Answer:
a) vertical position at 2s: 50.68 m
b) horizontal position at 2s: 13.05 m
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being the equations to find the position as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=7.2 m/s[/tex] is the rock's initial speed
[tex]\theta=25\°[/tex] is the angle at which the rock was thrown
[tex]t=2 s[/tex] is the time
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=25 m[/tex] is the initial height of the rock
[tex]y[/tex] is the height of the rock at 2 s
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin with the answers:
a) Vertical position at 2 s:
[tex]y=25 m+(7.2 m/s)sin(25\°) (2 s)-\frac{(9.8m/s^{2})(2)^{2}}{2}[/tex] (3)
[tex]y=25 m+6.085 m-19. 6 m[/tex] (4)
[tex]y=50.68 m[/tex] (5) This is the vertical position at 2 s
b) Horizontal position at 2 s:
[tex]x=(7.2 m/s) cos(25\°) (2 s)[/tex] (5)
[tex]x=13.05 m[/tex] (6) This is the horizontal position at 2 s
Explanation:
It is given that,
Position of the man, h = 25 m
Initial velocity of the rock, u = 7.2 m/s
The rock is thrown upward at an angle of 25 degrees.
(a) Let y is the vertical position of the rock at t = 2 seconds. Using the equation of kinematics to find y as :
[tex]y=-\dfrac{1}{2}gt^2+u\ sin\theta\ t+h[/tex]
[tex]y=-\dfrac{1}{2}\times 9.8(2)^2+7.2\ sin(25)\times 2+25[/tex]
y = 11.48 meters
(b) Let x is the horizontal position of the rock at that time. It can be calculated as :
[tex]x=u\ cos\theta \times t[/tex]
[tex]x=7.2\ cos(25)\times 2[/tex]
x = 13.05 meters
Hence, this is the required solution.
The Jurassic Park ride at Universal Studios theme park drops 25.6 m straight down essentially from rest. Find the time for the drop and the velocity at the bottom.
Answer:
V=22.4m/s;T=2.29s
Explanation:
We will use two formulas in order to solve this problem. To determine the velocity at the bottom we can use potential and kinetic energy to solve for the velocity and use the uniformly accelerated displacement formula:
[tex]mgh=\frac{1}{2}mv^{2}\\\\X= V_{0}t-\frac{gt^{2}}{2}[/tex]
Solving for velocity using equation 1:
[tex]mgh=\frac{1}{2}mv^{2} \\\\gh=\frac{v^{2}}{2}\\\\\sqrt{2gh}=v\\\\v=\sqrt{2*9.8\frac{m}{s^2}*25.6m}=22.4\frac{m}{s}[/tex]
Solving for time in equation 2:
[tex]-25.6m = 0\frac{m}{s}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\-51.2m=-9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{\frac{51.2m}{9.8\frac{m}{s^{2}}}}=2.29s[/tex]
At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she travel in the next 6.30 s? (Assume the deceleration of 1.70 m/s2 is constant over the full 6.30 s.)
Answer:
x=22.33m
Explanation:
Kinematics equation for constant deceleration:
[tex]x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m[/tex]
A stone is dropped into a river from a bridge 44.0 m above the water. Another stone is thrown vertically down 1.72 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
Answer:
u₀ = 17.14 m/s
Explanation:
given,
bridge height = 44 m
initial speed of the first stone = 0 m/s
initial speed of the second stone = ?
difference after which the second stone is thrown = 1.72 s
for stone 1
[tex]h = ut + \dfrac{1}{2}gt^2[/tex]
[tex]h =\dfrac{1}{2}gt_1^2[/tex]
for stone 2
[tex]h = u_0 (t_1-t) + \dfrac{1}{2}g (t_1-t) ^2[/tex]
[tex]t_1 =\sqrt{\dfrac{1}{2}gh}[/tex]
[tex]t_1 = \sqrt{\dfrac{1}{2}\times 9.81\times 44}[/tex]
t₁ = 14.69 s
[tex]44 = u_0 \times 1.72 + \dfrac{1}{2}g\times 1.72 ^2[/tex]
u₀ = 17.14 m/s
An astronaut must journey to a distant planet, which is 211 light-years from Earth. What speed will be necessary if the astronaut wishes to age only 15 years during the trip? (Give your answer accurater to five decimal places.) Hint: The astronaut will be traveling at very close to the speed of light. Therefore, approximate the dilated trip time At to be 211 years
Answer:
The speed necessary is 0.99747 cExplanation:
We know that the equation for time dilation will be:
[tex]\Delta t = \frac{\Delta t'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
where Δt its the time difference measured from Earth, and Δt' is the time difference measured by the astronaut.
Lets work a little the equation
[tex] \sqrt{1-\frac{v^2}{c^2}} = \frac{\Delta t'}{\Delta t}[/tex]
[tex] 1-\frac{v^2}{c^2}= (\frac{\Delta t'}{\Delta t})^2[/tex]
[tex] \frac{v^2}{c^2}= 1 - (\frac{\Delta t'}{\Delta t})^2[/tex]
[tex] \frac{v}{c}= \sqrt{ 1 - (\frac{\Delta t'}{\Delta t})^2 }[/tex]
[tex] v = \sqrt{ 1 - (\frac{\Delta t'}{\Delta t})^2 } c [/tex]
So, we got our equation. Knowing that Δt=211 years and Δt'=15 years
then
[tex] v = \sqrt{ 1 - (\frac{15 \ y}{211 \ y})^2 } c [/tex]
[tex] v = 0.99747 c [/tex]
To calculate the necessary speed, we can use the time dilation formula. Given the time experienced by the astronaut and the time experienced on Earth, we can solve for the velocity using the Lorentz factor. Using the approximate trip time of 211 years, we can calculate the necessary speed with the given equation.
Explanation:To calculate the speed necessary, we can use the time dilation formula:
Δt' = Δt / γ
Where Δt' is the time experienced by the astronaut, Δt is the time experienced on Earth, and γ is the Lorentz factor given by γ = 1 / √(1 - (v² / c²)), where v is the velocity of the astronaut and c is the speed of light.
Given that the astronaut wishes to age only 15 years during the trip, we can approximate the dilated trip time to be 211 years. Substituting these values, we have:
15 = 211 / γ
Simplifying the equation, we find:
γ = 211 / 15
Using this value of γ, we can calculate the velocity of the astronaut:
v = √((1 - (1 / γ²)) * c²)
Substituting the value of γ, we have:
v = √((1 - (1 / (211 / 15)²)) * c²)
A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a steady force P. The water is leaking out of the bucket at a steady rate such that the bucket is empty after a time T. Find the velocity of the bucket at the instant it becomes empty. Express your answer in terms of P, M, m, T, and g, the acceleration due to avily. Constant Rate Leak"
Answer:
[tex]V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT[/tex]
Explanation:
Given that
Constant rate of leak =R
Mass at time T ,m=RT
At any time t
The mass = Rt
So the total mass in downward direction=(M+Rt)
Now force equation
(M+Rt) a =P- (M+Rt) g
[tex]a=\dfrac{P}{M+Rt}-g[/tex]
We know that
[tex]a=\dfrac{dV}{dt}[/tex]
[tex]\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g[/tex]
[tex]\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt[/tex]
[tex]V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT[/tex]
[tex]V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT[/tex]
This is the velocity of bucket at the instance when it become empty.
The velocity of the bucket when it becomes empty can be found by setting up an equation using Newton's second law and the equation for final velocity. When the bucket becomes empty, its final velocity is equal to the force exerted by the rope multiplied by the time taken for the bucket to become empty, divided by the initial mass of the bucket.
Explanation:When the bucket becomes empty, the force exerted by the rope pulling the bucket up will all be used to accelerate the bucket. We can set up an equation using Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration: F_net = M * a. Since there are no other forces acting on the bucket except for the force exerted by the rope, we can equate the force to the force exerted by the rope: P = M * a.
Since the bucket is initially at rest, its initial velocity is zero. As the bucket becomes empty, the mass of the bucket decreases, but the force exerted by the rope remains constant. The acceleration of the bucket will therefore increase, resulting in an increasing velocity. When the bucket becomes empty, its final velocity can be determined 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 taken for the bucket to become empty. Rearranging the equation, we have v = 0 + Pt/M.
Suppose you made 5 measurements of the speed of a rocket:10.2 m/s, 11.0 m/s, 10.7 m/s, 11.0 m/s and 10.5 m/s. From these measurements you conclude the rocket is traveling at a constant speed. Calculate the mean, standard deviation, and error on the mean.
Answer:
mean = 10.68 m/s
standard deviation 0.3059
[/tex]\sigma_m = 0.14[/tex]
Explanation:
1) [tex]Mean = \frac{ 10.2+11+10.7+11+10.5}{5}[/tex]
mean = 10.68 m/s
2 ) standard deviation is given as
[tex]\sigma = \sqrt{ \frac{1}{N} \sum( x_i -\mu)^2}[/tex]
N = 5
[tex]\sigma =\sqrt{ \frac{1}{5} \sum{( 10.2-10.68)^2+(11-10.68)^2 + (10.7- 10.68)^2+ (11- 10.68)^2++ (10.5- 10.68)^2[/tex]
SOLVING ABOVE RELATION TO GET STANDARD DEVIATION VALUE
\sigma = 0.3059
3) ERROR ON STANDARD DEVIATION
[tex]\sigma_m = \frac{ \sigma}{\sqrt{N}}[/tex]
[tex]= \frac{0.31}{\sqrt{5}}[/tex]
[tex]\sigma_m = 0.14[/tex]
Answer:
Mean = = 10.68 m/s
Standard deviation = σ = 0.342 m/s
Error = 0.153 .
Explanation:
The data has 5 readings.
Let each of the readings be Y
Take average and find the mean X = (10.2+11+10.7+11+10.5)/5 = 53.4/5 = 10.68 m/s.
Take the difference between the data values and the mean and square them individually.
(10.2 - 10.68)² =(-0.48)² = 0.23
(11 - 10.68)² = 0.32² = 0.102
(10.7 - 10.68)² = (-0.02)² = 0.0004
(11-10.68)² =0.32² = 0.102
(10.5-10.68)² = (-0.18)² = 0.0324
Standard deviation = [tex]\sigma = \sqrt{\frac{\sum(Y-X)^2 }{n-1}}[/tex]
= [tex]\sqrt{(0.23+0.102+0.0004+0.102+0.0324)/(5-1)}[/tex]
= [tex]\sqrt{0.1167}[/tex] = 0.342 m/s
Error = Standard deviation / [tex]\sqrt{n}[/tex] = 0.342/5 = 0.153 .
An airplane undergoes the following displacements: First, it flies 40 km in a direction 30° east of north. Next, it flies 56 km due south. Finally, it flies 100 km 30° north of west. Using analytical methods, determine how far the airplane ends up from its starting point.
Answer:
Distance from start point is 72.5km
Explanation:
The attached Figure shows the plane trajectories from start point (0,0) to (x1,y1) (d1=40km), then going from (x1,y1) to (x2,y2) (d2=56km), then from (x2,y2) to (x3,y3) (d3=100). Taking into account the angles and triangles formed (shown in the Figure), it can be said:
[tex]x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)[/tex]
Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:
[tex]d=\sqrt{x3^{2} +y3^{2} }[/tex]
Replacing the given values in the equations, the distance is calculated.
Final Answer:
The airplane ends up approximately 72.53 km from its starting point.
Explanation:
To determine how far the airplane ends up from its starting point after these displacements, we can use vector addition to find the resultant displacement. Since the movements are given in terms of directions relative to north, we can use a coordinate system where north corresponds to the positive y-axis, and east corresponds to the positive x-axis.
Let's start with the first displacement:
1. The airplane flies 40 km in a direction 30° east of north.
We can resolve this displacement into x and y components:
- The x-component (eastward) is 40 km * sin(30°) because the angle is measured from the north (y-axis).
- The y-component (northward) is 40 km * cos(30°) because the angle is with respect to the vertical (north direction).
Using the fact that sin(30°) = 1/2 and cos(30°) = √3/2:
- x1 = 40 km * 1/2 = 20 km
- y1 = 40 km * √3/2 ≈ 40 km * 0.866 = 34.64 km
Now for the second displacement:
2. The airplane flies 56 km due south.
This movement is along the negative y-axis.
- x2 = 0 km (no movement east or west)
- y2 = -56 km (southward)
For the third displacement:
3. The airplane flies 100 km 30° north of west.
- The x-component (westward) will be -100 km * cos(30°) because we are measuring the angle from the north and going west is negative in our coordinate system.
- The y-component (northward) will be 100 km * sin(30°).
Using the trigonometric values found earlier:
- x3 = -100 km * √3/2 ≈ -100 km * 0.866 = -86.6 km
- y3 = 100 km * 1/2 = 50 km
Having found the components for each displacement, we can now sum them up to find the total displacement.
Total x-component (x_total) = x1 + x2 + x3 = 20 km + 0 km - 86.6 km = -66.6 km (westward)
Total y-component (y_total) = y1 + y2 + y3 = 34.64 km - 56 km + 50 km = 28.64 km (northward)
Now, we can determine the magnitude of the resultant displacement vector using the Pythagorean theorem:
R = √(x_total^2 + y_total^2)
R = √((-66.6 km)^2 + (28.64 km)^2)
R = √(4440.96 km^2 + 820.5696 km^2)
R = √(5261.5296 km^2)
R ≈ 72.53 km
So, the airplane ends up approximately 72.53 km from its starting point.
Which of the following is not a fundamental quantity? A. time B. weight C. mass D. length E. electric charge
Answer:
Option B
Explanation:
The fundamental quantities are the basic quantities from which other quantities are derived.
There are a total of 7 fundamental quantities, namely Length, Mass, Time, Luminous Intensity, Amount of substance, Electric Current, Temperature.
From these 7 quantities, the other quantities are derived.
Weight is the product of its mass and the gravitational acceleration 'g' which acts on that mass.
Where mass is the fundamental quantity and 'g' is also derived and hence weight is a derived quantity.
On October 21, 2001, Ian Ashpole of the United Kingdom achieved a record altitude of 3.35 km (11 000 ft) powered by 600 toy balloons filled with helium. Each filled balloon had a radius of about 0.54 m and an estimated mass of 0.27 kg. (a) Estimate the total buoyant force on the 600 balloons.
