Answers
The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have
• net horizontal force
[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R[/tex]
where [tex]m=4.10\,\rm kg[/tex], [tex]v=2.85\frac{\rm m}{\rm s}[/tex], and [tex]R[/tex] is the radius of the circular path.
As shown in the diagram, we can see that
[tex]\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)[/tex]
where [tex]r=1.69\,\rm m[/tex], so that
[tex]F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}[/tex]
The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have
• net vertical force
[tex]F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}[/tex]
Solve for [tex]\theta[/tex] :
[tex]\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0[/tex]
Complete the square:
[tex]\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}[/tex]
Plugging in the known quantities, we end up with
[tex]\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27[/tex]
The second case has no real solution, since [tex]-1\le\cos(\theta)\le1[/tex] for all [tex]\theta[/tex]. This leaves us with
[tex]\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}[/tex]
Related Questions
Two football players collide head-on in midair while chasing a pass. The first player has a 92.5 kg mass and an initial velocity of 5.50 m/s, while the second player has a 116 kg mass and initial velocity of -6 m/s. What is their velocity (in m/s) just after impact if they cling together? (Indicate the direction with the sign of your answer.)
Answers
Their velocity (in m/s) just after impact if they cling together is -0.9 m/s.
Final speed of the two players
Apply the principle of conservation of linear momentum;
Pi = Pf
where;
Pi is initial momentum of the playersPf is final momentum of the players(92.5)(5.5) + (116)(-6) = v(92.5 + 116)
508.75 - 696 = v(208.5)
-187.25 = 208.5v
v = -187.25/208.5
v = -0.9 m/s
Thus, their velocity (in m/s) just after impact if they cling together is -0.9 m/s.
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Andrew and Warren sit on opposite sides of a table of mass 28 kg. Andrew pushes the table to the right with a force of 277 N, and Warren pushes the table to the left with a force of 215 N. (Assume there is no friction.)
a. Draw the free-body diagram of the table. (You do not need to draw it perfectly to scale. Just make sure the directions are correct.)
b. Write the expression for the net force on the table along the y-axis.
c. Write the expression for the net force on the table along the x-axis.
d. What is the normal force acting on the table?
e. What is the net force on the table along the x-axis, including direction?
f. What is the acceleration of the table, including direction?
Answers
The acceleration is 2.2 m/s^2 and the net force is 62N towards the right
What is the net force?The net force is the effective force that acts on a body in a give direction.
1. The net force in the y axis is; ∑Fy = 0
2. The net force along the y axis is; ∑Fx = 277 N - 215 N
3. The normal reaction is given by; 28 kg * 9.8 m/s^2 = 274.4 N
4. The net force in the x axis is 277 N - 215 N = 62N towards the right
5. The acceleration of this force is obtained from;
F = ma
62 = 28 a
a = 62/28
a = 2.2 m/s^2
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A proton moves 0.10 m along the direction of an electric field of magnitude 3.0 V/m. What is the change in kinetic energy of the proton? (e = 1.60 × 10-19 C)
Answers
The change in kinetic energy of the proton is 4.8 x 10⁻²⁰.
Change in kinetic energy of the protonThe change in kinetic energy of the proton is calculated as follows;
ΔK.E = W = eV
where;
V is the potential differencee is charge of electronV = Ed
V = 3 V/m x 0.1 m = 0.3 V
ΔK.E = 1.6 x 10⁻¹⁹ x 0.3 = 4.8 x 10⁻²⁰ J
Thus, the change in kinetic energy of the proton is 4.8 x 10⁻²⁰ J.
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HELP ASAP PLEASE!!!
When a 20-Ohms resistor is connected across the terminals of a 12V battery, the terminal voltage of the battery falls to 10V. What is the internal resistance of the battery?
A. 4-ohms
B. 2-ohms
C. 10-ohms
Answers
A car moves with an average speed of 75 kmh^-1 from town P to town Q in 2 hours. By using information, you may calculate the distance between two towns. state a derived quantity and its S.I unit
Answers
Answer:
patron
Explanation:
The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.
The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.
Explanation:
patron
Explanation:
The patron is something or someone who defends some cause or point of view. In the art field, for example, the patron may be considered a sponsor, that is, someone who is known for defending a particular group of people or specific situation.
