frac{m}{s}}=0.2777\ldots s[/tex]During that time, the sparrow falls vertically according to the equation:
[tex]h=\frac{1}{2}gt^2[/tex]Where g is the gravitational acceleration on Earth:
[tex]g=9.81\frac{m}{s^2}[/tex]Replace the value of g and t=0.2777...s to find how far does the sparrow fall:
[tex]h=\frac{1}{2}(9.81\frac{m}{s^2})(0.2777\ldots s)^2=0.378472\ldots m[/tex]If the initial speed is increased, it will take less time to travel 0.500m, and then, the falling time will be decreased, and so the distance of fall would decrease. However, the horizontal speed does not affect the distance of fall if the fall time is the same (but the horizontal displacement would increase).
Therefore, the answers are:
a) The sparrow falls a distance of 0.378m
b) The distance of fall would decrease since it would take less time to travel 0.500m.
24.A projectile is launched upward at an angle of 30° from thehorizontal with an initial velocity of 145 meters per second. Howfar does the projectile travel horizontally?(a) 1.86 x 10^3 m(b) 2.28 x 10^3 m(c) 4.55 x 10^3 m(d) 9.10 x 10^3 m
ANSWER:
(a) 1.86 x 10^3 m
STEP-BY-STEP EXPLANATION:
The movement that the projectile takes is a parabolic movement, therefore, the horizontal range is determined as follows:
[tex]x=\frac{v^2_0\cdot\sin (2\theta)}{g}[/tex]We know the initial velocity, the angle and the gravity, therefore, we replace and calculate the value of the horizontal distance:
[tex]\begin{gathered} x=\frac{145^2\cdot\sin (2\cdot30)}{9.8} \\ x=1857.97\cong1.86\cdot10^3 \end{gathered}[/tex]The horizontal distance is equal to approximately 1860 meters
Determine the image distance and image height for a 5.00 cm tall object placed 20.0 cm from a double convex lens with a focal length of 15.0 cm.
Given:
the height of the object is
[tex]h_0=5\text{ cm}[/tex]The distance of the object is
[tex]d_0=-20\text{ cm}[/tex]The focal length of the lens is
[tex]f=15\text{ cm}[/tex]Required: the distance of the image and height of the image.
Explanation:
the lens formula is given by
[tex]\frac{1}{f}=\frac{1}{d_i}-\frac{1}{d_0}[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} \frac{1}{15\text{ cm}}=\frac{1}{d_i}-\frac{1}{-20\text{ cm}} \\ \frac{1}{d_i}=\frac{1}{15\text{ cm}}-\frac{1}{20\text{ cm}} \\ \frac{1}{d_i}=\frac{4-3}{60\text{ cm}} \\ d_i=60\text{ cm} \end{gathered}[/tex]Thus, the distance of the image is 60 cm.
now calculate the height of the image
we know that
[tex]\frac{h_i}{h_0}=\frac{d_i}{d_0}[/tex]substitute all the values in the above relation, we get:
[tex]\begin{gathered} h_i=5\text{ cm}\times\frac{60}{-20} \\ h_i=-15\text{ cm} \end{gathered}[/tex]Thus, the height of the image is 15 cm.
how does gravity affect the range of a projectile?
Gravity has an effect on a projectile because it reduces the height that it can achieve.
The object's upward movement is stopped by gravity, which pulls it back to earth, limiting the projectile's vertical component. As a projectile moves through the air it is slowed down by air resistance.
I take my calculator from the top of a building and toss it straight upward at 9 m/s from a 23 m tall building.A. what is the speed of the Calculator right before it hits the ground? B. What is the acceleration of the calculator at peak height? C. How much time does it take for the calculator to reach peak height?
Given,
The initial velocity with which the calculator was thrown, u=9 m/s
The height of the building, h=23 m
A.
When the calculator reaches the peak height, its velocity will become zero. That is v=0 m/s
And while it is going up the acceleration due to gravity will be acting downward. Thus the acceleration due to gravity will be a negative value.
From the equation of motion,
[tex]v^2-u^2=2gs_{}[/tex]Where s is the peak height reached by the calculator and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 0-9^2=2\times-9.8\times s \\ \Rightarrow s=\frac{-9^2}{2\times-9.8} \\ =4.13\text{ m} \end{gathered}[/tex]Thus the total height reached by the calculator from the ground is
[tex]\begin{gathered} H=h+s \\ =23+4.13 \\ =27.13 \end{gathered}[/tex]After reaching the peak height, the calculator starts descending. This descent starts with the initial velocity of v=0 m/s. And the acceleration due to gravity will be in the direction of motion of the calculator. Thus it will be a positive value.
From the equation of motion,
[tex]v^2_0-v^2=2gH[/tex]where v₀ is the velocity of the calculator right before it hits the ground.
On substituting the known values,
[tex]\begin{gathered} v^2_0-0=2\times9.8\times27.13 \\ \Rightarrow v_0=\sqrt[]{2\times9.8\times27.13}^{} \\ =23.06\text{ m/s} \end{gathered}[/tex]Thus the speed of the calculator right before it hits the ground is 23.06 m/s
B.
The acceleration of the calculator is a constant value. It is always equal to the acceleration due to gravity.
Thus the acceleration of the calculator at peak height is 9.8 m/s²
C.
From the equation of motion,
[tex]v=u+gt[/tex]Where t is the time it takes for the calculator to reach the peak height.
