Answer:
mouse speed is 5 m/s
Explanation:
given data
belt roll b = 3 m/s
head straight h = 4 m/s
to find out
mouse speed s
solution
we will apply here pythagorean theorem
that is
s = [tex]\sqrt{b^{2} +h^{2} }[/tex] ...........................1
put here value in equation 1 as b = 3 and h = 4
so
s = [tex]\sqrt{b^{2} +h^{2} }[/tex]
s = [tex]\sqrt{3^{2} +4^{2} }[/tex]
s = [tex]\sqrt{9 + 16 }[/tex]
s = [tex]\sqrt{25 }[/tex]
s = 5
so mouse speed is 5 m/s
The mouse's speed relative to the factory floor is calculated using the Pythagorean theorem, taking into account both the speed of the mouse and the conveyor belt. The correct answer is 5 m/s, option E.
Explanation:The speed of the mouse relative to the factory floor is calculated using the Pythagorean theorem given that the mouse's motion and the conveyor belt's motion form two sides of a right triangle. In this case, the speed of the mouse is 4 m/s and the speed of the conveyor belt is 3 m/s. The relative speed will therefore be the square root of (mousespeed)² + (conveyorbelt speed)². So, √(4² + 3²) = √(16 + 9) = √25 = 5 m/s. Therefore, the mouse's speed relative to the factory floor is 5 m/s, option E.
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A car is driving at a speed of 59.0 km/h toward an intersection just as the light changes from green to yellow. The driver has a reaction time of 0.750 s and the magnitude of the braking acceleration of the car is 5.50 m/s2. Find the minimum distance xmin the car travels before coming to a stop after the light changes. (Express to three sig figs)
Answer:
The car travels 36.8 m before coming to stop after the light changes
Explanation:
The car moves at a constant speed of 59.0 km/h for 0.750 s before the driver hits the brake.
The equation for the position of an object moving at constant speed is:
x = x0 + v t
where:
x = position at time t
x0 = initial position
v = speed
t= time
Let´s consider the initial position as the position at which the driver sees the traffic light and decides to brake. That will make x0 = 0. Then, the position after 0.750 s will be:
x = 59.0 km/h * 0.750 s (1 h /3600 s) = 0.0123 km (1000 m / 1 km) = 12.3 m
while braking, the car has a negative acceleration, then, the speed is not constant. The position of the car will be given by the following equation:
x = x0 + v0 t + 1/2 a t² ( where a = acceleration and v0 = initial speed)
and the speed can be expressed as follows:
v = v0 + a t
from this equation, we can calculate how much time it takes the car to stop (v = 0):
0 = v0 + a t
-v0 = a t
-v0 / a = t
v0 is the speed of the car as the driver hits the brake (59.0 km/h) and "a" is the acceleration (5.50 m/s²) that will be negative because the car is losing speed. Then:
-59.0 km/h (1000 m / 1 km) (1 h / 3600 s) / (-5.50 m/s²) = 2.98 s
Now, we can calculate the position at this time to know the minimum distance the car travels before coming to stop:
x = x0 + v0 t + 1/2 a t²
now x0 will be the distance traveled after the driver sees the light but before braking ( 12.3 m). v0 will be the speed before braking, 59.0 km / h or 16.4 m/s. Then:
x = 12.3 m + 16.4 m/s * 2.98 s +1/2 (-5.50 m/s²) * (2.98 s)²
x = 36.8 m
Final answer:
The minimum distance the car travels before coming to a stop is 1225 m.
Explanation:
To find the minimum distance the car travels before coming to a stop, we need to consider the distance traveled during the driver's reaction time and the distance traveled while the car is decelerating.
First, we calculate the distance traveled during the reaction time using the equation:
distance = speed × time
distance = 59.0 km/h × 0.750 s = 44.3 m
Next, we calculate the distance traveled while decelerating using the equation:
distance = (initial velocity² - final velocity²) / (2 × acceleration)
distance = (59.0 km/h)² / (2 × 5.50 m/s²) = 1181 m
The minimum distance the car travels before coming to a stop is the sum of the distances traveled during the reaction time and while decelerating:
minimum distance = 44.3 m + 1181 m = 1225 m
George uses crayons to draw a model of the solar system on a sheet of paper. What is a limitation of this model? A. George cannot place the planets in the correct order from the Sun. B. The model cannot show that the planets differ in size. C. The model cannot show that the planets differ in color. D. George cannot move the planets around the Sun.
If 300. mL of water are poured into the measuring cup, the volume reading is 10.1 oz . This indicates that 300. mL and 10.1 oz are equivalent. How many milliliters are in a fluid ounce based on this data?
Answer : 29.7 mL are in a fluid ounce based on this data.
Explanation :
As we are given that 300 mL and 10.1 oz are equivalent. That means,
300 mL = 10.1 oz
or,
10.1 oz = 300 mL
Now we have to determine the volume of fluid in milliliters.
