Final answer:
The problem involves a ballistic pendulum scenario to determine the initial speed of a bullet, requiring an analysis that combines conservation of momentum and energy. However, the twist in this problem is that the bullet exits the block, requiring indirect methods to deduce the initial conditions.
Explanation:
The question involves finding the initial speed of a bullet by analyzing a ballistic pendulum scenario. This is a physics problem that integrates concepts of momentum and energy conservation. The solution requires two steps: first, finding the velocity of the bullet-block system right after the bullet exits, and second, using energy conservation to link this velocity to the height reached by the block.
Use the conservation of momentum to find the system's velocity right after the bullet exits the block. The exiting speed of the bullet and the mass details are essential for this step. However, an important observation is that the information given does not allow for the direct application of momentum conservation in the traditional sense, since the bullet exits the block, unlike the typical scenario where it remains lodged in.
Next, use the conservation of energy principle to relate the kinetic energy of the block right after the bullet exits to the potential energy at the maximum height. The height given allows calculation of the velocity of the block right after the bullet exits, which indirectly informs the system's dynamics at impact.
This problem has a twist as the bullet does not remain in the block, which complicates the direct application of the conservation of momentum to find the initial speed of the bullet. Instead, it involves a more nuanced approach using the given exit velocity of the bullet and the rise of the block to infer the initial conditions.
The initial speed of the bullet is approximately 23.68 m/s.
To solve this problem, we can use the principle of conservation of momentum and conservation of energy.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
[tex]\[m_b v_b = (m_b + m_p)v'\][/tex]
Where:
[tex]\(m_b = 7.0 \, \text{g} = 0.007 \, \text{kg}\)[/tex] (mass of the bullet)
\(v_b\) = initial speed of the bullet (which we need to find)
[tex]\(m_p = 1.5 \, \text{kg}\)[/tex] (mass of the ballistic pendulum)
\(v'\) = final velocity of the bullet and pendulum system
2. Conservation of energy:
The initial kinetic energy of the bullet is equal to the sum of the final kinetic energy of the bullet and the kinetic energy of the pendulum, plus the gravitational potential energy gained by the pendulum.
[tex]\[ \frac{1}{2} m_b v_b^2 = \frac{1}{2} (m_b + m_p) v'^2 + m_p g h\][/tex]
Where:
[tex]\(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)[/tex]
[tex]\(h = 12 \, \text{cm} = 0.12 \, \text{m}\) (maximum height reached by the pendulum)[/tex]
Let's solve these equations step by step:
Step 1: Conservation of momentum:
[tex]\[0.007 \, \text{kg} \times v_b = (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'\][/tex]
[tex]\[0.007 \, \text{kg} \times v_b = 1.507 \, \text{kg} \times v'\][/tex]
[tex]\[v_b = \frac{1.507}{0.007} \times v' \][/tex]
[tex]\[v_b = 215.28 \times v'\][/tex] (Equation 1)
Step 2: Conservation of energy:
[tex]\[ \frac{1}{2} \times 0.007 \, \text{kg} \times v_b^2 = \frac{1}{2} \times (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'^2 + 1.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.12 \, \text{m}\][/tex]
[tex]\[0.5 \times 0.007 \, \text{kg} \times v_b^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]
Now, substitute [tex]\(v_b = 215.28 \times v'\)[/tex] from Equation 1 into the above equation:
[tex]\[0.5 \times 0.007 \, \text{kg} \times (215.28 \times v')^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]
Simplify and solve for \(v'\):
[tex]\[0.5 \times 0.007 \times (215.28)^2 \times v'^2 = 0.5 \times 1.507 \times v'^2 + 1.47\][/tex]
[tex]\[ 160.49 \times v'^2 = 0.7535 \times v'^2 + 1.47\][/tex]
[tex]\[159.7365 \times v'^2 = 1.47\][/tex]
[tex]\[v'^2 = \frac{1.47}{159.7365}\][/tex]
[tex]\[v' = \sqrt{\frac{1.47}{159.7365}}\][/tex]
[tex]\[v' \approx 0.11 \, \text{m/s}\][/tex]
Finally, we can use this value of \(v'\) to find \(v_b\) using Equation 1:
[tex]\[v_b = 215.28 \times 0.11\][/tex]
[tex]\[v_b \approx 23.68 \, \text{m/s}\][/tex]
So, the initial speed of the bullet is approximately [tex]\(23.68 \, \text{m/s}\).[/tex]
Dry air will break down if the electric field exceeds 3.0*10^6 V/m. What amount of charge can be placed on a parallel-plate capacitor if the area of each plate is 73 cm^2?