On substituting the known values,
[tex]\begin{gathered} 0=9+(-9.8)t \\ \Rightarrow t=\frac{-9}{-9.8} \\ =0.92\text{ s} \end{gathered}[/tex]Thus it takes 0.92 s for the calculator to reach the peak height.
Use g = 10 m/s/s for this problem. A force is applied to a block through tension in a rope, at an angle to the horizontal as shown. The block is being pulled across a rough surface. The mass of the block is 71 kg. The tension in the rope is 1197 N. The angle from horizontal is 15 degrees. The coefficient of friction μ is 0.3.
Answer:
14.6 m/s²
Explanation:
First, we will make the free body diagram
Since the net vertical force is equal to 0 because the block is not moving up, we can write the following equation
[tex]\begin{gathered} F_{nety}=F_n+F_{Ty}-mg=0 \\ F_n+F_T\sin15-mg=0 \end{gathered}[/tex]Then, we can solve for the normal force and replace Ft = 1197 N, m = 71 kg and g = 10 m/s²
[tex]\begin{gathered} F_n=mg-F_T\sin15 \\ F_n=(71\text{ kg\rparen\lparen10 m/s}^2)-(1197N)(0.26) \\ F_n=400.2\text{ N} \end{gathered}[/tex]Now, we can write the following equation for the net horizontal force
[tex]\begin{gathered} F_{net}=F_{Tx}-F_f=ma \\ F_T\cos15-\mu F_n=ma \end{gathered}[/tex]Where μ is the coefficient of friction and a is the acceleration of the block. Solving for a and replacing Ft = 1197 N, m = 71 kg, Fn = 400.2 N and μ = 0.3, we get
[tex]\begin{gathered} a=\frac{F_T\cos15-\mu F_n}{m} \\ \\ a=\frac{(1197\text{ N\rparen}\cos15-0.3(400.2N)}{71\text{ kg}} \\ \\ a=14.6\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration of the block is 14.6 m/s²
Why are metals the best conductors of electricity? Why are metals the best magnetisers ? What is the key characteristic of metals that allows us to answer these two apparently different questions?
Question: What is the key characteristic of metals that allows us to answer these two apparently different questions?
Answer:
Explanation:
The atoms of a metal contains at least one free electron. Conduction of electricity is due to the these free electrons.
Magnetism is also affected by the availability of free electrons.
Thus, metals are good conductors of electricity and also the best magnetisers because of the availability of free electrons in metals
AM radio signals are broadcast at frequencies between 550 kilohertz (kHz) and 1600 kilohertz. FM radio signals are broadcast at frequencies between 88.0 megahertz (MHz) and 108.0 megahertz. a. What is the wavelength of an AM station with a frequency of 790 kilohertz? Include units in your answer.b. What is the wavelength of an FM station with a frequency of 98.7 megahertz? Include units in your answer.
a)
From the information given,
Lowest frequency = 550 khz = 550 x 10^3 hz
Highest frequency = 1600 khz = 1600 x 10^3 hz
The formula for calculating wavelength is expressed as
wavelength = velocity/frequency
Recall, radio signals travel at the speed of light and speed of light = 3 x 10^8 m/s
Thus, for a frequency of 790 khz which is equal to 790 x 10^3 hz,
the wavelength of the AM station would be
3 x 10^8/790 x 10^3
wavelength of the AM station = 379.75 m
b) From the information given,
Lowest frequency = 88 mhz = 88 x 10^6 hz
Thus, for a frequency of 98.7 mhz = 98.7 x 10^6 hz,
wavelngth of FM station would be
3 x 10^8/98.7 x 10^6
wavelength of the FM station = 3.04 m
Point A is on the ground, and points B and C are h = 20 meters above the ground. Point B is directly above point A, and point C is L = 40 meters away from point A as shown.(a) How much work must be done by an external agent to move a 2 kg object from rest at point A to rest at point B?(b) How much work must be done to move the same object from rest at point A to rest at point C?
Given the mass of the object, m = 2 kg
Distance between point A and B, h= 20 m
(a) To find work done in moving the object from A to B
Work done is
[tex]\begin{gathered} W1=Force\times displacement\text{ } \\ =mgh \end{gathered}[/tex]Here, g is the acceleration due to gravity whose value is 9.8 m/s^2.
Substituting the values, we get
[tex]\begin{gathered} W1=2\times9.8\times20 \\ =\text{ 392 J} \end{gathered}[/tex](b) To find work done in moving the object from A to C
Gravitational force is a conservative force and work done depends only on the initial and final position and not on the path.
So, the work done will be
[tex]\begin{gathered} W2=\text{mgh} \\ =2\times9.8\times20 \\ =\text{ 392 J} \end{gathered}[/tex]Let f(x) = 5* Let g(x) = 54 - 7 = - g Which statement describes the graph of g(x)with respect to the graph of f(x)? O g(x)is translated 7 units right from f(x). O g(x)is translated 7 units down fromf (x). O g(x)is translated 7 units left fromf (x). O g(x)is translated 7 units up fromf (x).
Answer:
[tex]\text{Second Option}[/tex]Explanation: The two functions are:
[tex]\begin{gathered} f(x)=5^x \\ g(x)=5^x-7 \end{gathered}[/tex]The graphs of these two functions are as follows:
From this we can infer that the answer is: