Swimming downstream and upstream in a river with a current of 0.52 m/s results in approximately an 86% increase in time compared to swimming the same distance in still water at a speed of 1.73 m/s.
1. Understanding the problem:
You can swim 1.73 m/s in still water.
The river current flows at 0.52 m/s.
We need to find the percentage increase in time to swim downstream and back upstream compared to swimming the same distance in still water.
2. Effective speeds:
Downstream: Your speed increases with the current.
Effective speed = swimming speed + current speed = 1.73 m/s + 0.52 m/s = 2.25 m/s
Upstream: Your speed decreases against the current.
Effective speed = swimming speed - current speed = 1.73 m/s - 0.52 m/s = 1.21 m/s
3. Calculating time for each leg:
Let D be the distance for the entire round trip.
Time downstream: D / 2.25 m/s
Time upstream: D / 1.21 m/s
4. Finding the time difference:
Time difference = Time upstream - Time downstream
Time difference = (D / 1.21 m/s) - (D / 2.25 m/s)
5. Calculating percentage increase:
To find the percentage increase in time compared to still water, divide the time difference by the time taken to swim downstream and multiply by 100.
Percentage increase = [(D / 1.21 m/s) - (D / 2.25 m/s)] / (D / 2.25 m/s) * 100
6. Simplifying the equation:
After simplifying, you'll find the percentage increase to be approximately 86%.
Therefore, it takes approximately 86% longer to complete the round trip compared to swimming the same distance in still water.
A water gun is fired horizontally from a 4 meter tall hill and lands 7 meters from the base of the hill. (a) How long is the water in the air? (b) What is the initial velocity of the water?
Answer:
(a) 0.9 s
(b) 7.78 m/s
Explanation:
height, h = 4 m
Horizontal distance, d = 7 m
Let it takes time t to reach the ground and u be the initial velocity of the jet.
(a) Use second equation of motion in vertical direction
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
In vertical direction, uy = 0 m/s, a = g = - 9.8 m/s^2, h = - 4 m
By substituting the values, we get
[tex]-4 = 0 - \frac{1}{2}\times 9.8\times t^{2}[/tex]
t = 0.9 second
Thus, the time taken by water jet in air is 0.9 second.
(b) Use
Horizontal distance = horizontal velocity x time
d = u t
7 = u x 0.9
u = 7.78 m/s
Thus, the initial velocity of water jet is 7.78 m/s.
The duration the water is in the air is found using the formula for the motion under gravity, which depends on the vertical distance and the acceleration due to gravity. After finding the time, the initial velocity of the water is calculated using the horizontal distance and the fraction of time the water was in motion.
Explanation:To determine how long the water is in the air (a), and the initial velocity of the water (b) when a water gun is fired horizontally from a hill, we can use the principles of projectile motion. The time a projectile is in the air is solely determined by its vertical motion. Since the water gun is fired horizontally, it has an initial vertical velocity of 0 m/s.
For part (a), the time (t) it takes for the water to reach the ground can be calculated using the formula for the motion under gravity, which is y = 0.5 * g * t2, where y is the vertical distance (4 meters in this case) and g is the acceleration due to gravity (approximated to 9.81 m/s2). Solving for t gives us the time the water is in the air.
For part (b), once we have the time, we can use the horizontal distance (7 meters) to find the initial velocity (v0) using the formula x = v0 * t. This provides the initial horizontal velocity of the water gun's jet. The overall process involves solving for time first and then using that time to find the initial velocity.
A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.8 m/s, from a height of 0.8 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. A)What is the maximum height the blue ball reaches?
B)What is the height of the blue ball 1.8 seconds after the red ball is thrown?
C)How long after the red ball is thrown are the two balls in the air at the same height?
Answer:
A) The maximum height of the blue ball is 29.7 m above the ground.
B) The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m
C) The balls are at the same height 1.41 s after the red ball is thrown
Explanation:
A) At maximum height, the velocity of the blue ball is 0 because for that instant, the ball does not go up nor down.
The equation for velocity for an accelerated object moving in a straight line is:
v = v0 + a*t
where
v = velocity.
v0 = initial velocity.
a = acceleration, in this case, it is the acceleration of gravity, g, 9.81 m/s².
t = time.
Then:
0 = v0 + g * t (if the origin of the reference system is the ground, then g is negative)
0 = 23.8 m/s - 9.81 m/s² * t
-23.8 m/s / -9.81 m/s² = t
t = 2.43 s
With this time, we can calculate the position of the blue ball. The equation for position is:
y = y0 + v0 * t + 1/2 * g * t²
y = 0.8 m + 23.8 m/s * 2.43 s - 1/2 * 9.81 m/s² * (2.43 s)²
y = 29.7 m
the maximum height of the blue ball is 29.7 m above the ground.
B) 1.8 s after throwing the red ball, the blue ball was in the air for (1.8 s - 0.6) 1.2 s. Then, using the equation for the position of the blue ball:
y = 0.8 m + 23.8 m/s * 1.2 s - 1/2 * 9.81 m/s² * (1.2 s)² = 22.3 m
The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m
C) Now, we have to find the time at which both positions are equal. Notice that the time of the blue ball is not the same as the time for the red ball. The time for the blue ball is the time of the red ball minus 0.6 s:
t blue = t red - 0.6 s
Then:
position red ball = position blue ball
y0 + v0 * t + 1/2 * g * t² = y0 + v0 * (t- 0.6) + 1/2 * g * (t-0.6s)²
25 m + 1.2 m/s * t -1/2 * 9.81 m/s² * t² = 0.8 m + 23.8 m/s * (t-0.6 s) - 1/2 * 9.81 m/s² * (t-0.6 s)²
24.2 m + 1.2 m/s * t -4.91 m/s² * t² = 23.8 m/s * t - 14.28 m - 4.91 m/s² * (t-0.6 s)²
38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² (t² - 1.2 s * t + 0.36 s²)
38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² * t² + 5.89 m/s * t - 1.77 m
40.3 m - 28.5 m/s * t = 0
t = -40.3 m / -28.5 m/s
t = 1.41 s
The balls are at the same height 1.41 s after the red ball is thrown and 0.81 s after the blue ball is thrown.
A charge of 5 µC is on the y axis at y = 7 cm, and a second charge of -5 µC is on the y axis at y = -7 cm. Find the force on a charge of 2 µC on the x axis at x = 3 cm.
Answer:
The force on X Fx=0 N
The force on Y Fy=-2.18 N
Explanation:
We have an array of charges, we will use the coulomb's formula to solve this:
[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]
but we first have to find the distance and the angule of the charge respect the charges on the Y axis:
[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]
we can notice that it is the same distance from both charges on Y axis.
we can find the angle with:
[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]
for the charge of 5µC [tex]\alpha =-66.80^o[/tex]
for the charge of -5µC [tex]\alpha =66.80^o[/tex]
the net force on the X axis will be:
[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]
and
[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]
So the net force on X will be Zero.
for the force on Y we have:
[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]
and
[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]
Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]
So the net force on Y is Fy=-2.18N
To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. The force is calculated to be 5.12 N.
Explanation:To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * |q1 * q2| / r^2
Where F is the force, k is the electrostatic constant (k ≈ 8.99 * 10^9 N * m² / C²), q1 and q2 are the magnitudes of the charges, and r is the distance between them.
In this case, one charge is 5 µC and the other charge is -5 µC, with a distance of 14 cm between them (7 cm on the positive y-axis and 7 cm on the negative y-axis). The charge we're interested in is 2 µC at x = 3 cm. Plugging these values into the formula:
F = (8.99 * 10^9 N * m² / C²) * |(5 µC) * (2 µC)| / (0.14 m)²
F = 5.12 N
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Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is their effective spring constant? 004393 Submit Answer Incorrect. Tries 4/99 Previous Tres If a mass of 0.31 kg is attached what will be the frequency of oscillation?
Answer:
[tex]K_e_q=22.75878093\frac{N}{m}[/tex]
[tex]f=1.363684118Hz[/tex]
Explanation:
In order to calculate the equivalent spring constant we need to use the next formula:
[tex]\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}[/tex]
Replacing the data provided:
[tex]\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}[/tex]
[tex]K_e_q=22.75878093\frac{N}{m}[/tex]
Finally, to calculate the frequency of oscillation we use this:
[tex]f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }[/tex]
Replacing m and k:
[tex]f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz[/tex]
A tennis ball is hit straight up at 20 m/s from the edge of a sheer cliff. Some time later, the ball passes the original height from which it was hit. (Ignore the effects of air resistance.) (a) How fast is the ball moving at that time? m/s (b) If the cliff is 50 m high, how long will it take the ball to reach the ground level? s (c) What total distance did the ball travel? m
The speed of the ball when passing its original height on its way down is 20 m/s. The time it takes to reach the ground level is 3.19 seconds. The ball travels a total distance of 70 meters.
Explanation:The motion of the tennis ball can be analyzed by using the principles of Physics, particularly the laws of motion and the concept of gravity.
(a) The ball will have the same speed when it passes its original height on its way down, which is 20 m/s. This is based on the principle of conservation of energy. Since air resistance is ignored, the speed when it passes the original height on its way down should be the same as its initial speed.
(b) Calculating the total time it takes the ball to reach the ground involves the formula for time in free fall, which is t = sqrt(2h/g), wherein h is the height and g is the gravity. Thus plug in the values: t = sqrt((2*50)/9.8) = 3.19 seconds.
(c) The total distance traveled by the ball includes it going up and falling back down. The ball rises to a height of 20 m/s before falling from that height plus an additional height corresponding to the height of the cliff, total distance is 70 meters.
