Final answer:
The final momentum of the car pushed by two men with a net force of 690 N for 6.6 seconds is 4554 N·s. This is calculated using the formula for impulse, which is the product of the net force and the time over which the force is applied.
Explanation:
The question is asking to calculate the final momentum of a car after two men have pushed it with a net force of 690 N for 6.6 seconds. According to Newton's second law of motion, the change in momentum (also known as impulse) is equal to the net force multiplied by the time over which the force is applied. The formula for impulse is:
Impulse (J) = Force (F) × Time (t)
By applying this formula, we can find the final momentum:
Force (F) = 690 NTime (t) = 6.6 secondsTherefore, the impulse is:
J = 690 N × 6.6 s = 4554 N·s
Since the car was initially at rest and assuming there is no external resistance, the final momentum of the car will be equal to the impulse given to it:
Final momentum = 4554 N·s
The final momentum of the car, with an applied force of +690 N for 6.6 seconds, is 4554 Ns. The initial momentum is zero since the car starts from rest.
To determine the final momentum of the car, we will utilize the relationship between force, time, and momentum. According to the impulse-momentum theorem, the impulse applied to an object is equal to the change in its momentum.
Impulse is given by the formula:
Impulse (J) = Force (F) × Time (t)
Given:
Force (F) = +690 NTime (t) = 6.6 sNow, calculate the impulse:
J = 690 N × 6.6 s = 4554 Ns
In this scenario, the car starts from rest, which means its initial momentum (p_initial) is 0. Hence, the final momentum ([tex]p_{final}[/tex]) after 6.6 seconds is equal to the impulse:
[tex]p_{final}[/tex] = 4554 Ns
The final momentum of the car is 4554 Ns.
A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a constant rate of 1.0 m/s^2 until it stops. How much time elapses from start to stop? We assume an answer in seconds
Answer:
33 seconds.
Explanation:
The equation for speed with constant acceleration at time t its:
[tex]V(t) \ = \ V_0 \ + \ a \ t[/tex]
where [tex]V_0[/tex] is the initial speed, and a its the acceleration.
First half of the problemStarting at rest, the initial speed will be zero, so
[tex]V_0 = 0[/tex]
the final speed is
[tex]V(t_{f1}) = 23 \frac{m}{s}[/tex]
and the acceleration is
[tex]a = 2.3 \frac{m}{s^2}[/tex].
Taking all this together, we got
[tex]V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}[/tex]
[tex]23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}[/tex]
[tex]\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}[/tex]
[tex]10 s = t_{f1}[/tex]
So, for the first half of the problem we got a time of 10 seconds.
Second half of the problemNow, the initial speed will be
[tex]V_0 = 23 \frac{m}{s}[/tex],
the acceleration
[tex]a=-1.0 \frac{m}{s^2}[/tex],
with a minus sign cause its slowing down, the final speed will be
[tex]V(t_{f2}) = 0[/tex]
Taking all together:
[tex]V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}[/tex]
[tex] 23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}[/tex]
[tex] \frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}[/tex]
[tex] 23 s = t_{f2}[/tex]
So, for the first half of the problem we got a time of 23 seconds.
Total time[tex]t_total = t_{f1} + t_{f2} = 33 \ s[/tex]
A charge located at (3,-5) in the x-y plane exerts an attractive force of 0.75N on another charge located at (2,7) (all distances are in cm) a) In what direction does this force point? The answer should be given as a counterclockwise angle with respect to the positive x-axis. b) Write down an expression for this force in terms of unit vectors i and (don't forget your units). c) If the charge at (3,-5) is +2pic, determine the magnitude and sign of the other charge.
Answer:
(a): [tex]247.77^\circ.[/tex]
(b): [tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}.[/tex]
(c): Magnitude = [tex]0.604\ C.[/tex]
Sign = negative.
Explanation:
(a):
Given that first charge is located at (3,-5) and second charge is located at (2,7).
The electrostatic force between two charges acts long the line joining the two charges, therefore, the direction of electrostatic force of attraction between these two charges is along the position vector of charge at (2,7) with respect to the position of charge at (3,-5).
Assuming, [tex]\hat i,\ \hat j[/tex] are the unit vectors along positive x and y axes respectively.
Position vector of charge at (3,-5) with respect to origin is
[tex]\vec r_1 = 3\hat i+(-5)\hat j[/tex]
Position vector of charge at (2,7) with respect to origin is
[tex]\vec r_2 = 2\hat i+7\hat j[/tex]
The position vector of at (2,7) with respect to the position of charge at (3,-5) is
[tex]\vec r = \vec r_2-\vec r_1\\=(3\hat i-5\hat j)-(2\hat i+7\hat j)\\=1\hat i-12\hat j.[/tex]
If [tex]\theta[/tex] is the angle this position vector is making with the positive x axis then,
[tex]\rm \tan\theta =\dfrac{y\ component\ of\ (\hat i-12\hat j)}{x\ component\ of\ (\hat i-12\hat j}=\dfrac{-12}{1} = -12.\\\theta =\tan^{-1}(-12)=-85.23^\circ.[/tex]
The negative sign indicates that this position vector is [tex]85.23^\circ.[/tex] below the positive x axis, therefore, in counterclockwise direction from positive x axis, its direction = [tex]360^\circ-85.23^\circ = 247.77^\circ.[/tex]
The force is also along the same direction.
(b):
According to Coulomb's law, the expression of this force is given by
[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \hat r[/tex]
where,
[tex]q_1,\ q_2[/tex] are the charges.[tex]\vec r[/tex] = position vector of one charge with respect to another charge.[tex]\hat r[/tex] = unit vector along the direction of [tex]\vec r[/tex].[tex]k[/tex] = Coulomb's constant, which have value = [tex]\rm 9\times 10^9\ Nm^2/C^2[/tex].We have,
[tex]\vec r = \hat i-12\hat j\\\therefore |\vec r| = \sqrt{1^2+(-12)^2}=\sqrt{145}.\\\Rightarrow \hat r = \dfrac{\vec r}{|\vec r|} = \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]
Putting this value, we get,
[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]
(c):
If charge at (3,-5) is [tex]+2\ pC[/tex], then the magnitude of the force between the two charges is given by
[tex]F=k\ \dfrac{q_1q_2}{|\vec r|^2}[/tex]
where, we have,
[tex]F=0.75\ N\\|\vec r| = \sqrt{145}\ cm = \sqrt{145}\times 10^{-2}\ m.\\q_1 = +2\ pC = +2\times 10^{-12}\ C.[/tex]
Putting all these values,
[tex]0.75=9\times 10^9\times \dfrac{(+2\times 10^{-12})\times q_2}{(\sqrt{145}\times 10^{-2})^2}\\\Rightarrow q_2 = \dfrac{0.75\times(\sqrt{145}\times 10^{-2})^2}{9\times 10^9\times 2\times 10^{-12}} =0.604\ C.[/tex]
Since, the electric force between the two spheres is given to be attractive therefore this charge must be negative as the other charge is positive.
A player kicks a football (from the ground) at an initial angle of 30°. The football is in the air for 2.4 s before it hits an opposing player. The opposing player was a horizontal distance x = 50 m away from where the football was kicked. What was the initial horizontal velocity component of the football?
Answer:
The initial horizontal velocity was 21 m/s
Explanation:
Please, see the figure for a better understanding of the problem.
The equation for the position of an object moving in a parabolic trajectory is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g ·t²)
Where:
r = vector position at time t
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity
Notice that at time t = 2.4 s the vector "r" is the one in the figure. We know that the x-component of that vector is 50 m. Then using the equation for the x-component of the vector "r", we can calculate the initial velocity:
x = x0 + v0 · t · cos α
Let´s place the center of the frame of reference at the point of the kick so that x0 = 0.
x = v0 · t · cos α
x/t = v0 · cos α
Notice in the figure that v0 · cos 30° = v0x which is the initial horizontal velocity. Remember trigonometry of right triangles:
cos α = adjacent / hypotenuse = v0x / v0
Then:
50 m/ 2.4 s = v0 · cos 30° = v0x
v0x = 21 m/s
The population mean on a statistics exam is 72, with a standard deviation of 12. The population average on the class project is 95, with a standard deviation of 4. If the exam is 70% of the final grade and the project 30%, what are the mean and standard deviation of final grades?
Answer:
mean = 54
standard deviation = 6.24
Explanation:
given data
mean a = 72
standard deviation σx = 12
average b = 95
standard deviation σy = 4
final grade Z = 70% = 0.70
project y = 30% = 0.30
to find out
mean and standard deviation
solution
final grade equation will be here
final grade = 0.70x + 0.30 y .....a
here x is statics grade and y is class project
so
mean will be 0.70x + 0.30 y
mean = 0.70 (a) + 0.30 (b)
mean = 0.70 (12) + 0.30 (95)
mean = 54
and
standard deviation = 0.70x + 0.30 y
standard deviation = 0.70(σx) + 0.30 (σy)
standard deviation = 0.70(12) + 0.30 (4)
standard deviation = 6.24
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 force of magnitude 33.73 lb directed toward the right is exerted on an object. What other force must be applied to the object so that the resultant force is zero?
An additional force of 33.73 lb must be applied to the left to cancel out the initial force and create a resultant force of zero, according to Newton's first law of motion.
Explanation:To ensure that the resultant force on the object is zero, another force of equal magnitude but in the opposite direction must be applied. In this case, since a force of 33.73 lb is directed toward the right, an additional force must be applied to the left with the same magnitude, i.e., 33.73 lb to the left. This balances out the forces acting upon the object, creating a state of equilibrium where the sum of the forces equals zero, thus satisfying Newton's first law of motion.
Now, considering the provided example where a man applies a force of +50 N, it's clear that the force of friction opposing the motion must also be 50 N but in the negative direction (-50 N) for the forces to cancel each other and fulfill the condition of Newton’s first law where Fnet = 0.
In the context of three forces acting on an object, discussing the necessary condition for an object to move straight down, the forces in the horizontal plane (to the right and left) must be balanced. This means that force A and force B, which act in opposite directions, must be equal in magnitude.
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 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 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.
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
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.
If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in meters.
Answer:
Height, H = 25.04 meters
Explanation:
Initially the ball is at rest, u = 0
Time taken to fall to the ground, t = 2.261 s
Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :
[tex]H=ut+\dfrac{1}{2}at^2[/tex]
Here, a = g
[tex]H=\dfrac{1}{2}gt^2[/tex]
[tex]H=\dfrac{1}{2}\times 9.8\times (2.261)^2[/tex]
H = 25.04 meters
So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.
Final answer:
Using the equation of motion under gravity, it was calculated that a ball dropped and taking 2.261 s to hit the ground was released from a height of approximately 25.07 meters.
Explanation:
To find out from what height H a ball was released if it takes 2.261 s to fall to the ground, we use the equation of motion under gravity, which is H = 0.5 * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth) and t is the time in seconds.
Substituting the given values in the equation, we get: H = 0.5 * 9.8 * (2.261)^2. After calculating, we find that the height H approximately equals 25.07 meters.
Therefore, the ball was released from a height of approximately 25.07 meters.
A 1.8-kg block is released from rest at the top of a rough 45° inclined plane. As the block slides down the incline, its acceleration is 2.0 m/s, down the incline. Determine the magnitude of the force of friction acting on the block. a. 16.1 N b. 8.9 N c. 12.5 N d. 2.3 N e. 17.6N
Answer:
Ff= 8.9N:Force of friction acting on the block.
Explanation
Box kinetics: in x₁-y₁ :We apply the second law of Newton
∑Fx₁=ma : second law of Newton Formula (1)
Where:
∑Fx₁:algebraic sum of forces in the direction of x1, positive in the direction of movement of the block, which is down and negative in the direction opposite to the movement of the block
m: is the mass of the block
The x₁ axis coincides with the plane of sliding of the block, that is, the x₁ axis forms 45 degrees with the horizontal
The total weight (W) of the block is in the vertical direction and the tip of the vector down and its magnitude is calculated as follows:
W=m*g
Where:
m: is the mass of the block
g: is the acceleration due to gravity
Calculation of the weight in the direction x₁
Wx₁=Wcos45°= m*g*cos 45 Equation( 1)
Data
m=1.8kg
g=9.8 m/s²
a=2m/s
Wx₁=m*g*cos 45°= 1.8*9.8 *[tex]\frac{\sqrt{2} }{2}[/tex]= 12.47 N
Friction force (Ff ) calculation
We apply formula (1)
∑Fx₁=m*a
Wx₁ - Ff= m*a
Ff=Wx₁ - m*a
We replace data
Ff= 12.47 - 1.8*2 =8.87
Ff= 8.9N
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]
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.
A fighterjet is launched from an aircraft carrier with the
aidof its own engines and a steam powered catapult. The thurst of
itsengine2.3*10^5N. In being launched from rest it moves through
adistance 87m and has KE 4.5*10^7J at lift off.What is the work
doneon the jet by the catapult.
Answer:
The work done on the jet is [tex]W_{jet} = 2.49\times 10^{7} J[/tex]
Given:
Force, F = [tex]2.3\times 10^{5} N[/tex]
Distance moved by the jet, x = 87 m
Kinetic Energy, KE = [tex]4.5\times 10^{7} J[/tex]
Solution:
Now, to calculate the work done on the fire fighter jet by the catapult, we find the difference between the KE of the jet at lift off and the work done by the engines.
This difference provides the amount of work done on the jet and is given by:
[tex]W_{jet} = KE - W_{engines}[/tex] (1)
Now, the work done by the engines is given by:
[tex]W_{engines} = Fx = 2.3\times 10^{5}\times 87 = 2.001\times 10^{7} J[/tex]
Now, using eqn (1):
[tex]W_{jet} = 4.5\times 10^{7} - 2.001\times 10^{7} = 2.49\times 10^{7} J[/tex]
The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430m in a direction 42 degrees north of east. Vector Y is along a measured trail that goes 2200m due south. The hiker then follows vector Z and ends up back where they started, so that X + Y + Z = 0. Find the magnitude of Z and an angle that specifies its direction.
Answer:
magnitude : 1635.43 mAngle: 130°28'20'' north of eastExplanation:
First, we will find the Cartesian Representation of the [tex]\vec{X}[/tex] and [tex]\vec{Y}[/tex] vectors. We can do this, using the formula
[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] its the magnitude of the vector and θ the angle. For [tex]\vec{X}[/tex] we have:
[tex]\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )[/tex]
[tex]\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )[/tex]
where the unit vector [tex]\hat{i}[/tex] points east, and [tex]\hat{j}[/tex] points north. Now, the [tex]\vec{Y}[/tex] will be:
[tex]\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )[/tex]
Now, taking the sum:
[tex]\vec{X} + \vec{Y} + \vec{Z} = 0[/tex]
This is
[tex]\vec{Z} = - \vec{X} - \vec{Y}[/tex]
[tex](Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )[/tex]
[tex](Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )[/tex]
[tex](Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )[/tex]
Now, for the magnitude, we just have to take its length:
[tex]|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}[/tex]
[tex]|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}[/tex]
[tex]|\vec{Z}| = 1635.43 m[/tex]
For its angle, as the vector lays in the second quadrant, we can use:
[tex]\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m}) [/tex]
[tex]\theta = 180\° - arctan( -1.1720) [/tex]
[tex]\theta = 180\° - 45\°31'40'' [/tex]
[tex]\theta = 130\°28'20'' [/tex]
A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal voltage of the battery drops to 10 V when the engine is cranked over with the starter motor. Determine the internal resistance of the 12 volt battery.
Answer:
[tex]r=\frac{1}{75}[/tex] Ω≈0.01333Ω
Explanation:
You can check the internal resistance model in the image I attached you.
Now using Kirchhoff's voltage law:
ε=[tex]Ir+IR[/tex] (1)
Where:
[tex]IR=V_R=LoadVoltage=10V[/tex]
ε=12V
[tex]I=150A[/tex]
Evaluating the data provide in (1)
[tex]r=\frac{12-10}{150} =\frac{1}{75}[/tex]≈0.01333Ω
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
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]
A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50 m tall, you observe that it takes the ball 0.210 s to traverse the length of the window. 1)Determine how high above the top of your window the ball was dropped. Ignore the effects of air resistance. (Express your answer to three significant figures.)
Answer:1.902 m
Explanation:
Given
height of apartment=1.5 m
It takes 0.21 sec to reach the bottom from apartment
So
[tex]s=u_1t+\frac{gt^2}{2}[/tex]
[tex]1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}[/tex]
[tex]u_1=6.11 m/s[/tex]
i.e. if ball is dropped from top its velocity at window is 6.11 m/s
So height of upper floor above window
[tex]v^2-u^2=2as[/tex]
where s= height of upper floor above window
here u=0
[tex]6.11^2=2\times 9.81\times s[/tex]
[tex]s=\frac{6.11^2}{2\times 9.81}[/tex]
s=1.902 m
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 8 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 36 feet from the wall?
The rate at which the area of the triangle formed by the wall, the ground, and the ladder is changing, at the instant the bottom of the ladder is 36 feet from the wall, is -216 square feet per second.
Explanation:This question revolves around the concept of related rates in calculus. The area of the triangle formed by the ground, the wall, and the ladder is given by the formula ½ * base * height. In this situation, the base is x, which represents the distance from the wall to the base of the ladder, and the height is the vertical reach of the ladder onto the wall, represented by y (which can be found using the Pythagorean theorem). Differentiating this equation with respect to time gives us dA/dt = ½ (x dy/dt + y dx/dt).