Answer:
The charge on each plate of the capacitor is [tex]19.38 \mu C[/tex]
with one plate positive and one negative, i.e., [tex]\pm 19.38 \mu C[/tex]
Solution:
According to the question:
Critical value of Electric field, [tex]E_{c} = 3.0\times 10^{6} V/m[/tex]
Area of each plate of capacitor, [tex]A_{p} =73 cm^{2} = 73\times 10^{- 4} m^{2}[/tex]
Now, the amount of charge on the capacitor's plates can be calculated as:
Capacitance, C = [tex]\frac{epsilon_{o}\times Area}{Distance, D}[/tex] (1)
Also, Capacitance, C = [tex]\frac{charge, q}{Voltage, V}[/tex]
And
Electric field, E = [tex]\frac{Voltage, V}{D}[/tex]
So, from the above relations, we can write the eqn for charge, q as:
q = [tex]\epsilon_{o}\times E_{c}\times A_{p}[/tex]
q = [tex]8.85\times 10^{- 12}\times 3.0\times 10^{6}\times 73\times 10^{- 4}[/tex]
[tex]q = 19.38 \mu C[/tex]
A Horizontal rifel is fired at a bull's-eye. The muzzle
speedof the bullet is 670 m/s. The barrel is pointed directly at
thecenter of the bull's-eye, but the bullet strikes the
target0.025m below the center. What is the horizontal distance
betweenthe end of the rifel and the bull's eye?
Answer:
The horizontal distance is 478.38 m
Solution:
As per the question:
Initial Speed of the bullet in horizontal direction, [tex]v_{x} = 670 m/s[/tex]
Initial vertical velocity of the bullet, [tex]v_{y} = 0 m/s[/tex]
Vertical distance, y = 0.025 m
Now, for the horizontal distance, 'x':
We first calculate time, t:
[tex]y = v_{y}t - \frac{1}{2}gt^{2}[/tex]
(since, motion is vertically downwards under the action of 'g')
[tex]0.025 = 0 - \frac{1}{2}\times 9.8t^{2}[/tex]
[tex]t = \sqrt{0.05}{9.8} = 0.0714 s[/tex]
Now, the horizontal distance, x:
[tex]x = v_{x}t + \frac{1}{2}a_{x}t^{2}[/tex]
[tex]x = v_{x}t + \frac{1}{2}0.t^{2}[/tex]
(since, the horizontal acceleration will always be 0)
[tex]x = 670\times 0.714 = 478.38 m[/tex]
the driver of a car slams on her brakes to avoid collidingwith
a deer crossing the highway. what happens to the car's
kineticenergy as it comes to rest?
Answer:
Its dissipated by the brake.
Explanation:
Traditional brakes use friction to stop the wheels (or axis). This friction its dissipated in the form of heat. There are others mechanism to brake that don't dissipated the energy, they stored it. In electric cars (or hybrids), there are regenerative brakes, that store the kinetic energy as electrical energy.
As the driver slams the brakes, the kinetic energy of the car gets converted into other forms of energy until the car comes to a stop, at which point it becomes zero.
Explanation:When the driver of a car slams on her brakes to avoid a collision with a deer crossing the highway, the kinetic energy of the car decreases until the car comes to rest. This is due to the principle of energy conservation. Initially, when the car is moving, it has kinetic energy. However, when the brakes are applied, the kinetic energy gets converted into other forms of energy such as heat energy (due to friction between the brakes and the wheels) and potential energy (if the car is moving uphill). The kinetic energy keeps decreasing until the car comes to a complete stop. At that point, the kinetic energy of the car is zero because kinetic energy is associated with motion and the car is no longer moving.
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Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carrying a total charge of 4.95 μC .
Final answer:
To estimate the electric field near a uniformly charged wire, one must determine the linear charge density and integrate the contributions from infinitesimal charge elements along the wire, considering the symmetry that leads to the cancellation of horizontal components.
Explanation:
To estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire, we use the principle of superposition and integrate the contributions of the infinitesimally small charge elements along the wire.
First, we determine the linear charge density λ by dividing the total charge Q by the length L of the wire:
λ = Q / L
Then, we apply the formula for the electric field due to an infinitesimal charge element dQ at a distance r:
dE = (k * dQ) / r²
Where k is the Coulomb's constant (approximately 8.99 × 10⁹ Nm²/C²). Since we need to integrate over the length of the wire to find the total electric field, we use the linear charge density λ and a differential element of the wire's length dl:
dQ = λ * dl
The resulting integral, when evaluated, will give us the magnitude of the electric field E at the specified distance from the wire.
It is important to note that due to the symmetry of the problem, the horizontal components of the electric field due to charge elements on the wire will cancel out, leaving only the vertical components to contribute to the total electric field at the point of interest
Vesna Vulovic survived the longest fall on record without a parachute when her plane exploded and she fell 5 miles, 733 yards. What is this distance in meters?