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You go watch your friend as she runs at a cross country meet. The first time she passes you, your friend runs by going due north at a speed of 4.40 m/s. You move to a different place on the course and twelve minutes after you saw her the first time, your friend runs past you again. This time she is traveling at a speed of 3.80 m/s in the direction 30.0° W of S. What is the direction of her average acceleration between those same two times?
Answer:
76 degree south of west.
Explanation:
We shall represent velocities in vector form , considering east as x axes and west as Y axes.
V₁ = 4.4 j
V₂ = 3.8, 30 degree west of south
V₂ = - 3.8 sin 30 i - 3.8 cos 30 j
= - 1.9 i - 3.29 j
Change in velocity
= V₂ - V₁
= - 1.9 i - 3.29 j - 4.4 j
= - 1.9 i - 7.69 j
Acceleration
= change in velocity / time
(- 1.9 i - 7.69 j ) / 60 ms⁻² .
Direction of acceleration θ
Tan θ = 7.69 / 1.9 =4.047
θ = 76 degree south of west.
The Asian Silver Carp is an invasive species of fish which has infested a number of American rivers. The disturbance of passing boats has been known to make these fish leap large distances out of the water. One such fish leaps with an initial velocity at the water's surface of 11.0 m/s at an angle of 34.0° above the horizontal. How high above the water's surface, in meters, does the fish reach?
Answer:
1.93 m
Explanation:
Initial velocity of fish u = 11 m /s
Angle of jump = 34 degree.
Vertical component of its velocity = u sin34 = 11 x.5592 = 6.15 m /s
Considering its motion in vertical direction ,
u = 6.15 m/s
g = - 9.8 m /s
maximum height attained = h
From the formula
v² = u² - 2gh
0 = 6.15x 6.15 - 2 x 9.8 h
h = (6.15 x 6.15 )/ 2 x 9.8
= 1.93 m .
What is the diameter of a 12lb shot if the specific gravity is of the shot iron in the shot is 6.8, the density of fresh water 62.4lb/ft
Answer:
The diameter is 0.378 ft.
Explanation:
Given that,
Mass of shot = 12 lb
Density of fresh water = 62.4 lb/ft
Specific gravity = 6.8
We need to calculate the volume of shot
[tex]V = \dfrac{4}{3}\pi r^3\ ft^3[/tex]
The density of shot is
Using formula of density
[tex]\rho = \dfrac{m}{V}[/tex]
Put the value into the formula
[tex]\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}[/tex]
We need to calculate the radius
Using formula of specific gravity
[tex]specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}[/tex]
Put the value into the formula
[tex]6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}[/tex]
[tex]r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}[/tex]
[tex]r^3=0.0067514[/tex]
[tex]r =(0.0067514)^{\frac{1}{3}}[/tex]
[tex]r=0.1890\ ft[/tex]
The diameter will be
[tex]d = 2\times r[/tex]
[tex]d =2\times0.1890[/tex]
[tex]d =0.378\ ft[/tex]
Hence, The diameter is 0.378 ft.
An amplitude modulation transmitter radiates 10 KW power with the modulation percentage of 75 %. Find the magnitude of the carrier power?
Answer:
carrier power is 7.8 kW
Explanation:
given data
power = 10 kW
modulation percentage = 75 %
to find out
carrier power
solution
we will use here power transmitted equation that is
power = [tex]carrier power * ( 1+ \frac{modulation}{2})[/tex] .................1
put here value in equation 1 we get carrier power
10 = [tex]carrier power *(1+ \frac{0.75}{2})[/tex]
carrier power = 7.8
so carrier power is 7.8 kW
A stone thrown off a bridge at 20 m above a river has an initial velocity of 12 m/s at an angle of 45° above the horizontal. What is the range of the stone? At what velocity does the stone strike the water?
Answer:
a) [tex]x=26m[/tex]
b) [tex]v_{y}=-21.5m/s[/tex]
Explanation:
From the exercise we know initial velocity, initial height
[tex]y_{o}=20m[/tex]
[tex]v_{o} =12m/s[/tex] [tex]\beta =45[/tex]º
a) The range of the stone is defined by how far does it goes. From the theory of free falling objects, we have:
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
The stone strike the water at y=0
[tex]0=20+12sin(45)t-\frac{1}{2}(9.8)t^{2}[/tex]
Solving for t, using the quadratic formula
[tex]t=-1.33s[/tex] or [tex]t=3.06s[/tex]
Since time can't be negative, the answer is t=3.06s
Now, we can calculate the range of the stone
[tex]x=v_{o}t=(12cos(45))m/s(3.06s)=26m[/tex]
b) We can calculate the velocity were the stone strike the water using the following formula
[tex]v_{y}=v_{oy}+gt=12sin(45)m/s-(9.8m/s^{2})(3.06s)=-21.5m/s[/tex]
The negative sign indicates that the stone is going down
Answer:
(a) x= 26m
(b) vf= 23.15m/s
Explanation:
Given data
h=20m
Ɵ=45°
to find
(a) range of stone=x=?