The patron is one who advocates, advises and directs. In the military, patrons are heroic figures who are chosen to defend a military unit, for example.
D = (75)(2)
D = 150 km.
Therefore, the distance between town p and town q is 150 km.
In a game at a carnival, a contestant rolls a ball up the slope with an initial speed vi. The object of the game is to roll the ball in such a way that it will get “stuck” in the depression at B and not return back down the slope. This will happen if the ball’s speed when it gets to point A is essentially zero. (The speed of the ball at point A really has to be greater than zero in order for the ball to make it past point A, but
the speed at point A must be greater than zero only by an arbitrarily small amount, so that we can say that the condition for the ball not to return is essentially that the speed at A must be zero.) Assuming that the ball rolls without slipping and that energy losses due to friction are negligible, find the initial speed vi required to make the
speed of the ball at point A zero. Let the mass of the ball be called M and let the radius of the ball be called R. Treat the ball as a solid sphere.
Answers
Initial speed of ball will be 2.7m/s.
When a body is rotating about an axis, then it has kinetic energy.
And this energy is called rotational kinetic energy.
It is given as - R.K.E. = 1/2 Iω²
And if a ball is rolling without slipping.
Then the moment of inertia of the solid ball is written as -
I = 25MR²
Vi = Rω
Here it is given in the problem that-
height(h) = 0.53m
Now by the conservation of energy we can write the equation as -
1/2MVi² + 1/2Iω² = Mgh
so that -
(1/2)MVi² + (1/2)×(2/5MR²) ×(Vi/R)² = Mgh(1/2)Vi² + (1/5)Vi²
= gh(7/10)Vi² = 9.8 × 0.53
Vi = 2.7 m/s
So that the initial velocity of ball came out to be 2.7m/s after applying all concepts or rotational motion.
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16. What was the purpose of the first personality tests?
O A. To determine jobs for people
O B. To match individuals for relationships
O C. To place people in jobs
O D. To determine which soldiers during WWI would be more prone to PTSD, or "sh
O Mark for review (Will be highlighted on the review page)
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Answers
Answer:
A. To determine jobs for people
What is the difference in the weight of a 117 kg person as measured at sea level and at the top of the Vinson Massif, the highest peak in Antarctica? The product GMEarth = 3.99 x 10^14 N x m2 x kg-1, the radius of the Earth is 6.38 × 10^6 m, and the height above sea level of the Vinson Massif is 5.14 × 10^3 m. Neglect the flattening of the Earth at the poles.
Answers
The difference in weight of a 117 kg person as measured at sea level and at the top of the Vinson Massif, the highest peak in Antarctica is -1.847 N
How to find the difference in weight?
To find the difference in weight, we find the change in acceleration due to gravity.
What is acceleration due to gravity?
The acceleration due to gravity is the acceleration exerted by the force of gravity on an object. It is given by
g = GM/r² where
GM = 3.99 × 10¹⁴ Nm²/kg and r = radius of earth = 6.38 × 10⁶ mThe change in acceleration due to gravityNow, to find the change in g, we differentiate g with respect to r,
So, dg/dr = d(GM/r²)
= -2GM/r³
So, the change in acceleration due to gravity, Δg is
Δg = dg/dr × Δr
= -2GM/r³ × Δr
Now, Δr = height above sea level of Vinson Massif = 5.14 × 10³ m
Substituting the values of the variables into the equation, we have
Δg = -2GM/r³ × Δr
Δg = [(-2 × 3.99 × 10¹⁴ Nm²/kg)/(6.38 × 10⁶ m)³] × 5.14 × 10³ m
Δg =[ (-7.98 × 10¹⁴ Nm²/kg)/259.69 × 10¹⁸ m³] × 5.14 × 10³ m
Δg = -0.03073 × 10⁻⁴ N/mkg × 5.14 × 10³ m
Δg = -0.1579 × 10⁻¹ N/kg
Δg = -0.01579 N/kg
The difference in weight of the 117 kg personSince weight W = mg where
m = mass of person and g = acceleration due to gravityThe difference in weight is dW = mdg
Since
m = 117 kg and dg = -0.01579 N/kgdW = mdg
= 117 kg × -0.01579 N/kg
= -1.847 N
So, the difference in weight of a 117 kg person as measured at sea level and at the top of the Vinson Massif, the highest peak in Antarctica is -1.847 N
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1. A football is kicked with a velocity. of 20m/s at an angle of 45° with the horizontal i. Find the time taken by the ball to strike the ground ii. Find the maximum height 11. How far from the kick does it hit the ground Take gravity = 10 m/s
Answers
(a) The maximum height reached by the projectile is 10.2 m and
(b) The horizontal distance traveled by the projectile is 40.8 m.