Answer:
8717 meters.
Explanation:
We need to know the conversion factors. We know that:
1 mile = 1609.34 meters
1 yard = 0.9144 meters
This means that:
[tex]\frac{1609.34 meters}{1 mile}=1[/tex]
[tex]\frac{0.9144 meters}{1 yard}=1[/tex]
It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.
We will convert first the miles, then the yards, and add them up.
[tex]5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters[/tex]
[tex]733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters[/tex]
So total distance is the sum of these, 8717 meters.
An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway was used?
Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet
To find the runway distance used by an aircraft accelerating to 140 knots in 28 seconds, we convert knots to feet per second, find the acceleration, and then apply the kinematic formula for distance to obtain approximately 3308 feet.
Explanation:To calculate how many feet of runway an aircraft used to accelerate to 140 knots in 28 seconds, we need to convert the speed to consistent units and use the kinematic equations for uniformly accelerated motion. Knots need to be converted to feet per second (since the answer is required in feet). Since 1 knot = 1.68781 feet per second, 140 knots is equivalent to 236.293 feet per second. Now, we can use the formula for distance d when given initial velocity vi, final velocity vf, and acceleration a:
d = (vi + vf) / 2 * t
Here, as the plane is starting from a standstill, vi is 0, vf is 236.293 feet/second, and t is 28 seconds. We first need to find the acceleration:
a = (vf - vi) / t = 236.293 / 28 = 8.439 feet/second2
Now we can calculate the distance:
d = (0 + 236.293) / 2 * 28 = 3308.1 feet
The aircraft used approximately 3308 feet of runway.
A box moving on a horizontal surface with an initial velocity of 20 m/s slows to a stop over a time period of 5.0 seconds due solely to the effects of friction. What is the coefficient of kinetic friction between the box and the ground?
Answer:
μ= 0.408 : coefficient of kinetic friction
Explanation:
Kinematic equation for the box:
[tex]a=\frac{v_{f} -v_{i} }{t}[/tex] Formula( 1)
a= acceleration
v_i= initial speed =0
v_f= final speed= 20 m/s
t= time= 5 s
We replace data in the formula (1):
[tex]a=\frac{0-20}{5}[/tex]
[tex]a= -\frac{20}{5}[/tex]
a= - 4m/s²
Box kinetics: We apply Newton's laws in x-y:
∑Fx=ma : second law of Newton
-Ff= ma Equation (1)
Ff is the friction force
Ff=μ*N Equation (2)
μ is the coefficient of kinetic friction
N is the normal force
Normal force calculation
∑Fy=0 : Newton's first law
N-W=0 W is the weight of the box
N=W= m*g : m is the mass of the box and g is the acceleration due to gravity
N=9.8*m
We replace N=9.8m in the equation (2)
Ff=μ*9.8*m
Coefficient of kinetic friction ( μ) calculation
We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):
-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation
9.8*μ=4
μ=4 ÷ 9.8
μ= 0.408
If a beam passes from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 at an angle of 25 degrees (from normal), what is the angle of refraction of the beam? Is the beam bent towards normal or away from it? Sketch a diagram of this problem with rays, angles, and the interface labeled.
Answer:
18.86° , it will bend towards normal.
Explanation:
For refraction,
Using Snell's law as:
[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]
Where,
[tex]{\theta_i}[/tex] is the angle of incidence ( 25.0° )
[tex]{\theta_r}[/tex] is the angle of refraction ( ? )
[tex]{n_r}[/tex] is the refractive index of the refraction medium (n=1.7)
[tex]{n_i}[/tex] is the refractive index of the incidence medium ( n=1.3)
Hence,
[tex]1.3\times {sin\ 25.0^0}={1.7}\times{sin\theta_r}[/tex]
Angle of refraction = [tex]sin^{-1}0.3232[/tex] = 18.86°
Since, the light ray is travelling from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 or lighter to denser medium, it will bend towards normal.
The diagram is shown below:
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:
[tex]|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82[/tex]
If you are playing soccer and you always hit the left goal post instead of scoring, are you: a. accurate? b. precise? c. both? or d. neither?
Answer:
b. precise
Explanation:
Accuracy means how close a measurement is to the true value (in our example the true value is score a goal )
Precision refers how close two or more measurements to each other.
So if the playing soccer hit always the left goal post, he is not accurate (e doesn't score any goals ) but he is very precise
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height (h) from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 m/s when they reach the bottom of the ramp. You determine that for a 85.0 kg skier with good form, friction and air resistance will do total work of magnitude 4000 J on him during his run down the slope.What is the maximum height (h) for which the maximum safe speed will not be exceeded?
Answer:
h = 50.49 m
Explanation:
Data provided:
Speed of skier, u = 2.0 m/s
Maximum safe speed of the skier, v = 30.0 m/s
Mass of the skier, m = 85.0
Total work = 4000 J
Height from the starting gate = h
Now, from the law of conservation of energy
Total energy at the gate = total energy at the time maximum speed is reached
[tex]\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2[/tex]
where, g is the acceleration due to the gravity
on substituting the values, we get
[tex]\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2[/tex]
or
170 + 833.85 × h = 4000 + 38250
or
h = 50.49 m
A steel rope has a 25 mm diameter and density of 5000 kg/m^3. The largest tension it can withstand is 320,000 N. What is the longest length of of this rope in m that can hang without breaking?
Answer:
longest length is 13304.05 m
Explanation:
given data
diameter = 25 mm
so radius r = 12.5 × [tex]10^{-3}[/tex] m
density = 5000 kg/m³
tension = 320000 N
to find out
longest length of of this rope
solution
we know that here tension is express as
T = m × g
and m is mass and g is acceleration due to gravity
here mass = density × volume
and volume = π×r²×l
here r is radius and l is length
so T = density c volume × g
put here value
320000 = 5000 × π×12.5² × [tex]10^{-3}[/tex] × l × 9.8
solve it we get length l
l = 13304.05
so longest length is 13304.05 m
The output of an ac generator connected to an RLC series combination has a frequency of 12 kHz and an amplitude of 28 V. If R = 4.0 Ohms, L = 30 μH, and C = 8 μF, find a. The impedance
b. The amplitude for current
c. The phase difference between the current and the emf of the generator
Please show all steps and units. Thank you.
Answer:
(a) 4.04 ohm
(b) 6.93 A
(c) 8.53°
Explanation:
f = 12 kHz = 12000 Hz
Vo = 28 V
R = 4 ohm
L = 30 micro Henry = 30 x 10^-6 H
C = 8 micro Farad = 8 x 10^-6 F
(a) Let Z be the impedance
[tex]X_{L} = 2\pi fL=2\times3.14\times12000\times30\times10^{-6}= 2.26 ohm[/tex]
[tex]X_{c} = \frac{1}{2\pi fC}=\frac{1}{2\times3.14\times12000\times8\times10^{-6}}= 1.66 ohm[/tex]
[tex]Z = \sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{4^{2}+\left ( 2.26-1.66 \right )^{2}}[/tex]
Z = 4.04 Ohm
(b) Let Io be the amplitude of current
[tex]I_{o}=\frac{V_{o}}{Z}[/tex]
[tex]I_{o}=\frac{28}{4.04}[/tex]
Io = 6.93 A
(c) Let the phase difference is Ф
[tex]tan\phi = \frac{X_{L}-X_{C}}{R}[/tex]
[tex]tan\phi = \frac{2.26-1.66}{4}[/tex]
tan Ф =0.15
Ф = 8.53°
Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle of the hill?What was the vertical component of Jill's velocity?
Answer:[tex]\theta =41.409 ^{\circ}[/tex]
Explanation:
Given
Jack and Jill ran up the hill at 2.8 m/s
Horizontal component of Jill's velocity vector was 2.1 m/s
Let [tex]\theta [/tex]is the angle made by Jill's velocity with it's horizontal component
Therefore
[tex]2.8cos\theta =2.1[/tex]
[tex]cos\theta =\frac{2.1}{2.8}[/tex]
[tex]cos\theta =0.75[/tex]
[tex]\theta =41.409 ^{\circ}[/tex]
Vertical velocity is given by
[tex]V_y=2.8sin41.11=1.85 m/s[/tex]
The question can be solved using the Pythagorean theorem to calculate the vertical velocity component and apply trigonometry to find the angle of the hill. Remember the conversion from radians to degrees.
Explanation:The question asks about the angle of the hill that Jack and Jill climbed, as well as the vertical component of Jill’s velocity. We know that Jill ran up the hill at a velocity of 2.8 m/s and the horizontal component of her velocity was 2.1 m/s. These two components form a right angle triangle where the hypotenuse is the total velocity (2.8 m/s), one side is the horizontal velocity (2.1 m/s), and the other side, which we're finding, is the vertical component of the velocity.
We can find the angle of the hill, θ, using the tangent function of trigonometry. Tan θ = opposite side / adjacent side. In this case, the 'opposite side' is the vertical velocity component we're after, and the 'adjacent side' is the horizontal component (2.1 m/s). To find the vertical velocity component, you can use the Pythagorean theorem, which states that: (Hypotenuse)² = (Adjacent Side)² + (Opposite Side)² or (2.8 m/s)² = (2.1 m/s)² + (vertical velocity)².
Once you solve for the vertical velocity through the Pythagorean theorem, you then insert it into the tangent equation to solve for the angle θ. Remember that when you use the arctangent function on a calculator to find the angle, the answer will likely be in radians, so you might need to convert them into degrees.
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Tarzan, in one tree, sights Jane in another tree. He grabs
theend of a vine with length 20 m that makes an angle of 45
degreeswith the vertical, steps off his tree limb, and swings down
andthen up to Jane's open arms. When he arrives, his vine makesan
angle of 30 degrees with the vertical. determine whetherhe gives
her a tender embrace or knocks her off her limb bycalculating
Tarzan's speed just before he reaches Jane. Youcan ignore air
resistace and the mass of the vine.
Answer:
He knocks her. V = 7.97m/s
Explanation:
Let A be Tarzan's starting position and B Jane's position (Tarzan's final position).
Since there is no air resistance, energy at position A must be equal to energy at B.
At position A, Tarzan's speed is zero and since the 45° of the vine is greater than the final 30°, Tarzan will have potential energy at point A.
[tex]E_{A}=m_{T}*g*L*(cos(30)-cos(45))[/tex]
At point B, it is the lowest point (between A and B), so it has no potential energy. It will only have kinetic energy (ideally zero, for Jane's sake, but we don't know).
[tex]E_{B}=\frac{m_{T}*V^{2}}{2}[/tex]
Because of energy conservation, we know that Ea=Eb, so:
[tex]m_{T}*g*L*(cos(30)-cos(45))=\frac{m_{T}*V^{2}}{2}[/tex] Solving for V:
[tex]V=\sqrt{2*g*L*(cos(30)-cos(45))}=7.97m/s[/tex]
Which of the following statements about electric field lines are true? (choose all that are true) a) They are only defined for positive charges.
b) They are always tangent to electric field vectors.
c) They are always perpendicular to charged surfaces.
d) They are a simple way to visualize the electric field vectors. e) None of the above.
Answer: b) TRUE and d) TRUE
Explanation: a) FALSE the electric field lines are used to represent the charges postives and negatives.
b) TRUE it is the definition of electric field lines , they are tangent to the electric field vector.
c) FALSE the electric field lines only for conductor are perpendicular to the surface in other any situation there is a tangencial electric field components.
d) TRUE since it is a way to describe and imagine the effect of vectorial fields.
e) FALSE
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of O 100 m/s. 60 m/s. 50 m/s 80 m/s. 110 m/s.
Answer:
The initial velocity of the ball should be 50 m/s.
Explanation:
Since the trip of the ball shall consist of upward ascend and the downward descend and since the ascend and the descend of the ball is symmetrical we infer that the upward ascend of the ball shall last for a time of 5 seconds.
Now since the motion of ball is uniformly accelerated we can find the initial speed of the ball using first equation of kinematics as
[tex]v=u+gt[/tex]
where,
'v' is final velocity of the ball
'u' is initial velocity of the ball
'g' is acceleration due to gravity
't' is the time of motion
Now we know that the ball will continue to ascend until it's velocity becomes zero hence to using the above equation we can write
[tex]0=u-9.81\times 5\\\\\therefore u=9.81\times 5=49.05m/s\approx 50m/s[/tex]
A barge is carrying a load of gravel along a river.
Itapproaches a low bridge, and the captain realizes that the top
ofthe pile of gravel is not going to make it under the bridge.
Thecaptain orders the crew to quickly shovel gravel from the pile
intothe water. Is this a good decision? Explain.
Answer:
Explanation:
Shoveling gravel into the water will increase the buoyancy of the barge, which will make it float higher. Without data it is hard to tell if it will raise the barge enough to be counterproductive, but in any case throwing away the payload is not a good idea. Adding some weights would make the barge float lower, and then maybe the plie can make it under the bridge.
Newton's second law: Rudolph the red nosed reindeer is pulling a 25 kg sled across the snow in a field. The coefficient of kinetic friction is .12 The rope that is pulling the sled is coming off at a 29 degree angle above the horizontal. Find the force in the rope when the acceleration is .12 m/s^2.
Answer:
[tex]F=39,68N[/tex]
Explanation:
Data:
Mass [tex]m=25 Kg[/tex]
Coefficient of kinetic friction [tex]\mu=0.12[/tex]
Angle = [tex]29^{0}[/tex]
Acceleration = [tex]0.12 \frac{m}{s^{2} }[/tex]
Solution:
By Newton's first law we know that for the x-axis:
[tex]F_{rope_x}-F_f=F_R[/tex] Where [tex]F_R[/tex] is the resulting force, and [tex]F_f[/tex] is the friction force.