(b) velocity=vf=?
Solution
For part (a)
You need to solve for time first using
yf = yi + visinƟt + 1/2gt^2
0 = 20m + 12sin45t + 1/2(-9.8)t^2
and use the quadratic equation to solve for t
t = 3.064 sec
To solve for the distance traveled use
x = xi + vicosƟt + 1/2at^2 there is no acceleration in the x direction so that cancels
x = 12cos(45)(3.064)
x= 26m
For part(b)
For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.
Theres no change in the velocity in the x direction so its just vfx = vixcosƟ = 12cos45 = 8.49m/s
For the y direction its vfy^2 = viy^2 + 2g(Δy)
vfy = sqrt((12sin(45))^2 + 2(-9.8)(0-20m)) = 21.54m/s
The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(8.49^2 + 21.54^2)
vf= 23.15m/s
An object moving with constant acceleration changes its speed from 20 m/s to 60 m/s in 2.0 s. What is the acceleration? (b) How far did it move in this time?
Answer:
acceleration, a = [tex]20 m/s^{2}[/tex]
distance, d = 80 m
Given:
Initial velocity of object, v = 20 m/s
Final velocity of object, v' = 60 m/s
Time interval, [tex]\Delta t = 2.0 s[/tex]
Solution:
(a) Acceleration is the rate at which velocity changes and constant acceleration is when the velocity changes by equal amount in equal intervals of time.
Thus
acceleration, a = [tex]\frac{v' - v}{\Delta t}[/tex]
a = [tex]\frac{60 - 20}{2.0} = 20 m/s^{2}[/tex]
(b) Now, the distance covered is given by:
[tex]d = vt + \frac{1}{2}at^{2}[/tex]
[tex]d = 20\times 2.0 + \frac{1}{2}20\times 2^{2} = 80 m[/tex]
A mass of 156.7g of Helium gas at an initial temperature of 35.73°C and at an initial absolute pressure of 3.55 atm undergoes an isothermal expansion until its volume increases by a factor of 1.75. (a) What is the final pressure? (b) How much work is done on the gas? (c) How much heat does the gas absorb? (d) What is the change in the total internal energy of the gas? (a) Pa Answer part (a) (b) Joules Answer part (b) Joules Joules Answer part (C) Answer part (d) (d) Submit
Answer:
Explanation:
Moles of helium ( n ) = 156.7 / 4 = 39.175
Temperature T₁ = 35.73 +273 = 308.73 K
Volume V₁ = V
Pressure P₁ = 3.55 atm
V₂ =1.75 V
a ) For isothermal change
P₁ V₁ = P₂V₂
P₂ = P₁ V₁ / V₂
= 3.55 X V / 1.75 V
= 2.03 atm.
b ) Work done by the gas = nRT ln(V₂/V₁)
= 39.175 X 8.321 X 308.73 X ln 1.75
= 56318.8 J
Work done on the gas = - 56318.8 J
c ) Since there is no change in temperature , internal energy of gas is constant
Q = ΔE + W
ΔE = 0
Q = W
Work done by gas = heat absorbed
heat absorbed = 56318.8 J
d ) Change in the internal energy of gas is zero because temperature is constant.
Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500m3/sat a location 90 m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.
Answer:
mechanical energy per unit mass is 887.4 J/kg
power generated is 443.7 MW
Explanation:
given data
average velocity = 3 m/s
rate = 500 m³/s
height h = 90 m
to find out
total mechanical energy and power generation potential
solution
we know that mechanical energy is sum of potential energy and kinetic energy
so
E = [tex]\frac{1}{2}[/tex]×m×v² + m×g×h .............1
and energy per mass unit is
E/m = [tex]\frac{1}{2}[/tex]×v² + g×h
put here value
E/m = [tex]\frac{1}{2}[/tex]×3² + 9.81×90
E/m = 887.4 J/kg
so mechanical energy per unit mass is 887.4 J/kg
and
power generated is express as
power generated = energy per unit mass ×rate×density
power generated = 887.4× 500× 1000
power generated = 443700000
so power generated is 443.7 MW
The total mechanical energy per unit mass of the river water is the sum of the potential and kinetic energy. The potential energy is calculated using the height and the gravitational constant, while kinetic energy is calculated using velocity. The power generation potential of the river is the total mechanical energy multiplied by the volume flow rate.
Explanation:The total mechanical energy per unit mass of the river water at this location can be determined using the principle of mechanical energy conservation and the knowledge of potential and kinetic energy. The total mechanical energy (E) equals the potential energy (PE) plus the kinetic energy (KE) per unit mass.
The potential energy is given by: PE = mgh, where m is the mass, g is the acceleration due to gravity (~9.81 m/s^2), and h is the height (90 m above lake surface). But since we are looking for energy per unit mass, m cancels out and we get PE = gh.
The kinetic energy is given by: KE = 0.5mv^2, where v is the velocity (3 m/s). In per unit mass terms, this simplifies to KE = 0.5v^2.
Therefore, E = PE + KE = gh + 0.5v^2.
The power generation potential can be calculated using the equation: P = E * volume flow rate (in our case, 500 m^3/s).
So, P = (gh + 0.5v^2) * 500.
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A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads share the 28 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 N . Assume that A has a greater charge. What is the charge qA and qB on the beads?
Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C
The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.
We have to use Coulomb's Law and the principle of charge conservation.
1. Conservation of Charge:
The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:
[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]
2. Coulomb's Law:
The force between two charges is given by Coulomb's Law:
[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]
Let's set up the equation:
[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]
Rearrange to solve for [tex]\( q_A q_B \)[/tex]:
[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]
Now, we have two equations:
1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]
2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]
To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:
[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]
[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]
[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]
Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]
This gives us two solutions for [tex]\( q_A \)[/tex]:
1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]
2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]
Since bead A has the greater charge, we take:
[tex]\[q_A = 21.907 \, \text{nC}\][/tex]
And the charge on bead B:
[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]
So, the charges on the beads are:
[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The box experiences a friction force of 30 N as it slows to a stop. How long does it take the box to stop?
Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:
[tex]F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}[/tex]
Given mass of box = 5.0 kg
Frictional force = 30 N
thus
[tex]acceleration=\frac{30}{5}=6m/s^{2}[/tex]
Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero
[tex]v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds[/tex]
Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.
Jacob is traveling at 5.00 m/s North. Jacob throws a ball with a velocity of 5.00 m/s South. Jacob throws the ball from a height of 1.45 m above the ground. Assuming gravitational acceleration is 9.81 m/s^2, how long does the ball stay in the air after it is thrown? Describe the motion of the ball relative to the ground. Describe the motion of the ball relative to Jacob.
Answer:t=0.54 s
Explanation:
Given
Jacob is traveling 5 m/s in North direction
Jacob throw a ball with a in south direction with a velocity of 5 m/s
Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0
Time taken by ball to reach ground
[tex]s=ut+\frac{gt^2}{2}[/tex]
[tex]1.45=0+\frac{9.81\times t^2}{2}[/tex]
[tex]t^2=frac{2\times 1.45}{9.81}[/tex]
t=0.54 s
Motion of ball will be straight line
What force will give a 40.0 kg grocery cart an acceleraticn of 2.4 m/s^2?
Answer:
force is 96 N
Explanation:
given data
mass = 40 kg
acceleration = 2.4 m/s²
to find out
force
solution
we know force is mass time acceleration so
we will apply here force formula that is express as
force = m × a ..............1
here m is mass and a is acceleration so
put here value in equation 1 we get force
force = 40 × 2.4
force = 96
so force will be 96 N
When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force is at right angles to the lever arm, how much torque does the wood exert on the blade? Does this torque make the blade turn faster or slower?
Answer:
Explanation:
When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.
So torque exerted by wood on the blade
= force x perpendicular distance from the axis of rotation
= 104 x .128
=13.312 Nm.
Since this torque opposes the movement of blade , it turns the blade slower.
What is the forward velocity of a 250 lb stone projectile that leaves a medieval catapult 55 degrees from the ground at 50 m/s?
Answer:
28.5 m/s
Explanation:
The speed has two orthogonal components, horizontal and vertical. To decompose the speed into these components we can use these trigonometric equations:
Vh = V * cos(a)
Vv = V * sin(a)
The forwards velocity is the horizontal component, so we use
Vh = V * cos(a)
Vh = 50 * cos(55)
Vh = 50 * 0.57
Vh = 28.5 m/s
At what distance from a long straight wire carrying a
currentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?
Answer:
The distance is 2 cm
Solution:
According to the question:
Magnetic field of Earth, B_{E} = [tex]5.0\times 10^{- 5} T[/tex]
Current, I = 5.0 A
We know that the formula of magnetic field is given by:
[tex]B = \farc{\mu_{o}I}{2\pi d}[/tex]
where
d = distance from current carrying wire
Now,
[tex]d = \frac{\mu_{o}I}{2\pi B}[/tex]
[tex]d = \frac{4\pi\times 10^{- 7}\times 5.0}{2\pi\times 5.0\times 10^{- 5}}[/tex]
d = 0.02 m 2 cm
The distance from the wire where the magnitude of the magnetic field equals the Earth's magnetic field is approximately 0.4 meters.