As, 10.1 oz of fluid = 300 mL
So, 1 oz of fluid = [tex]\frac{1oz}{10.1oz}\times 300mL[/tex]
= 29.7 mL
Therefore, 29.7 mL are in a fluid ounce based on this data.
Based on the given data, there are approximately 29.7 milliliters in one fluid ounce. Conversion takes place from ounce to millimeters.
To find the number of milliliters in a fluid ounce based on the given data, we can set up a proportion using the information provided:
300 mL corresponds to 10.1 oz.
300 mL / 10.1 oz = x mL / 1 oz
300 × 1 = 10.1 × x
300 = 10.1 × x
Dividing both sides by 10.1:
300 / 10.1 = x
x ≈ 29.7 mL
Therefore, based on the given data, there are approximately 29.7 milliliters in one fluid ounce.
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Creation of a proton gradient by the electron transport chain represents
Answer:
Potential Energy
Explanation:
The electron transport chain is a series of proteins and organic molecules located in the inner membrane of the mitochondria.During electron transfer and proton pumping electrons are moved from a higher energy level to a lower one and in the process release energy.This energy is used to make ATP.It is stored in the electrochemical gradient of protons and it has to be released for electron transport to continue.
Creation of a proton gradient by the electron transport chain represents a pivotal process in cell respiration known as oxidative phosphorylation, in which the energy from electrons is used to form a gradient of protons. This gradient powers ATP production, converting the energy of electrons to a form suitable for cellular work.
Explanation:The creation of a proton gradient by the electron transport chain represents a crucial step in cell respiration called oxidative phosphorylation. In the mitochondria, the electron transport chain uses energy from electrons to pump protons (hydrogen ions) from the mitochondrial matrix into the intermembrane space, forming a gradient. This proton gradient is the potential energy that drives ATP production when the protons flow back across the membrane via ATP synthase. Therefore, the electron transport chain converts the energy of electrons into a useable form for the cell to perform work.
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An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.4 s. A passenger in the elevator is holding a 3.3 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates?
Answer:
35.71 N
Explanation:
The elevator starts from the rest means its initial velocity is zero.
Given that, the height achieved by the elevator in 1.4 s will be, [tex]S=1m[/tex]
Given that the mass of the bundle which is hold by passenger is, [tex]m=3.3 kg[/tex]
Now according to second equation of motion.
[tex]S=ut+\frac{1}{2}at^{2}[/tex]
Here, S is the height, u is the initial velocity, t is the time taken, and a is the acceleration.
Now initial velocity is zero therefore,
[tex]S=\frac{1}{2}at^{2}\\a=\frac{2S}{t^{2} }[/tex]
According to the free body diagram tension and acceleration in upward direction and weight is in downward direction.
So,
[tex]ma=T-mg\\T=m(g+a)[/tex]
Put the value of a from the above
[tex]T=m(g+\frac{2S}{t^{2} })[/tex]
Put all the variables.
[tex]T=3.3(9.8+\frac{2\times 1}{1.4^{2} })\\T=3.3(9.8+1.02)\\t=35.71N[/tex]
This the required tension.
Final answer:
To determine the tension in the cord as the elevator accelerates, the acceleration of the elevator is first calculated using the kinematic equation and is found to be approximately 1.02 m/s². Then, the tension is calculated using the formula T = mg + ma and is determined to be approximately 36.04 N.
Explanation:
To calculate the tension in the cord as the elevator accelerates, we first need to determine the acceleration of the elevator. Using the distance traveled and the time it took, we can apply the kinematic equation s = ut + 0.5at2 to find the acceleration 'a', where s is the distance (1 m), u is the initial velocity (0 m/s), and t is the time (1.4 s). After finding 'a', we can calculate the tension (T) in the cord using the formula T = mg + ma, where m is the mass of the bundle (3.3 kg), g is the acceleration due to gravity (9.81 m/s2), and a is the acceleration of the elevator.
First, we find the acceleration:
s = u t + 0.5at2
1 = 0 + 0.5a(1.4)2
a ≈ 1.02 m/s2
Next, we calculate the tension:
T = mg + ma
T = 3.3 x 9.81 + 3.3 x 1.02
T ≈ 32.67 + 3.366
T ≈ 36.04 N
The density of mercury is 13.5939 g/cm3. Calculate the mass in kilograms of a drum full of mercury with the following dimensions: height = 1.100 meters, diameter = 0.492 meters. Assume pi has a value of 3.14.
Answer:
Mass, m = 0.00284 kg
Explanation:
Given that,
Density of mercury, [tex]d=13.5939\ g/cm^3[/tex]
Height of mercury column, h = 1.1 m
Diameter of mercury, d = 0.492 meters
Radius of mercury column, r = 0.246 m
We need to find the mass of a drum. The density is given by :
[tex]d=\dfrac{m}{V}[/tex]
V is the volume of mercury column
[tex]d=\dfrac{m}{\pi r^2h}[/tex]
[tex]m=d\times \pi r^2h[/tex]
[tex]m=13.5939\times 3.14\times (0.246)^2\times 1.1[/tex]
m = 2.84 grams
or
m = 0.00284 kg
So, the mass of a drum full of mercury is 0.00284 kg. Hence, this is the required solution.