And for the y-axis:
[tex]F_{rope_y}+N=W[/tex], where N is the normal force, and W is the weight of the sled.
We know that the resulting force's acceleration is [tex]0.12 \frac{m}{s^{2} }[/tex], and by using Newton's second law, we obtain:
[tex]F=m.a[/tex]
[tex]F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N[/tex] .
Now, the horizontal component of the force in the rope will be given by
[tex]F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f[/tex], since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.
To obtain the friction force, we must know the normal force:
[tex]F_f=\mu. N[/tex]
Clearing N in the y-axis equation:
[tex]N=W-F_{rope_y}=W-F_{rope}.sin(29^0)[/tex]
So we can express the x-axis equation as follows:
[tex]F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))[/tex]
Finally, solving for F we get
[tex]F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))[/tex]
[tex]F=39,68N[/tex]
You are on a train traveling east at speed of 28 m/s with respect to the ground. 1) If you walk east toward the front of the train, with a speed of 1.5 m/s with respect to the train, what is your velocity with respect to the ground? (m/s east)
2) If you walk west toward the back of the train, with a speed of 2.1 m/s with respect to the train, what is your velocity with respect to the ground? (m/s, east)
3) Your friend is sitting on another train traveling west at 22 m/s. As you walk toward the back of your train at 2.1 m/s, what is your velocity with respect to your friend? (m/s, east)
Answer:
A) 29.5m/s
B) 25.9m/s
C) 47.9 m/s
Explanation:
This is a relative velocity problem, which means that the velocity perception will vary from each observer at a different reference point.
We can say that the velocity of the train respect the ground is 28m/s on east direction If I walk 1.5m/s respect the train, the velocity of the person respect to the ground is the sum of both velocities:
Vgp=Vgt+Vtp
where:
Vgp=velocity of the person respect the ground
Vgt=Velocity of the train respect the ground
Vtp=Velocity of the person respect the train
1) Vgp=28m/s+1.5m/s=29.5m/s
2) here the velocity of the person respect the train is negative because it is going backward
Vgp=28m/s-2.1m/s=25.9m/s
3) the velocity of the train respect with your friend will be the sum of both velocities(Vft).
Vfp=Vft+Vtp
Vfp=velocity of the person respect the friend
Vft=Velocity of the train respect the friend
Vtp=Velocity of the person respect the friend
Vfp=(22m/s+28m/s)+(-2.1m/s)=47.9 m/s
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the hand is 2.30 m above the ground. How long is the ball in the air before it hits the ground?
Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known, and identify its value. Then, identify the unknown and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long a time is the dolphin in the air? Neglect any effects resulting from his size or orientation.
(a) Initial velocity (v₀) = 0 m/s, Final velocity (v) = 15.0 m/s.
(b) The dolphin rises approximately 11.5 meters above the water.
(c) The dolphin is in the air for approximately 1.5 seconds.
(a)
Initial velocity (v₀) = 0 m/s (since the dolphin jumps straight up from the water, its initial velocity is zero)
Final velocity (v) = 15.0 m/s (given)
(b)
To solve for the height, we can use the kinematic equation that relates the final velocity (v), initial velocity (v₀), acceleration (a), and displacement (Δy):
v² = v₀² + 2aΔy
v² = v₀² + 2gh
v² = 2gh
Solving for h:
h = (v²) / (2g)
h = (15.0)² / (2 × 9.8)
h ≈ 11.5 m
Therefore, the dolphin rises approximately 11.5 meters above the water.
(c) To determine the time the dolphin is in the air, we can use the equation for time (t) derived from the kinematic equation:
Δy = v₀t + (1/2)at²
Δy = (1/2)at²
t = √((2Δy) / a)
t = √((2 × 11.5) / -9.8)
t ≈ 1.5 s
Therefore, the dolphin is in the air for approximately 1.5 seconds.
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The dolphin rises 8.64 meters above the water and is in the air for 1.33 seconds.
Explanation:(a) The knowns in this problem are:
Initial velocity of the dolphin: 13.0 m/sAcceleration due to gravity: -9.8 m/s²(b) To find the height above the water, we need to find the time taken by the dolphin to reach that height and use the equation:
Final velocity = Initial velocity + Acceleration × Time
Solving for time, we have:
Time = (Final velocity - Initial velocity) / Acceleration
Substituting the known values into the equation, we get:
Time = (0 m/s - 13.0 m/s) / -9.8 m/s² = 1.33 s
Next, we can use the formula for height:
Height = Initial velocity × Time + 0.5 × Acceleration × Time²
Substituting the known values, we have:
Height = 13.0 m/s × 1.33 s + 0.5 × -9.8 m/s² × (1.33 s)² = 8.64 m
(c) The time the dolphin is in the air is equal to the time calculated above, which is 1.33 seconds.