To estimate the total buoyant force on the 600 balloons filled with helium, we use Archimedes' principle. By calculating the volume of each balloon and using the formula for buoyant force, we can determine the force exerted by each balloon. Multiplying this force by the number of balloons gives us the total buoyant force.
Explanation:To estimate the total buoyant force on the 600 balloons, we need to calculate the buoyant force on each balloon and then multiply it by the number of balloons. The buoyant force on a balloon can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. In this case, the fluid is the surrounding air.
The volume of each balloon can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the balloon. In this case, r = 0.54 m. Using this formula, we can calculate the volume of each balloon to be approximately 0.653 m^3.
The buoyant force on each balloon can be calculated using the formula F = ρVg, where F is the buoyant force, ρ is the density of the fluid, V is the volume of the fluid displaced, and g is the acceleration due to gravity. Since the density of air is approximately 1.225 kg/m^3 and g is approximately 9.8 m/s^2, we can calculate the buoyant force on each balloon to be approximately 7.93 N.
Finally, to calculate the total buoyant force on the 600 balloons, we can multiply the buoyant force on each balloon by the number of balloons: F_total = F_per_balloon * number_of_balloons. Plugging in the values, we get F_total ≈ 7.93 N * 600 = 4,758 N.
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The total buoyant force on the 600 balloons that Ian Ashpole used can be calculated using Archimedes' Principle and the principle of Buoyancy. The resulting force is the difference between the weight of the air displaced by the balloons and the weight of the balloons themselves, multiplied by the number of balloons.
Explanation:This is a classic problem of Archimedes' Principle and Buoyancy, principles in Physics. In simple terms, the buoyant force on a body submerged in a fluid is equal to the weight of the fluid displaced by the body. In the case of a balloon, the buoyant force can be calculated as the difference between the weight of the air displaced by the balloon and the weight of the balloon itself. For a single balloon, this would be:
FB = (weight of the air displaced) - (weight of balloon). But we have 600 balloons, so, we multiply this force by 600 to get the total buoyant force on all the balloons. Given Ian Ashpole used 600 balloons, each with an estimated mass of 0.27 kg and a radius of about 0.54 m, we can calculate the total buoyant force on these balloons.
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Interstellar space is filled with blackbody radiation that has a distribution peaking a his radiation is considered to be a remnant of the "big bang. wavelength of 970 um. What is the corresponding blackbody temperature of this radiation?
Final answer:
The corresponding blackbody temperature of the radiation with a wavelength of 970 um is approximately 2.725 K (Kelvin). The blackbody radiation observed in interstellar space is known as the cosmic microwave background (CMB). This temperature represents the average temperature of the universe at the time the CMB was emitted.
Explanation:
The corresponding blackbody temperature of the radiation with a wavelength of 970 um is approximately 2.725 K (Kelvin).
The blackbody radiation observed in interstellar space is known as the cosmic microwave background (CMB). It is the afterglow of the Big Bang and fills all of space. The blackbody spectrum of the CMB has a temperature of 2.725 K, as determined from observations.
This temperature represents the average temperature of the universe at the time the CMB was emitted. It provides valuable insights into the early universe and supports the idea of the expanding universe.
which of the following best describes the objectives section of a lesson plan
A. The specific learning behaviors you are expecting from the lesson
B. How will you simplify the lesson in the event that the content is too advanced for some learners?
C. How you will go about determining the childrens strengths and weaknesses once the lesson is taught.
D. The specific steps to be taken in carrying out the lesson plan.
Answer:
A. The specific learning behaviors you are expecting from the lesson
Explanation:
Lesson plan is systematic way of approaching subject learning in schools and colleges. A Lesson plan has various section among them there is section of Objectives.
Objectives are defined precise and focused goals of the learning from the particular topic that the student must learn. It is a goal oriented method where aim is already known before its accomplishment at the end of the chapter. Hence option A seem most appropriate answer.
Wes stands on the roof of a building, leans over the edge, and drops a rock. Lindsay waits 1.20 s after Wes releases his rock and throws her own rock straight down at 21.0 m/s. Both rocks hit the ground simultaneously. 1) Calculate the common height from which the rocks were released. Ignore the effects of air resistance. (Express your answer to three significant figures.)
To calculate the common height from which the rocks were released, use the equations of motion. Substitute the given values and solve for the height using the equations h = (1/2)gt^2 and h = v0t + (1/2)gt^2.
Explanation:To calculate the common height from which the rocks were released, we need to use the equations of motion. Let's assume the common height is h. For Wes, the time taken to reach the ground is given as 1.20 s. Using the equation h = (1/2)gt^2, where g is the acceleration due to gravity, we can substitute the values and solve for h. For Lindsay, the time taken to reach the ground is the same, 1.20 s. Using the equation h = v0t + (1/2)gt^2, where v0 is the initial velocity, we can substitute the values and solve for h. By calculating the common height from these two equations, we can determine the height from which the rocks were released.