Explanation:To find the distance from the wire where the magnitude of the magnetic field is equal to the strength of the Earth's magnetic field, we can use the equation:
B = μ0 * I / (2π * r)
Where B is the magnetic field, μ0 is the permeability of free space (4π x [tex]10^-7[/tex]A), I is the current, and r is the distance from the wire.
Plugging in the given values, we have:
B_wire = μ0 * 5.0A / (2π * r) and B_earth = 5.0 x [tex]10^-5 T[/tex]
Setting B_wire equal to B_earth and solving for r:
5.0 x[tex]10^-5[/tex]* 5.0A / (2π * r)
Solving for r, we find that the distance from the wire where the magnitude of the magnetic field is equal to the Earth's is approximately 0.4 meters.
a speeding car is travelling at a constant 30.0 m/s when it passes a stationary police car. If the police car delays for 1.00 s before starting, what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m?
(A) 6.00 m/s2
(B) 3.00 m/s2
(C) 7.41 m/s2
(D) 1.45 m/s2
(E) 3.70 m/s2
Answer:
option (C)
Explanation:
Speed of car = 30 m/s
Let the time taken by the police car to catch the speeding car is t
The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t
Distance traveled by the police car in time t
[tex]s=ut + 0.5 at^{2}[/tex] .... (1)
Distance traveled by the speeding car in t + 1 second
s = 30 (t + 1) = 300
t + 1 = 10
t = 9 s
Put the value of t in equation (1), we get
300 = 0 + 0.5 x a x 9 x 9
a = 7.41 m/s^2
C. 7.41 meters per square second.
In this question, the car is travelling at constant speed, whereas the police car accelerates uniformly after some delay to catch the car, the respective kinematic formulas are shown below:
Car
[tex]x_{C} = x_{o} + v_{C}\cdot t[/tex] (1)
Police car
[tex]x_{P} = x_{o} + \frac{1}{2}\cdot a_{P}\cdot (t-t')^{2}[/tex] (2)
Where:
[tex]x_{o}[/tex] - Initial position, in meters. [tex]x_{C}[/tex] - Final position of the car, in meters.[tex]x_{P}[/tex] - Final position of the car, in meters. [tex]t[/tex] - Time, in seconds.[tex]t'[/tex] - Delay time, in seconds.[tex]a_{P}[/tex] - Acceleration of the police car, in meters per square seconds.[tex]v_{C}[/tex] - Speed of the car, in meters per second.If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{C} = x_{P} = 300\,m[/tex], [tex]t' = 1\,s[/tex] and [tex]v_{C} = 30\,\frac{m}{s}[/tex], then we have the following system of equations:
[tex]300 = 30\cdot t[/tex] (1)
[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (t-1)^{2}[/tex] (2)
By (1):
[tex]t = 10[/tex]
Then we find that acceleration of the police car must be:
[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (10-1)^{2}[/tex]
[tex]a_{P} = 7.407\,\frac{m}{s^{2}}[/tex]
Therefore, the correct choice is C.
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A small particle has charge -3.90 μC and mass 1.80×10^−4 kg . It moves from point A, where the electric potential is VA= 130 V , to point B, where the electric potential VB = 500 V. The electric force is the only force acting on the particle. The particle has a speed of 4.90 m/s at point A. What is its speed at point B? Express your answer in meters per second to three significant figures.
Final answer:
By applying the conservation of energy principle, accounting for the work done by the electric field as change in potential energy, we calculate the change in kinetic energy to find the speed of the charged particle at a different electric potential.
Explanation:
To calculate the speed of the particle at point B, we need to consider the change in electric potential energy and convert it to kinetic energy. The work done on the particle by the electric field is equal to the change in potential energy, which is given by W = q(VB - VA), where q is the charge of the particle, VA and VB are the electric potentials at points A and B respectively. Since no other forces act on the particle, the work done by the electric field causes a change in kinetic energy, which is defined by ½mv22 - ½mv12 = q(VB - VA), where m is the mass of the particle and v1 and v2 are the speeds at points A and B.
Rearranging the equation to solve for v2 gives us v2 = √(2q(VB - VA)/m + v12). Substituting in the given values, we can find the speed at point B.
After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand?
The speed of the ball when it leaves Sarah's hand is 8.2 m/s.
Given that;
The ball leaves Sarah's hand at a distance of 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head.
For the speed of the ball when it leaves Sarah's hand, use the equations of motion and consider the vertical motion of the ball.
Since the ball is thrown vertically upward and then comes back down, the time taken to reach the maximum height is half of the total time of flight.
Therefore, the time to reach the maximum height is,
t/2 = 1.619 s / 2.
So, the time to reach the maximum height is 0.8095 s.
Now, let's find the initial vertical velocity of the ball using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
At the maximum height, the final vertical velocity is 0 m/s because the ball momentarily stops.
The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downward).
Using the equation, we have:
0 m/s = u + (-9.8 m/s²) × 0.8095 s
Simplifying the equation:
u = 7.93 m/s
So, the initial vertical velocity of the ball is 7.93 m/s.
Since the ball travels in a parabolic path, the time taken to reach Julie's horizontal position is the same as the time taken to reach the maximum height, which is 0.8095 s.
Now, let's calculate the initial horizontal velocity of the ball, using the equation:
s = ut
where s is the horizontal distance travelled, u is the initial horizontal velocity, and t is the time.
The horizontal distance travelled is equal to the horizontal distance between Sarah and Julie, which we don't have.
However, since we are only interested in the initial horizontal velocity, we can assume that the horizontal distance travelled is equal to the distance between Sarah and Julie.
Therefore, s = 1.5 m.
Using the equation, we have:
1.5 m = u × 0.8095 s
u = 1.5 m / 0.8095 s
Calculating u, we find:
u ≈ 1.853 m/s
So, the initial horizontal velocity of the ball is 1.853 m/s.
Finally, the speed of the ball when it leaves Sarah's hand by combining the horizontal and vertical components of velocity using the Pythagorean theorem:
speed = √(horizontal velocity² + vertical velocity²)
speed = √(1.853 m/s)² + (7.951 m/s)²
Calculating the speed, we find:
speed ≈ 8.2 m/s
Therefore, the speed of the ball when it leaves Sarah's hand is 8.2 m/s.
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Final answer:
Using the kinematic equations for vertical motion, the initial speed of the ball Sarah threw is calculated to be approximately 7.94 m/s, based on the given maximum height and time to reach Julie's horizontal position.
Explanation:
To find the initial speed of the ball when it leaves Sarah's hand, we need to use the information about the ball's motion under gravity. We know it reaches a maximum height of 8 m and takes 1.619 s to get directly over Julie's head, starting from a height of 1.5 m.
The motion of the ball can be divided into two segments: ascending and descending. During the ascending part, the ball slows down due to gravity until it reaches its maximum height. In the descending part, it accelerates back down. Since the motion is symmetrical, the time to reach the maximum height is half of the total time, which is 1.619 s / 2 = 0.8095 s.
To find the initial velocity (v_i), we can use the kinematic equation for vertical motion:
v_i = v_f - g*t
where v_f is the final velocity (0 m/s at the maximum height), g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]), and t is the time to reach maximum height.
Plugging in the values, we get:
v_i = 0 m/s - (-9.81 m/[tex]s^2[/tex] * 0.8095 s) = 7.94 m/s
Therefore, the initial speed of the ball when it leaves Sarah's hand is about 7.94 m/s.
Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C, A, B) = 0
Answer:
Three linear operators A,B, and C will satisfy the condition [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].
Explanation:
According to the question we have to prove.
[tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex]
Now taking Left hand side of the equation and solve.
[tex][[A, B],C] + [[B,C), A] + [[C, A], B][/tex]
Now use commutator property on it as,
[tex]=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0[/tex]
Therefore, it is proved that [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].
What is the difference between Reynolds equation set and Navier Stokes equation?
Answer:
Navier Stokes equation
( 1 ) it is a partial differential equation that is describe the flow of incompressible fluids
Reynolds equation
(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication
Explanation:
Navier Stokes equation
( 1 ) it is a partial differential equation that is describe the flow of incompressible fluids
(2) Navier Stokes equation is used to model weather and ocean current and water flow in the pipe and air flow around wing
( 3) equation is
[tex]\nabla .\overrightarrow{v} = 0[/tex] momentum equation
[tex]\rho \frac{d\overrightarrow{v}}{dt} = \nabla p + \rho \overrightarrow{g} + \mu \nabla ^2 v^2[/tex]
here [tex]\nabla p[/tex] is pressure gradient and [tex]\rho \overrightarrow{g}[/tex] is body force and [tex]\mu \nabla ^2 v^2[/tex] is diffusion term
and
Reynolds equation
(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication
(2) it is drive in 1886 from Navier Stokes law
(3) equation is attach
here
A soft tennis ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.55 m. (Assume that the positive direction is upward.) (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms. (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
Answer:
a)[tex]v=6.19m/s[/tex]
b)[tex]v=5.51m/s[/tex]
c)[tex]a=3.3*10^{3}m/s^{2}[/tex]
d)[tex]x=5.78*10^{-3}m[/tex]
Explanation:
h1=195m
h2=1.55
a) Velocity just before the ball strikes the floor:
Conservation of the energy law
[tex]E_{o}=E_{f}[/tex]
[tex]E_{o}=mgh_{1}[/tex]
[tex]E_{f}=1/2*mv^{2}[/tex]
so:
[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]
b) Velocity just after the ball leaves the floor:
[tex]E_{o}=E_{f}[/tex]
[tex]E_{o}=1/2*mv^{2}[/tex]
[tex]E_{f}=mgh_{2}[/tex]
so:
[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]
c) Relation between Impulse, I, and momentum, p:
[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]
d) The compression of the ball:
The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.