Draw, as best you can, what a velocity graph would look like as a function of time if the acceleration of the object is negative.
Answer:
Depends on the specific relation of acceleration as function of time, but it would always look like decresing or increasing negatively.
Explanation:
Acceleration is the derivative of velocity with respect to time. That means that it is the change. The fact that is negative means that it is decreasing, or increasing in the negative direction (i.e. going backwards faster).
Attached is the graph with constant negative acceleration, using kinematics relations, it would be a linear equation with negative slope.
Answer:a straight line with negative slope
Explanation:
Hydroelectric dams use ------- to produce electricity. gravitational potential energy of falling water
nuclear energy
geothermal energy
fossil fuel energy
solar energy
Answer:
Gravitational potential energy
Explanation:
Hydroelectric dams are the power plants which generates electricity by using the energy of falling water from a great height.
A the water is stored in big reservoirs and at a great height, it contain lot of potential energy due to the height. As it falls downwards, the potential energy is converted into kinetic energy of the water. this kinetic energy of the falling water is used to run the turbine, and then the electric energy is generated.
So, in hydroelectric power stations, the potential energy of water is converted into the electric energy.
A hot air ballo0n is ascending straight up at a constant
speedof 7.0 m/s. When the balloon is 12.0 m above the ground, agun
fires a pellet straight up from ground level with an initialspped
of 30.0 m/s. Along the paths of the ballon and thepellet, there are
two places where each of them has the altitude atthe same time. How
far above ground level are theseplaces?
Answer: The two places altitudes are: 16.17 m and 40.67 m
Explanation:
Hi!
Lets call z to the vertical direction (z= is ground) . Then the positions of the balloon and the pellet, using the values of the velocities we are given, are:
[tex]z_b =\text{balloon position}\\z_p=\text{pellet position}\\z_b=(7\frac{m}{s})t\\z_p=30\frac{m}{s}(t-t_0)-\frac{g}{2}(t-t_0)^2\\g=9.8\frac{m}{s^2}[/tex]
How do we know the value of t₀? This is the time when the pellet is fired. At this time the pellet position is zero: its initial position. To calculate it we know that the pellet is fired when the ballon is in z = 12m. Then:
[tex]t_0=\frac{12}{7}s[/tex]
We need to know the when the z values of balloon and pellet is the same:
[tex]z_b=z_p\\(7\frac{m}{s})t =30\frac{m}{s}(t-\frac{12}{7}s)-\frac{g}{2}(t-\frac{12}{7}s)^2[/tex]
We need to find the roots of the quadratic equation. They are:
[tex]t_1=2.31s\\t_2=5.81[/tex]
To know the altitude where the to objects meet, we replace the time values:
[tex]z_1=16,17m\\z_2=40,67m[/tex]
A hot-air balloon is descending at a rate of 2.3 m/s when a pas- senger drops a camera. If the camera is 41 m above the ground when it is dropped, (a) how much time does it take for the cam- era to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.
Answer:
a) time taken = 2.66 s
b) v = 28.34
Explanation:
given,
rate of descending = 2.3 m/s
height of camera above ground = 41 m
using equation of motion
[tex]h = u t + \dfrac{1}{2}gt^2[/tex]
[tex]41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2[/tex]
4.9 t² + 2.3 t - 41 =0
t = 2.66 ,-3.13
time taken = 2.66 s
b) v² = u² + 2 g h
v² = 0 + 2× 9.8 × 41
v = 28.34
A charge of 7.0 μC is to be split into two parts that are then separated by 9.0 mm. What is the maximum possible magnitude of the electrostatic force between those two parts?
Answer:
[tex]F = 1361.1 N[/tex]
Explanation:
As we know that a charge is split into two parts
so we have
[tex]q_1 + q_2 = 7\mu C[/tex]
now we have
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know
[tex]F = \frac{kq_1(7\mu C - q_1)}{r^2}[/tex]
here we have
r = 9 mm
now to obtain the maximum value of the force between two charges
[tex]\frac{dF}{dq_1} = 0[/tex]
so we have
[tex]7 \mu C - q_1 - q_1 = 0[/tex]
so we have
[tex]q_1 = q_2 = 3.5 \mu C[/tex]
so the maximum force is given as
[tex]F = \frac{(9\times 10^9)(3.5 \mu C)(3.5 \mu C)}{0.009^2}[/tex]
[tex]F = 1361.1 N[/tex]
An airplane flies 200 km due west from city A to city B and then 275 km in the direction of 26.0° north of west from city B to city C. (a) In straight-line distance, how far is city C from city A?
Relative to city A, in what direction is city C?