An object is thrown vertically upward and has a speed of 32.6 m/s when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.
Answer:
The maximum height is 162.67 m.
Explanation:
Suppose the total height is h.
And at the height of 2/3h the speed of an object is,
[tex]u=32.6m/s[/tex]
And the remaining height will be,
[tex]h'=h-\frac{2}{3}h\\ h'=\frac{1}{3}h[/tex]
So, according to question the initial speed is,
[tex]u=32.6m/s[/tex]
Acceleration in the upward direction is negative,
[tex]a=-9.8m/s^{2}[/tex]
And the final speed will be v m/s which is 0 m/s.
Now according to third equation of motion.
[tex]v^{2} =u^{2} -2as[/tex]
Here, v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement.
[tex]0^{2} =32.6^{2} +2(-9.8)\dfrac{h}{3} \\h=\dfrac{3\times 1062.76}{2\times 9.8}\\h=162.67 m[/tex]
Therefore, the maximum height is 162.67 m.
If the world population grows at a constant rate of 1.8% per annum, how many years will it take to double? A) 17.7 years
B) 23.4 years
C) 35.0 years
D) 38.8 years
E) 69.7 years
Answer:
after 38.8 years it will double
correct option is D 38.8 years
Explanation:
given data
population grows rate = 1.8%
to find out
how many years will it take to double
solution
we consider here initial population is x
so after 1 year population will be = (100% + 1.8% ) x = 1.018 x
and after n year population will be = [tex]1.018^{n} x[/tex]
so it will double
2x = [tex]1.018^{n} x[/tex]
take log both side
log 2 = n log (1.018)
n = [tex]\frac{log2}{log1.018}[/tex]
n = 38.853
so after 38.8 years it will double
correct option is D 38.8 years
A rocket is used to place a
synchronoussatellite in orbit about the earth. What is the speed of
thesatelliet in orbit?
Answer:
3073 m/s
Explanation:
The key is that the period of the satellite is 24 hours because it is synchronized with the rotation of the Earth. You can use Kepler's Third law to find the radius of the orbit:
[tex]4\pi^2 r^3 = G MT^2[/tex]
[tex]4\pi^2 r^3= 6.67\times10^{-11}\times6\times10^{24}\times86400^2[/tex]
r = 4.225 x 10^7 meters
So one complete orbit is a distance of
[2πr = 2× π × 4.225 x 10^7 = 26.55 x 10^7 meters
So the speed is
distance / time = 26.55 x 10^7 meters / 86400 seconds =
= 3073 m/s
A ball thrown by a pitcher on a women’s softball team is timed at 56.9 mph. The distance from the pitching rubber to home plate is 47.9 ft. In major league baseball the corresponding distance is 60.5 ft. If the batter in the softball game and the batter in the baseball game are to have equal times to react to the pitch, with what speed must the baseball be thrown? Assume the ball travels with a constant velocity. [Hint: There is no need to convert units; set up a ratio.]
Answer:[tex]v_b=71.86 mph[/tex]
Explanation:
Given
Velocity of soft ball is 56.9 mph
Distance between Pitching rubber to home plate is 47.9 ft
In major league distance is 60.5 ft
Let velocity of baseball is [tex]v_b[/tex]
Let t be the time for ball to reach to batter and its reaction time
since t is same for both case
[tex]\frac{47.9}{56.9}=\frac{60.5}{v_b}[/tex]
[tex]v_b=56.9\times \frac{60.5}{47.9}[/tex]
[tex]v_b=71.86 mph[/tex]
To determine the speed at which the baseball must be thrown to allow for equal reaction times between the softball and baseball batters, we can set up a ratio using the distances from the pitching rubber to home plate in both sports and solve for the desired speed.
Explanation:
To determine the speed at which the baseball must be thrown, we can set up a ratio using the distances from the pitching rubber to home plate in softball and baseball. Since the times for the batters to react should be equal, the distance ratio is equal to the speed ratio. Therefore, we can write the proportion:
(56.9 mph)/(47.9 ft) = x/(60.5 ft)
Where x represents the speed of the baseball. To solve for x, we can cross-multiply and solve for x:
x = (56.9 mph * 60.5 ft) / 47.9 ft
Calculating the right-hand side of the equation gives us the speed of the baseball:
x ≈ 71.92 mph
Therefore, the baseball must be thrown at approximately 71.92 mph to allow for equal reaction times between the softball and baseball batters.
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A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her piston is 19.0 cm in diameter, what is the diameter of the football players' piston?
Answer:
The diameter of the piston of the players equals 55.136 cm.
Explanation:
from the principle of transmission of pressure in a hydraulic lift we have
[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}[/tex]
Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get
[tex]\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}[/tex]
Solving for [tex]D_{2}[/tex] we get
[tex]D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm[/tex]
The diameter of the football players' piston is [tex]\({55.4 \text{ cm}} \).[/tex]
To solve for the diameter of the football players' piston in the hydraulic lift, we start by using the principle of hydraulic systems, which states that pressure is constant throughout the fluid.