Maximum height reached by the projectile
H = u²sin²θ/2g
H = (20²(sin(45))² / (2 x 9.8)
H = 10.2 m
Time of motionT = (2u sinθ)/g
T = (2 x 20 x sin(45) )/(9.8)
T = 2.886 s
Horizontal distance traveled by the projectilex = (v cos(45) x t
x = (20 x cos(45)) x 2.886
x = 40.8 m
Thus, the maximum height reached by the projectile is 10.2 m and the horizontal distance traveled by the projectile is 40.8 m.
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describe 5 steps you would take when trouble shooting a test kit
Answers
Answer:
1. Information Gathering
2. Analysis and Planning.
3. Implementation of a solution.
4. Assessment of the effectiveness of the solution.
5. Documentation of the incident.
Assume the three resistances (R1, R2, R3) in the problem 4 are now connected in parallel. (A) What is the total resistance of the parallel resistors? (B) What is the current in the overall circuit? (C) What is the current through each resistance?
Answers
(A)The total resistance of the parallel resistors is: [tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]
(B) The current in the overall circuit is: [tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]
(C) The current through each resistance is as follows:
[tex]I_{1} = \frac{V}{R_{1}}[/tex][tex]I_{2} = \frac{V}{R_{2}}[/tex][tex]I_{3} = \frac{V}{R_{3}}[/tex]What is the the total resistance for resistors in parallel?For the three resistances connected in parallel, R1, R2, R3 , the total resistance, [tex]R_{T}[/tex] is calculated as follows:
[tex]R_{T} = \frac{R_{1} \times R_{2} \times R_{3}}{R_{2}R_{3}+R_{1}R_{3}+R_{1}R_{2}}[/tex]
The current in the overall circuit is calculated using the formula:
[tex]I_{T} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}[/tex]
The current through each resistance is given as follows;
Current through R1, [tex]I_{1} = \frac{V}{R_{1}}[/tex]
Current through R2; [tex]I_{2} = \frac{V}{R_{2}}[/tex]
Current through R3; [tex]I_{3} = \frac{V}{R_{3}}[/tex]
In conclusion, the voltage across resistances in parallel is the same but the current varies.
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(10 pts.) A car of mass m is on an icy driveway inclined at an angle .
(A)Find the acceleration of the car, assuming the driveway is frictionless.
(B) Suppose the car is released from rest at the top of the incline and the distance from the car’s front
bumper to the bottom of the incline is d. How long does it take the front bumper to reach the
bottom of the hill, and what is the car’s speed as it arrives there?
Answers
The acceleration and time is mathematically given as
a=gsinθ
[tex]t=\frac{2d}{gsinθ}[/tex]
What is the acceleration of the car, assuming the driveway is frictionless, and how long does it take the front bumper to reach the bottom of the hill, and what is the car’s speed as it arrives there?Generally, the equation for the Force of the car is mathematically given as
[tex]N = mg cos\theta[/tex]
Therefore
[tex]ma=mgsin\theta[/tex]
a=gsinθ
The acceleration of the car is mathematically given as
a=gsinθ
Generally, the equation for the final velocity of the car is mathematically given as
[tex]v^2−u^2=2as[/tex]
Therefore
[tex]v 2 −(0) 2 =2(gsin\theta)d[/tex]
[tex]v= \sqrt {2gsin\theta.d}[/tex]
Generally, the equation for the motion of the car is mathematically given as
v−u=at
[tex]2gsin\thetad =(gsin\theta)t[/tex]
[tex]t=\frac{2d}{gsinθ}[/tex]
In conclusion, the time it takes the front bumper to reach the bottom of the hill is
[tex]t=\frac{2d}{gsinθ}[/tex]
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A person throws a 0.21-kg ball straight up into the air. It reaches a height of
10 m. What is the force on the ball as it begins to fall? (The acceleration due
to gravity is 9.81 m/s².)