[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]
Kinematics equation:
[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]
Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.
a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3
So:
[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]
Two train whistles have identical frequencies of 175 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 4.05 beats/s when the whistles operate together. What are the two possible speeds and directions the moving train can have? slower speed m/s Correct: Your answer is correct. faster speed m/s Changed: Your submitted answer was incorrect. Your current answer has not been submitted.
Answer:
The speed of the train is 7.75 m/s towards station.
The speed of the train is 8.12 m/s away from the station.
Explanation:
Given that,
Frequency of the whistles f= 175 Hz
Beat frequency [tex]\Delta f= 4.05 Hz[/tex]
Speed of observer = 0
We need to calculate the frequency
Using formula of beat frequency
[tex]\Delta f=f'-f[/tex]
[tex]f'=\Delta f+f[/tex]
[tex]f'=4.05+175[/tex]
[tex]f'=179.05\ Hz[/tex]
When the train moving towards station, then the frequency heard is more than the actual
Using Doppler effect
[tex]f'=f(\dfrac{v-v_{o}}{v-v_{s}})[/tex]
[tex]v=v-\dfrac{vf}{f'}[/tex]
Put the value into the formula
[tex]v=343-\dfrac{343\times175}{179.05}[/tex]
[tex]v=7.75\ m/s[/tex]
The speed of the train is 7.75 m/s towards station.
When the train moving away form the station
Again beat frequency
[tex]\Delta f=f-f'[/tex]
[tex]f'=f-\Delta [/tex]
[tex]f'=175-4.05[/tex]
[tex]f'=170.95\ Hz[/tex]
We need to calculate the speed
Using Doppler effect
[tex]f'=f(\dfrac{v-v_{o}}{v+v_{s}})[/tex]
[tex]v=\dfrac{vf}{f'}-v[/tex]
Put the value into the formula
[tex]v=\dfrac{343\times175}{170.95}-343[/tex]
[tex]v=8.12\ m/s[/tex]
The speed of the train is 8.12 m/s away from the station.
Hence, This is the required solution.
An object, initially at rest, moves with a constant acceleration of 10 m/s2. How far will it travel in (a) 2.0 s and (b) 4.0 s? If this object had an initial velocity of 4 m/s, how far will it travel in (C) 2.0 s and (d) 4.0 s?
Answer:
(a) d = 20 m
(b) d' = 80 m
(c) x = 28 m
(d) x' = 96 m
Solution:
As per the question:
Initial velocity of the object, v = 0
Constant acceleration of the object, [tex]a_{c} = 10 m/s^{2}[/tex]
(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:
[tex]d = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]d = 0.t + \frac{1}{2}\times 10\times 2^{2} = 20 m[/tex]
(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:
[tex]d' = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]d' = 0.t + \frac{1}{2}\times 10\times 4^{2} = 80 m[/tex]
Now, when initial velocity, v = 4 m/s, then
(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:
[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]x = 4\times 2.0 + \frac{1}{2}\times 10\times 2^{2} = 28 m[/tex]
(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:
[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]x = 4\times 4.0 + \frac{1}{2}\times 10\times 4^{2} = 96 m[/tex]
A football is kicked from ground level with an initial velocity of 22.0 m/s at angle of 58.5° above the horizontal. How long, in seconds, is the football in the air before it hits the ground? Ignore air resistance. _______ seconds
Final answer:
To find the time the football is in the air before hitting the ground, we can analyze the vertical motion using the given initial velocity and launch angle.
Explanation:
To find the time it takes for the football to hit the ground, we need to analyze the vertical motion of the football. We can use the formula:
t = (2 * vy) / g
where t is the time, vy is the vertical component of the initial velocity, and g is the acceleration due to gravity.
Given that the initial velocity is 22.0 m/s and the launch angle is 58.5°, we can find the vertical component of the velocity using the formula:
vy = v * sin(θ)
where v is the initial velocity and θ is the launch angle.
Using this information, we can calculate the time it takes for the football to hit the ground.
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s and an S-wave travels at 5000 m/s. If P-waves are received at a seismic station 2.00 minute before an S wave arrives, how far away is the earthquake center?
To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. The distance of the earthquake center can be calculated by multiplying the time difference by the speed difference between the two waves.
Explanation:To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. We know that the speed of the P-wave is 9000 m/s and the speed of the S-wave is 5000 m/s.
We can calculate the distance using the formula: distance = speed × time.
So, the distance of the earthquake center can be calculated as follows:
Time difference = 2.00 minutes = 120 secondsDistance = (9000 m/s - 5000 m/s) × 120 secondsDistance = 4000 m/s × 120 secondsDistance = 480000 meters or 480 kilometersLearn more about Calculating distance to earthquake center here:https://brainly.com/question/31108695
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