Given data:
- Mass of the cheerleader, [tex]\( m_1 = 57.0 \)[/tex] kg
- Mass of four football players, [tex]\( m_2 = 4 \times 120 \)[/tex] kg
- Height difference, [tex]\( h = 1.10 \)[/tex] m
- Diameter of the cheerleader's piston, [tex]\( D_1 = 19.0 \)[/tex] cm
First, convert the diameter of the cheerleader's piston into meters:
[tex]\[ D_1 = 19.0 \text{ cm} = 0.19 \text{ m} \][/tex]
The force exerted by the cheerleader's piston is:
[tex]\[ F_1 = m_1 g \][/tex]
The force exerted by the football players' piston is:
[tex]\[ F_2 = m_2 g \][/tex]
According to Pascal's principle:
[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]
where [tex]\( A_1 \)[/tex] and [tex]\( A_2 \)[/tex] are the cross-sectional areas of the cheerleader's and football players' pistons, respectively.
The area [tex]\( A \)[/tex] of a piston is given by:
[tex]\[ A = \frac{\pi D^2}{4} \][/tex]
Now, calculate [tex]\( A_1 \)[/tex]:
[tex]\[ A_1 = \frac{\pi (0.19)^2}{4} \][/tex]
[tex]\[ A_1 = \frac{\pi \times 0.0361}{4} \][/tex]
[tex]\[ A_1 \approx 0.0284 \text{ m}^2 \][/tex]
Next, find [tex]\( F_1 \)[/tex]:
[tex]\[ F_1 = m_1 g = 57.0 \times 9.8 \][/tex]
[tex]\[ F_1 \approx 558.6 \text{ N} \][/tex]
Now, find [tex]\( A_2 \)[/tex]:
[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]
[tex]\[ A_2 = \frac{F_2 \times A_1}{F_1} \][/tex]
Calculate [tex]\( F_2 \)[/tex]:
[tex]\[ F_2 = m_2 g = 480 \times 9.8 \][/tex]
[tex]\[ F_2 = 4704 \text{ N} \][/tex]
Now, find [tex]\( A_2 \)[/tex]:
[tex]\[ A_2 = \frac{4704 \times 0.0284}{558.6} \][/tex]
[tex]\[ A_2 \approx 0.2397 \text{ m}^2 \][/tex]
Finally, solve for [tex]\( D_2 \)[/tex]:
[tex]\[ A_2 = \frac{\pi (D_2)^2}{4} \][/tex]
[tex]\[ (D_2)^2 = \frac{4 \times 0.2397}{\pi} \][/tex]
[tex]\[ D_2 \approx \sqrt{\frac{0.9588}{\pi}} \][/tex]
[tex]\[ D_2 \approx 0.554 \text{ m} \][/tex]
Convert [tex]\( D_2 \)[/tex] from meters to centimeters:
[tex]\[ D_2 \approx 55.4 \text{ cm} \][/tex]
Two spaceships A and B move in opposite directions from earth at the same speed as measured by an earthbound observer. If an observer on spaceship A measures the velocity of spaceship B to be 0.99c, what is the speed of each spaceship as seen from earth (in units of c)?
Answer:
the magnitude of each spaceships velocity is 0.495c
Explanation:
This is a relative velocity problem, that means that the velocity is relative to the observer, in this case, there is an observer on earth, from this reference point the velocity of each spaceship are the same, but for the observer on a spaceship would be different, we can see it as a velocity difference :
[tex]V=Va-Vb\\[/tex]
because the spaceship B is going on the opposite direction from A, the velocity is negative, the observer on earth says the velocity is the same for both o them, so:
[tex]0.99c=va-(-va)\\0.99c=2va\\va=0.495c[/tex]
vb=-va
A ball is hurled straight up at a speed of 15 m/s, leaving the hand of the thrower 2.00 m above the ground. Compute the times and the ball’s speeds when it passes an observer sitting at a window in line with the throw 10.0 m above the point of release.
Answer:
5.37 m/s
0.98 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s[/tex]
Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s[/tex]
Time taken by the ball to pass the observer sitting at a window is 0.98 seconds
Consider a uniform electric field of 50 N/C directed towards east. if the voltage measured relative to ground at a given point is 80 V what is the voltage at a point 1.0 m directly west of the point? assume a constant electric field?
In a uniform electric field, the electric potential or voltage increases or decreases linearly with distance based on the direction. Given initial voltage of 80V and field strength of 50 N/C, the voltage 1.0 m west increases by 50V to 130V.
Explanation:For the given scenario, we are dealing with a uniform electric field and we need to determine the voltage at a certain distance in the field.
The key relationship between the electric field (E) and voltage (V) in a uniform electric field is that E = ΔV / Δd, where, ΔV is the potential difference and Δd is distance.