A. 2.06 N
B. 4.32 N
C. 1.18 N
OD. 4.67 N
Answers
Answer:
A
Explanation:
As it begins to fall
F = ma a = 9.81
F = .21 * 9.81 = 2.06 N
The definition of force in physics is the push or pull on a massed object that changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. The correct option is A.
Force is a physical factor that alters or has the potential to alter an object's state of rest or motion as well as its shape. Newton is the SI unit of force.
One of the most fundamental types of physical entities is the force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction. Frictional force and contact force are the two categories under which all forces can be classified.
F = m × a
F = 0.21 × 9.81
F = 2.06 N
Thus the correct option is A.
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How is Behaviorism and Psychoanalysis theories are different?
Answers
Answer:
Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.
Explanation:
Answer: Behaviorists give prominence to the external behavior of individuals and believe that behavior is a response to external stimuli. On the other hand, psychoanalysis emphasizes the centrality of the human mind. They believe that the unconscious has the potential to motivate behavior.
Explanation:
In an experiment, the __________ is what researchers measure and expect to change as a result of manipulation.
Answers
Answer:
dependent variable
Explanation:
What does newtons 2nd law explain?
Answer asap for brainlist
Answers
Answer:
B. why you must push harder to move a car farther.
Explanation:
acceleration (gaining speed) happens when a force acts on a mass (object). FORCE=MASSxACCELERATION
A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.
(a)What is the maximum potential difference across the resistor (in V)?
(b)What is the maximum current through the resistor (in A)?
(c)What is the rms current through the resistor (in A)?
(d)What is the average power dissipated by the resistor (in W)?
Answers
(a) The peak voltage across the resistor is 339.41 V.
(b) The maximum current through the resistor is 0.23 A.
(c) The rms current through the resistor is 0.16 A.
(d) The average power dissipated by the resistor is 38.4 W.
Peak voltage or maximum potential differenceVrms = 0.7071V₀
where;
V₀ is peak voltageV₀ = Vrms/0.7071
V₀ = 240/0.7071
V₀ = 339.41 V
rms current through the resistorI(rms) = V(rms)/R
I(rms) = (240)/(1,540)
I(rms) = 0.16 A
maximum current through the resistorI₀ = I(rms)/0.7071
I₀ = (0.16)/0.7071
I₀ = 0.23 A
Average power dissipated by the resistorP = I(rms) x V(rms)
P = 0.16 x 240
P = 38.4 W
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A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs of the same constant K= 250 N/m, then the block is pulled a distance A and released. What's the speed V and angular velocity w of the block as it passes through the equilibrium when the two springs are arranged in:
a) parallel
b) series
Answers
a.
i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/sii. the angular velocity when the two springs are in parallel is 7.07 rad/sb.
i. the speed of the block of mass when the springs are connected in series is 11.2 A m/sii. the angular velocity when the two springs are in series is 11.2 rad/sa. i. How to calculate the velocity of the mass when the springs are connected in parallel?Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k
= 2k
= 2 × 250 N/m
= 500 N/m
Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA
Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where
k' = equivalent spring constant in parallel = 500 N/m, A = maximum displacement of spring, x = equilibrium position = 0 m, M = mass of block = 10 kg and v = speed of block at equilibrium positionMaking v subject of the formula, we have
v = √[k'(A² - x²)/M]
Substituting the values of the variables into the equation, we have
v = √[k'(A² - x²)/M]
v = √[500 N/m(A² - (0)²)/10]
v = √[50 N/m(A² - 0)]
v = [√50]A m/s
v = [5√2] A m/s
v = 7.07 A m/s
So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s
ii. The angular velocity of mass when the springs are in parallel
Since velocity of spring v = ω√(A² - x²) where
ω = angular velocity of spring, A = maximum displacement of spring and x = equilbrium position of spring = 0 mMaking ω subject of the formula, we have
ω = v/√(A² - x²)
Since v = 7.07 A m/s
Substituting the values of the other variables into the equation, we have
ω = v/√(A² - x²)