Given that dx/dt is 8 feet per second, and x=36 feet, to find dy/dt, we use the Pythagorean relationship (39² = 36² + y²), and solve for y to get y = 15 feet. Therefore, substituting all these values into the equation we find:** dA/dt = -216 sq.ft/sec**.
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The area of the triangle is changing at a rate of ( -285.6 ) square feet per second at the instant the bottom of the ladder is 36 feet from the wall. The negative sign indicates that the area is decreasing.
Let ( x ) be the distance of the bottom of the ladder from the wall, and ( y) be the height of the ladder on the wall. We are given that [tex]\( \frac{dx}{dt} = 8 \)[/tex] feet per second, and we want to find [tex]\( \frac{dA}{dt} \) when \( x = 36 \) feet.[/tex]
Using the Pythagorean theorem, we have [tex]\( x^2 + y^2 = 39^2 \).[/tex]Differentiating both sides with respect to time ( t ), we get:
[tex]\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \][/tex]
We can solve for [tex]\( \frac{dy}{dt} \):[/tex]
[tex]\[ 2y\frac{dy}{dt} = -2x\frac{dx}{dt} \][/tex]
[tex]\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \][/tex]
Now, we can express the area ( A ) in terms of ( x ) and ( y ):
[tex]\[ A = \frac{1}{2}xy \][/tex]
Differentiating ( A ) with respect to time ( t ), we get:
[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right) \][/tex]
Substituting [tex]\( \frac{dy}{dt} \)[/tex] from above, we have:
[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(x\left(-\frac{x}{y}\frac{dx}{dt}\right) + y\frac{dx}{dt}\right) \][/tex]
[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(-\frac{x^2}{y}\frac{dx}{dt} + y\frac{dx}{dt}\right) \][/tex]
[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(y - \frac{x^2}{y}\right)\frac{dx}{dt} \][/tex]
We know [tex]\( \frac{dx}{dt} = 8 \)[/tex] feet per second, and we need to find [tex]\( \frac{dA}{dt} \)[/tex]when ( x = 36 ) feet. To find ( y ) at this instant, we use the Pythagorean theorem:
[tex]\[ 36^2 + y^2 = 39^2 \][/tex]
[tex]\[ y^2 = 39^2 - 36^2 \][/tex]
[tex]\[ y^2 = 1521 - 1296 \][/tex]
[tex]\[ y^2 = 225 \][/tex]
[tex]\[ y = 15 \][/tex]
Now we can find [tex]\( \frac{dA}{dt} \):[/tex]
[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(15 - \frac{36^2}{15}\right) \times 8 \][/tex]
[tex]\[ \frac{dA}{dt} = 4\left(15 - \frac{1296}{15}\right) \][/tex]
[tex]\[ \frac{dA}{dt} = 4\left(15 - 86.4\right) \][/tex]
[tex]\[ \frac{dA}{dt} = 4\left(-71.4\right) \][/tex]
[tex]\[ \frac{dA}{dt} = -285.6 \][/tex]
So, the area of the triangle is changing at a rate of ( -285.6 ) square feet per second at the instant the bottom of the ladder is 36 feet from the wall. The negative sign indicates that the area is decreasing.
The sound of one student typing furiously on their History paper which is due at 8:00 am tomorrow morning is 60.0 dB. What will the decibel level be when three sleep deprived students are typing furiously in the same room?
Answer:
The the decibel level be when three sleep deprived students are typing furiously in the same room is 64.77 dB.
Explanation:
Given that,
Decibel level = 60.0 dB
Using formula of decibel level
[tex]\beta= 10 log\dfrac{I}{I_{0}}[/tex]
For one student,
[tex]60.0=10 log\dfrac{I}{I_{0}}[/tex]
For 3 students,
[tex]\beta'=10 log\dfrac{3I}{I_{0}}[/tex]
[tex]\beta'=10log 3+10 log\dfrac{I}{I_{0}}[/tex]
[tex]\beta'=4.77+60.0[/tex]
[tex]beta'=64.77\ dB[/tex]
Hence, The the decibel level be when three sleep deprived students are typing furiously in the same room is 64.77 dB.