Given the electric field strength E = 50 N/C, and the initial voltage V1 = 80 V, we want to find the potential V2 a distance d = 1.0 m to the west, against the direction of the field. Since the electric field is uniform and points towards decreasing potential, the potential a distance d against the direction of field would increase. Therefore, ΔV = E * d = 50 N/C * 1.0 m = 50 V. Thus, the voltage at the point 1.0 m west would be V2 = V1 + ΔV = 80V + 50V = 130V.
Learn more about Voltage in Uniform Electric Field here:https://brainly.com/question/26446532
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Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the total force exerted by the four charges on a charge Q located a distance b along a line perpendicular to the plane of the square and equidistant from the four charges?
Answer:
[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]
Explanation:
Since all the four charges are equidistant from the position of Q
so here we can assume this charge distribution to be uniform same as that of a ring
so here electric field due to ring on its axis is given as
[tex]E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}[/tex]
here we have
x = b
and the radius of equivalent ring is given as the distance of each corner to the center of square
[tex]R = \frac{d}{\sqrt2}[/tex]
now we have
[tex]E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]
so the force on the charge is given as
[tex]F = QE[/tex]
[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]
The magnitude of the force exerted by object W on object Z is F.
Explanation:The magnitude of the force exerted by object W on object Z can be determined by analyzing the geometry and the charges involved. Since objects X and Z are at the midpoints of the sides of the square, they are equidistant from all four charges. As a result, the magnitude of the force exerted by each charge on object Z will be the same as the magnitude of the force exerted by each charge on object X. Therefore, the magnitude of the force exerted by object W on object Z is also F.
discuss whether any work is being done by each of thefollowing
agents and if so, tell whether the work is positive
ornegative?
(a) a chicken scratching the ground
(b) a person studying
(c) a crane lifting a bucket of concrete
(d) the force of gravity on the bucket in part (c)
(e) the leg muscles of a person in the act of
sittingdown.
Answer:
Explained
Explanation:
a) the work done will be positive since the chicken is scratching the ground. Here displacement is along the direction of force.
b) A person studying does no work in the language of physics because there is no displacement.
C) the work is done on the bucket by the crane and work is positive and here the displacement is in the direction of force.
d) Gravitational force act on the bucket in downward direction, here the work done will be negative as the force displacement are opposite to each other.
E)Negative work will be done as the force applied by the muscle is in opposite to the displacement.
Consider a physical pendulum with length of 81.9 cm and mass of 165 g. If the pendulum was released from an angle less than 10°, then calculate the period of the pendulum. (g = 9.80 m/s^2)
Answer:
The period of the pendulum is 1.816 sec.
Explanation:
Given that,
Length = 81.9 cm
Mass = 165 g
Angle = 10°
We need to calculate the period of the pendulum
Using formula of period
[tex]T = 2\pi\sqrt{\dfrac{l}{g}}[/tex]
Where, l = length
g = acceleration due to gravity
Put the value into the formula
[tex]T =2\pi\sqrt{\dfrac{81.9\times10^{-2}}{9.80}}[/tex]
[tex]T=1.816\ sec[/tex]
Hence, The period of the pendulum is 1.816 sec.
A bird flies 2.0km south and then 1.5km 37° east of south. How far will it have to fly to get back to its original place if it flies in a straight line?
Answer:
3.324 km
Explanation:
d1 = 2 km south
d2 = 1.5 km at 37° east of south
Write the displacements in vector form
[tex]\overrightarrow{d_{1}}=-2\widehat{j}[/tex]
[tex]\overrightarrow{d_{2}}=1.5\left (Sin37\widehat{i}-Cos37\widehat{j} \right )=0.9\widehat{i}-1.2\widehat{j}[/tex]
The resultant displacement is given by
[tex]\overrightarrow{d} = \overrightarrow{d_{1}}+ \overrightarrow{d_{2}}[/tex]
[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -2-1.2 \right )\widehat{j}[/tex]
[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -3.2\right )\widehat{j}[/tex]
The magnitude of displacement is given by
[tex]d=\sqrt{0.9^{2}+\left ( -3.2 \right )^{2}}=3.324 km[/tex]
Thus, the bird has to travel 3.324 km in a straight line to return to its original place.
According to the Creative Curriculum model, the _____ sets the context for teaching and learning
A. structure of the classroom
B. children
C. Centers used in the classroom
D. teacher
Answer:
A. structure of the classroom
Explanation:
According to the Creative Curriculum model, the A. structure of the classroom sets the context for teaching and learning.
Creative curriculum model stands to enhance the skill of the students through a structured class room. Creative curriculum goes beyond the rote learning and focuses on big ideas and individual passion of each students.