ω = 7.07 A m/s/√(A² - 0²)
ω = 7.07 A m/s/√(A² - 0)
ω = 7.07 A m/s/√A²
ω = 7.07 A m/s/A m
ω = 7.07 rad/s
So, the angular velocity when the two springs are in parallel is 7.07 rad/s
b. i. How to calculate the velocity of the mass when the springs are connected in series?Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k
= 2/k
⇒ k" = k/2
k" = 250 N/m ÷ 2
= 125 N/m
Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA
Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where
k" = equivalent spring constant in series = 125 N/m, A = maximum displacement of spring, x = equilibrium position = 0 m, M = mass of block = 10 kg and v' = speed of block at equilibrium positionMaking v subject of the formula, we have
v = √[k"(A² - x²)/M]
Substituting the values of the variables into the equation, we have
v = √[k"(A² - x²)/M]
v = √[125 N/m(A² - (0)²)/10]
v = √[125 N/m(A² - 0)]
v = [√125]A m/s
v = [5√5] A m/s
v = 11.2 A m/s
So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s
ii. The angular velocity of the mass when the springs are in series
Since velocity of spring v = ω√(A² - x²) where
ω = angular velocity of spring, A = maximum displacement of spring and x = equilbrium position of spring = 0 mMaking ω subject of the formula, we have
ω = v/√(A² - x²)
Since v = 11.2 A m/s
Substituting the values of the other variables into the equation, we have
ω = v/√(A² - x²)
ω = 11.2 A m/s/√(A² - 0²)
ω = 11.2 A m/s/√(A² - 0)
ω = 11.2 A m/s/√A²
ω = 11.2 A m/s/A m
ω = 11.2 rad/s
So, the angular velocity when the two springs are in series is 11.2 rad/s
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A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hi=nge. The rod can rotate without friction. (See figure.)
Initially the rod is held at rest at an angle of Θ = 70.4 with respect to the horizontal surface. Then the rod is released.
What is the angular speed of the rod, when it lands on the horizontal surface?
What is the angular acceleration of the rod, just before it touches the horizontal surface?
Answers
(a) The angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.
(b) The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .
Angular acceleration of the rodThe angular acceleration of the rod is calculated as follows;
Apply the principle of conservation of angular momentum;
τ = Iα
where;
τ is torqueI is moment of inertial of the rodα is the angular accelerationτ = Fr = mg(L/2) cosθ
I = mL²/3
mg(L/2) cosθ = mL²/3(α)
g(1/2) cosθ = L/3(α)
(³/₂)g cosθ = L(α)
(³/₂ L)g cosθ = α
1.5Lg cosθ = α
1.5(2.1) cos (70.4) = α
1.06 rad/s² = α
Angular speed of the rodωf² = ω₁² + 2αθ
ωf² = 0 + 2(1.06)(70.4π/180)
ωf² = 2.605
ωf = √2.605
ωf = 1.61 rad/s
Thus, the angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.
The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .
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A pan-galactic gargle (an alien) whose mass is 85 kg exploring a large asteroid. What is the force of
gravity holding it down?
Answers
The acceleration due to gravity of the asteroid is 833.85 N.
Force of gravity holding the the alien down
F = mg
where;
m is mass of the alieng is acceleration due to gravityAssuming the acceleration due to gravity of the asteroid = 9.81 m/s²
F = 85 x 9.81
F = 833.85 N
Thus, the acceleration due to gravity of the asteroid is 833.85 N.
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Need help in showing how to get the final velocity
Answers
I don't think the provided solution is correct because it's not dimensionally consistent. The quantity [tex]8h-1[/tex] in particular doesn't make sense since it's mixing a distance with a dimensionless constant.
Here's how I think the proper answer should look:
The net force on the box acting perpendicular to the ramp is
[tex]\sum F_\perp = F_{\rm normal} - mg \cos(\theta) = 0[/tex]
where [tex]F_{\rm normal}[/tex] is the magnitude of the normal force due to contact with the ramp and [tex]mg\cos(\theta)[/tex] is the magnitude of the box's weight acting in this direction. The net force is zero since the box doesn't move up or down relative to the plane of motion.
The net force acting parallel the ramp is
[tex]\sum F_\| = mg\sin(\theta) - F_{\rm friction} = ma[/tex]
where [tex]mg\sin(\theta)[/tex] is the magnitude of the parallel component of the box's weight, [tex]F_{\rm friction}[/tex] is the magnitude of kinetic friction, and [tex]a[/tex] is the acceleration of the box.