A crane carries a 1600 kg car at 1.5 m/s^2 with a chain that has a negligible mass. If the coefficient of kinetic friction between the wheels of the car and the road is 0.3, what is the tension in the chain? a. 3.8 kN
b. 5.9 kN
c. 6.7 kN
d. 7.1 kN
Answer:
option (d) 7.1 kN
Explanation:
Given:
Mass of the car, m = 1600 kg
Acceleration of the car, a = 1.5 m/s²
Coefficient of kinetic friction = 0.3
let the tension be 'T'
Now,
ma = T - f .................(1)
where f is the frictional force
also,
f = 0.3 × mg
where g is the acceleration due to the gravity
thus,
f = 0.3 × 1600 × 9.81 =
therefore,
equation 1 becomes
1600 × 1.5 = T - 4708.8
or
T = 2400 + 4708.8
or
T = 7108.8 N
or
T = 7.108 kN
Hence,
The correct answer is option (d) 7.1 kN
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
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
A person walks 27.0° north of east for 4.00 km. How far due north and how far due east would she have to walk to arrive at the same location?
Answer:
Along North, 1.816 km
Along east , 3.564 km
Explanation:
d = 4 km 27° North of east
As we need to find the distance traveled due north and the distance traveled due east, they are the components along north and along east. The component along north is the vertical component and the distance traveled along east is the horizontal component.
Distance traveled due North
y = d Sin 27 = 4 Sin 27 = 1.816 km
Distance traveled due east
x = d cos 27 = 4 cos 27 = 3.564 km
If a car stops suddenly, you feel "thrown forward." We’d like to understand what happens to the passengers as a car stops. Imagine yourself sitting on a very slippery bench inside a car. This bench has no friction, no seat back, and there’s nothing for you to hold onto.(a) identify all of the forces action on you as thecar travels at a perfectly steady speed on level ground.(b) repeat part A with the car slowing down(c) describe what happens to you as the car slowsdown(d) suppose now that the bench is not slippery. as the carslows down, you stay on the bench and dont slide off. what force isresponsible for you deceleration?
The forces acting on you are:
The gravitational pull of the Earth (an others Celestial objects)The Normal Force that balance the gravitational pull and points upward.That is. There is not need for any other force, cause the car its going at constant speed, so the acceleration its zero, as the net force its mass multiplied by acceleration, the net force is zero.
B)The forces acting on you are:
The gravitational pull of the Earth (an others Celestial objects)The Normal Force that balance the gravitational pull and points upward.That is. Again. The car its slowing down. But, it cant apply a force in you against the direction of movement to slow you down.
C)As you can't be slowed down, you will go forward at the same speed you had when the car went steady, and will crash against whatever its in front of you.
D)
The friction force, will pull you against the direction of movement, and slow you down will the car slows down.
When a car moves at a steady pace, the forces acting on a passenger include gravity and the normal force from the bench. When the car begins to slow down, the passenger experiences inertia, which might give the effect of being 'pushed' forward. If the bench is not slippery, friction will be the force that aids in decelerating the passenger along with the car.
Explanation:(a) If you are sitting on a frictionless bench in a car that is moving at a perfectly steady speed, the forces acting on you are: the gravitational force directed down and the normal force from the bench acting upward. Since the car moves at a constant speed, there is no horizontal force.
(b) If the car starts slowing down, then in addition to the forces mentioned before, there is one more force in play, which is inertia. It's not a force like gravity or the normal force, but it is a consequence of Newton's first law of motion. You experience this as a force pushing you forward, even though the car is slowing down or stopping.
(c) Because of inertia, as the car slows down, you tend to keep moving. Without friction or anything to hold onto, you would slide off the bench and continue moving forward at the original speed of the car.
(d) If the bench is not slippery and you don't slide off as the car slows down, the force responsible for your deceleration is friction. Even though you can't feel it, friction between you and the bench affects you when the car decelerates.
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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.
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.6 cm and a current of 12 A. The bigger loop has a current of 20 A . The magnetic field at the center of the loops is found to be zero. What is the radius of the bigger loop?
Answer:
the radius of bigger loop = 6 cm
Explanation:
given,
two concentric current loops
smaller loop radius = 3.6 cm
]current in smaller loop = 12 A
current in the bigger loop = 20 A
magnetic field at the center of loop = 0
Radius of the bigger loop = ?
[tex]B_t = B_1 + B_2[/tex]
[tex]0 = \dfrac{\mu_0I_1}{2R_1} +\dfrac{\mu_0I_2}{2R_2}[/tex]
now, on solving
[tex]\dfrac{I_1}{R_1} = \dfrac{I_2}{R_2}[/tex]
[tex]R_2 = I_2\dfrac{R_1}{I_1}[/tex]
= [tex]20\times \dfrac{3.6}{12}[/tex]
= 6 cm
hence, the radius of bigger loop = 6 cm