Calculate the net charge on a substance consisting of a combination of 1.3 × 10^13 protons and 4.4x 10^13 electrons. The elemental charge is 1.6 × 10^-19 C. Answer in units of C
Answer:
The charge transferred equals [tex]-4.96\times 10^{-6}Columbs[/tex]
Explanation:
The charge quantizatrion principle states that the charge is always transferred in terms of product of fundamental charges.
Mathematically [tex]Q=ne[/tex]
where
'e' is the fundamental magnitude of charge
Thus charge transferred by [tex]4.4\times 10^{13}[/tex] electrons equals
[tex]Q_{1}=4.4\times 10^{13}\times -1.6\times 10^{-19}=-7.04\times 10^{-6}Columbs[/tex]
Similarly charge transferred by [tex]1.3\times 10^{13}[/tex] protons equals
[tex]Q_{2}=1.3\times 10^{13}\times 1.6\times 10^{-19}=2.08\times 10^{-6}Columbs[/tex]
Thus the net charge transferred equals
[tex]Q_{1}+Q_{2}=-7.04\times 10^{-6}+2.08\times 10^{-6}\\\\=-4.96\times 10^{-6}columbs[/tex]
The cable supporting a 1950 kg elevator has a maximum strength
of21920 N. What maximum upwardacceleration can it give the elevator
without breaking?
Answer:
1.43 m/s^2
Explanation:
Each time you see mass and force, you will probably be going to need to use Newton's second Law. This law basically shows the relationship between the force being applied on an object and its mass and acceleration:
[tex]F = m*a[/tex]
Now, the force that the cable exerts on the elevator, not only has to accelarate it, but it also has to counter gravity. The maximum tension of the cable minus the weigth of the elevator would give us the net force being applied on the elevator:
[tex]T_{cable} - W_{elevator} = m_{elevator}*a[/tex]
[tex]21920 N - 1950kg*9.81 m/s^2 = 1950 kg*a\\a = \frac{21920 N - 1950kg*9.81 m/s^2}{1950kg} = 1.43 m/s^2[/tex]
A gallon of gasoline contains about 1.3 x 108joules
of energy. A 2000 kg car traveling at 20 m/s skids to astop.
Estimate how much gasoline it will take to bring the car backto the
original speed? To complicate matters further, consider thefact
that only about 15% of the energy extracted from gasolineactually
propels the car. The rest gets exhausted as heat andunburnt
fuel.
Answer:
The amount of gasoline is [tex]2.105\times10^{-2}\ gallons[/tex].
Explanation:
Given that,
Energy contained in gasoline [tex]= 1.3\times10^{8}\ J[/tex]
Mass = 2000 kg
Speed = 20 m/s
Energy used propel the car[tex] E=15\%\ of 1.3\times10^{8}\ J[/tex]
[tex]E=\dfrac{15}{100}\times1.3\times10^{8}[/tex]
[tex]E=19500000 = 1.9\times10^{7}\ J[/tex]
[tex]E=1.9\times10^{7}\ J[/tex]
We need to calculate the work done by the frictional force to stop the car
Using formula of work done
[tex]W=\Delta KE[/tex]
[tex]W=\dfrac{1}{2}m(v_{f}^2-v_{0}^2)[/tex]
[tex]W=\dfrac{1}{2}\times2000\times(0-20^2)[/tex]
[tex]W=-4.0\times10^{5}\ J[/tex]
Therefore,
Work done to bring the car back to its original speed
[tex]W=4.0\times10^{5}\ J[/tex]
[tex]Amount\ of\ gasoline\ needed = \dfrac{W}{E}[/tex]
[tex]Amount\ of\ gasoline =\dfrac{4.0\times10^{5}}{1.9\times10^{7}}[/tex]
[tex]Amount\ of\ gasoline =2.105\times10^{-2}\ gallons[/tex]
Hence, The amount of gasoline is [tex]2.105\times10^{-2}\ gallons[/tex].
You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler has - 12 μC of charge. The tissue has 3 g of mass. What is the minimum charge required to pick up the tissue paper?
Answer:
9.81 × 10⁻¹⁰ C
Explanation:
Given:
Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m
Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C
Mass of the tissue = 3 g = 0.003 Kg
Now,
The force required to pick the tissue, F = mg
where, g is the acceleration due to gravity
also,
The force between (F) the charges is given as:
[tex]F=\frac{kQq}{r^2}[/tex]
where,
q is the charge on the tissue
k is the Coulomb's constant = 9 × 10⁹ Nm²/C²
thus,
[tex]mg=\frac{kQq}{r^2}[/tex]
on substituting the respective values, we get
[tex]0.003\times9.81=\frac{9\times10^9\times(-12\times10^{-6})\times q}{0.06^2}[/tex]
or
q = 9.81 × 10⁻¹⁰ C
Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C
Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way between the plates?
Answer: 8*10^-15 N
Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E
The electric field between the plates is given by:
E = ΔV/d = 500 V/0.01 m=5*10^3 N/C
the force applied to the electron is: F=e*E=8*10^-15 N