From the first equation, we find
[tex]F_{\rm normal} = mg \cos(\theta)[/tex]
and since [tex]F_{\rm friction} = \mu F_{\rm normal}[/tex], we get from the second equation
[tex]mg\sin(\theta) - \mu mg\cos(\theta) = ma[/tex]
and with [tex]\mu = 0.25[/tex] and [tex]\theta=60^\circ[/tex], we get
[tex]a = g\sin(60^\circ) - 0.25g \cos(60^\circ) = \left(\dfrac{\sqrt3}2 - \dfrac18\right) g[/tex]
Let [tex]x[/tex] be the length of the ramp, i.e. the distance that the box covers as it slides down it. Then the box attains a final velocity [tex]v[/tex] such that
[tex]v^2 = 2ax[/tex]
From the diagram, we see that
[tex]\sin(\theta) = \dfrac hx \implies x = \dfrac h{\sin(60^\circ)} = \dfrac{2h}{\sqrt3}[/tex]
and so
[tex]v^2 = \dfrac{4ah}{\sqrt3} = \left(2 - \dfrac1{2\sqrt3}\right) hg = \dfrac14 \left(8 - \dfrac2{\sqrt3}\right) hg[/tex]
[tex]\implies v = \dfrac12 \sqrt{\left(8 - \dfrac2{\sqrt3}\right) hg}[/tex]
Which job falls within the field of organizational psychology?
Job Recruiter
Employee Conflict Mediators
Psychiatrist
Receptionist
Answers
To win the game, a place kicker must kick a
football from a point 28 m (30.6208 yd) from
the goal, and the ball must clear the crossbar,
which is 3.05 m high. When kicked, the ball
leaves the ground with a speed of 18 m/s at
an angle of 47.3
◦
from the horizontal.
The acceleration of gravity is 9.8 m/s
2
.
By how much vertical distance does the ball
clear the crossbar?
Answers
The answer to the question is that the vertical distance the ball clear the cross bar is 1.465 metres.
We have horizontal distance as 28 m, and velocity as 18 m/s and the angle of projection 46.8 degree.
so, we can say that
In horizontal direction,
28 = (18 cos 46°) t
t = 2.2723 seconds
Now, in vertical direction,
Let be the height of the ball
h = (18 sin 46°)t - (1/2)gt²
h = (18 sin 46°)t - 4.9t²
h = ( (13.1214) × (2.2723) ) - ( (4.9) × (2.2723)² )
h = 4.5154 metre
Vertical distance the ball clear the cross bar = h - 3.05
Vertical distance the ball clear the cross bar = 4.5154 - 3.05
Vertical distance the ball clear the cross bar = 1.465
Thus, we can conclude that after solving and applying the concepts of projectile motion we find out the vertical distance the ball will clear the cross bar is 1.465 metres.
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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0 m. (It is not a very smart idea to play soccer on the roof of tall buildings.)
The ball is kicked with a speed of v0 = 15.10 m/s at an angle of θ = 74.1° with respect to the horizontal. The mass of a size 5 soccer ball is m = 450 g. When the ball lands on the other building, its speed is 19.89 m/s.
How much energy was lost to air friction? The ball is kicked without a spin.
37.71 J is incorrect.
Answers
The energy was lost to air friction is determined as 42.56 J.
Energy lost due to friction
The energy lost due to friction is calculated as follows;
ΔE = ΔK.E + ΔP.E
where;
ΔK.E is change in kinetic energyΔP.E is change in potential energyΔE = ¹/₂(0.45)(19.89² - 15.1²) + (0.45)(9.8)(12 - 30.2)
ΔE = 37.7 - 80.26
ΔE = -42.56 J
Thus, the energy was lost to air friction is determined as 42.56 J.
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A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre the beam may carry, if the bending stress is not to exceed 120N/mm2.
Answers
The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.
What is bending stress?When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.
The given data in the problem is;
Bending stress, σ = 120 N/mm2
Moment of inertia, I = 8.5 × 106 mm⁴
Depth of beam, y = d/2 = 200/2 = 100 mm
Length of beam, L = 8 m = 8000 mm
Width of beam, W = 300 mm
The maximum bending moment of the beam with UDL;
[tex]\rm W = \frac{wL^2}{8}[/tex]
From the bending equation;
[tex]\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm[/tex]
The maximum bending moment of the beam with UDL;
[tex]\rm M = \frac{wL^2}{8} \\\\ 10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \ N/mm[/tex]
Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.