(c) Why is the answer only approximately correct?
The straight-line distance from city A to city C is approximately 458.80 km. The direction from city A to city C is 14.9° north of west. The approximation is due to ignoring Earth's curvature.
Explanation:To solve this problem, we will use the laws of vector addition as airplanes movements are vector quantities as it involves both magnitude (distance) and direction. Initially, the airplane flies 200 km west. Then, it changes its direction 26.0° north of west and flies an additional 275 km.
When we decompose the second segment of the flight into its westward and northward components, we can use trigonometric principles. The horizontal (westward) distance is 275 km * cos(26.0°) = 245.56 km. Adding that to the 200 km the airplane flew initially gives us a total westward distance of 445.56 km. The vertical (northward) distance is 275 km * sin(26.0°) = 118.40 km.
To find the straight-line distance between City A and City C, we apply the Pythagorean theorem: √[(445.56 km)² + (118.4 km)²] = 458.80 km. To find the direction, we take the arctan of the northward distance over the westward distance: arctan(118.4 km / 445.56 km) = 14.9° north of west.
It's only approximately correct because we ignore the curvature of the Earth, which would slightly modify the distances and angles involved.
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The straight-line distance from city A to city C is approximately 460 km and the direction is 15° north of west. The answer is only an approximation because we've considered a flat plane, ignoring the spherical nature of the Earth.
Explanation:The question is asking for the linear distance and direction from city A to city C, which can be solved through vector addition and trigonometric calculations in physics. The displacement from A to B is 200 km west. The displacement from B to C is 275 km, 26.0° north of west. These can be seen as components of a right triangle, and by applying the Pythagorean theorem, the straight-line distance, or the hypotenuse, can be calculated.
First, we derive the north and west components of the 275 km using cosine and sine respectively as the motion is angled. The westward motion is 275cos(26) = 244 km. Add this to the 200 km westward motion to get a total westward motion of 444 km. The northward motion is 275sin(26) = 121 km.
Then, we apply Pythagorean theorem. √(444^2 + 121^2) ≈ 460 km. This is the straight-line distance from city A to city C.
For the direction, we find the angle, using tan^(-1)(121/444) ≈ 15° north of west.
The answer is only approximately correct because in real situation we have to take into account the spherical nature of the Earth, but here we've considered a flat plane for simplicity.
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a. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kinetic energy of the helium atoms?
c. What is the rms speed of the helium atoms?
Answer:
a) 3.39 × 10²³ atoms
b) 6.04 × 10⁻²¹ J
c) 1349.35 m/s
Explanation:
Given:
Diameter of the balloon, d = 29.6 cm = 0.296 m
Temperature, T = 19.0° C = 19 + 273 = 292 K
Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa
Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]
or
Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{0.296}{2})^3[/tex]
or
Volume of the balloon, V = 0.0135 m³
Now,
From the relation,
PV = nRT
where,
n is the number of moles
R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol
on substituting the respective values, we get
1.013 × 10⁵ × 0.0135 = n × 8.314 × 292
or
n = 0.563
1 mol = 6.022 × 10²³ atoms
Thus,
0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms
b) Average kinetic energy = [tex]\frac{3}{2}\times K_BT[/tex]
where,
Boltzmann constant, [tex]K_B=1.3807\times10^{-23}J/K[/tex]
Average kinetic energy = [tex]\frac{3}{2}\times1.3807\times10^{-23}\times292[/tex]
or
Average kinetic energy = 6.04 × 10⁻²¹ J
c) rms speed = [tex]\frac{3RT}{m}[/tex]
where, m is the molar mass of the Helium = 0.004 Kg
or
rms speed = [tex]\frac{3\times8.314\times292}{0.004}[/tex]
or
rms speed = 1349.35 m/s
Which of the following would describe a length that is 2.0×10^-3 of a meter? a: 2.0 kilometers
b: 2.0 megameters
c: 2.0 millimeters
d: 2.0 micrometers
Answer:
Option (c) [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]
Explanation:
Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :
1 meter = 0.001 kilometers
1 meter = 10⁻⁶ megameters
1 meter = 1000 millimeters
1 meter = 1000000 micrometers
From the given option, the correct one is (c) because, 1 meter = 1000 millimeters
So, [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]
Hence, the correct option is (c). Hence, this is the required solution.
2.0×10^-3 of a meter is equivalent to 2.0 millimeters, hence option c is the correct answer.
Explanation:The given length, 2.0×10^-3 of a meter, corresponds to a unit of length commonly used in the metric system. In this system, 1 meter is equal to 10^3 millimeters. Therefore, 2.0×10^-3 of a meter equates to 2.0 millimeters. This means the best answer from the provided options would be option c: 2.0 millimeters.
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