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The speed of a bus is reduced uniformly from 20 m/s to 10 m/s while traveling 60 m. (a)
Compute the acceleration. (b) How much farther will the bus travel before coming to rest,
provided the acceleration remains constant? (c) Draw a diagram showing the motion from start to
finish of the bus.
Answers
The bus will travel a further 20 m before coming to rest.
What is acceleration?The term acceleration has to do with a change in velocity with time. Now we have;
u = 20 m/s
v = 10 m/s
s = 60 m
Now;
v^2 = u^2 -2as
v^2 - u^2 = -2as
(10)^2 - (20)^2 = - 2 * a * 60
a = (10)^2 - (20)^2/ - 2 * 60
a = 100 - 400/ - 2 * 60
a = 2.5 m/s^2
At that time;
v = 0 m/s
u = 20 m/s
a = 2.5 m/s^2
s = ?
Hence;
u^2 = 2as
s = u^2/2a
s = (20)^2/ 2 * 2.5
s = 400/5
s = 80 m
Hence, the bus will travel a further 20 m before coming to rest.
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A ball is thrown directly downward with an initial speed of 7.30 m/s, from a height of 29.0 m. After what time interval does it strike the ground?
Answers
Time taken by the ball to reach the ground is 1.8s.
What is free falling?When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.
A ball is thrown directly downward with an initial speed of 7.30 m/s, from a height of 29.0 m
Using the second equation of motion,
s = ut + 1/2 at²
Plug the values, we get
29 = 7.3t + 1/2 x9.81t²
4.905t² + 7.3t -29 =0
t =1.79871 or −3.28699
As time can't be negative, time taken to strike the ground is approximately 1.8 s.
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Due to continental drift, Africa and South America are moving away from each other at a rate of 4 centimeters per year. The two coasts are currently separated by 5,000 km. Assuming this drift rate is constant, how long ago were the coasts touching. Answer in millions of years
Answers
The coasts of Africa and South America were touched 125 million years ago due to continental drift.
The time can be computed from the ratio of distance and speed.
Given:
Distance = 5, 000,000
Speed = 4 cm/year = 0.04 m/year
The time is computed as:
Time = Distance / Rate
Time = 5,000,000 meters / 0.04 meters per year
Time = 125,000,000 years
Hence, the coasts of Africa and South America were touched 125 million years ago due to continental drift.
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A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced. What is the upward force exerted on the board by the support?
Answers
The upward force exerted on the board by the support is mathematically given as
Fu= 764.8 N
What is the upward force exerted on the board by the support?Generally, the equation for is mathematically given as
Considering that the Net Force on the system is null
The weight of the children plus the weight of the board equals the upward force imposed on the support.
The upward force
Fu= 440 + 272 + 52.8 N
Fu= 764.8 N
In conclusion, he upward force exerted on the board by the support
Fu= 764.8 N
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A ball is thrown directly downward with an initial speed of 8.30 m/s, from a height of 29.2 m. After what time interval does it strike the ground?
Answers
The ball's height [tex]y[/tex] at time [tex]t[/tex] is given by
[tex]y = 29.2\,\mathrm m - \left(8.30\dfrac{\rm m}{\rm s}\right) t - \dfrac12 gt^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex].
Solve for [tex]t[/tex] when [tex]y=0[/tex]. Omitting the units, we have
[tex]29.2 - 8.30t - \dfrac g2 t^2 = 0[/tex]
I'll solve by completing the square.
[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t\right) = 0[/tex]
[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t + \dfrac{8.3^2}{g^2}\right) = -\dfracg2 \times \dfrac{8.3^2}{g^2}[/tex]
[tex]29.2 - \dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = -\dfrac{8.3^2}{2g}[/tex]
[tex]\dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = 29.2 + \dfrac{8.3^2}{2g}[/tex]
[tex]\left(t + \dfrac{8.3}g\right)^2 = \dfrac{58.4}g + \dfrac{8.3^2}{g^2}[/tex]
[tex]t + \dfrac{8.3}g = \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]
[tex]t = -\dfrac{8.3}g \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]
[tex]\implies t \approx -3.43 \text{ or } t \approx 1.74[/tex]
Ignore the negative solution; the ball hits the ground about 1.74 s after being thrown.