Alan is trying to throw a ball across a 50m wide river 100m below him. Alan knows physics so he throws the ball at a 45 degree upward angle at 30m/s for maximum distance. Will the ball cross the river?
Answer:
yes
Explanation:
u = 30 m/s
θ = 45°
h = - 100 m (below)
d = 50 m
g = - 9.8 m/s^2
Use second equation of motion in vertical direction
[tex]h=u_{y}t +\frac{1}{2}a_{y}t^{2}[/tex]
[tex]-100=30\times Sin45\times t -0.5\times 9.8t^{2}[/tex]
[tex]-100 = 21.21 \times t -4.9 \times t^{2}[/tex]
[tex]t=\frac{21.21\pm \sqrt{21.21^{2}+4\times4.9 \times 100}}{9.8}[/tex]
By solving we get
t = 7.17 s
The horizontal distance traveled in this time
= u Cos45 x t = 30 x 0.707 x 7.17 = 152.1 m
This distance is more than the width of the river, So the ball crosses the river.
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Water flows out from the bottom through a small hole.
Answer:
The velocity of water at the bottom, [tex]v_{b} = 28.63 m/s[/tex]
Given:
Height of water in the tank, h = 12.8 m
Gauge pressure of water, [tex]P_{gauge} = 2.90 atm[/tex]
Solution:
Now,
Atmospheric pressue, [tex]P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa[/tex]
At the top, the absolute pressure, [tex]P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa[/tex]
Now, the pressure at the bottom will be equal to the atmopheric pressure, [tex]P_{b} = 1 atm = 1.01\times 10^{5} Pa[/tex]
The velocity at the top, [tex]v_{top} = 0 m/s[/tex], l;et the bottom velocity, be [tex]v_{b}[/tex].
Now, by Bernoulli's eqn:
[tex]P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b} [/tex]
where
[tex]h_{t} - h_{b} = 12.8 m[/tex]
Density of sea water, [tex]\rho = 1030 kg/m^{3}[/tex]
[tex]\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}[/tex]
[tex]\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}[/tex]
[tex]v_{b} = 28.63 m/s[/tex]
Two cars approach each other; both cars are moving westward, one at 79 km/h, the other at 60 km/h. (a) What is the velocity of the first car relative to (in the reference frame of) the second car
Explanation:
Given that,
There are two cars say A and B. Both are approaching each other moving towards westward.
Speed of car A, [tex]v_A=79\ km/h[/tex]
Speed of car B, [tex]v_B=60\ km/h[/tex]
We need to find the velocity of the first car relative to (in the reference frame of) the second car. As both are moving in same direction, there relative velocity is given by :
[tex]v'=v_A-v_B[/tex]
[tex]v'=79\ km/h-60\ km/h[/tex]
v' = 19 km/h
So, the velocity of car A with respect to car B is 19 km/hr. Hence, this is the required solution.
A test rocket is fired straight up from rest. The net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?
Answer:
Maximum elevation, h = 160 meters
Explanation:
Initially, the rocket is at rest, u = 0
Acceleration of the rocket, [tex]a=20\ m/s^2[/tex]
Time, t = 4 s
We need to find the maximum elevation reached by the rocket. It can be calculated using second equation of motion as :
[tex]h=ut+\dfrac{1}{2}\times a\times t^2[/tex]
[tex]h=\dfrac{1}{2}\times 20\times (4)^2[/tex]
h = 160 meters
So, the maximum elevation reached by the rocket is 160 meters. Hence, this is the required solution.
Answer:
486.5 m
Explanation:
Initial velocity is zero as the rocket is fired from rest. u = 0.
Displacement of the rocket during this time:
s = ut +0.5 at²
s = 0+0.5 ×20×4²
s = 160 m
The final velocity at the end of 4 s is:
v = u + at
v = 0 + (20)(4)
v = 80 m/s
This will become the initial velocity for the next half of the motion.
At the maximum elevation, velocity is zero. v = 0
Acceleration due to gravity always acts downwards.
[tex]s=\frac{v^2-u^2}{2a}\\s=\frac{0-80^2}{2\times -9.8} = 326.5 m[/tex]
Thus, the maximum elevation that test rocket would reach is:
326.5 m + 160 m = 486.5 m
what are the components of vector c
A. Cx = -5.20m Cy= 3.00m
B. Cx=5.20m Cy= 3.00m
C. Cx= 3.00m Cy=5.20m
D. Cx= -3.00m Cy= -5.20m
Answer:
The answer to your question is: Cx = -3.0 m
Cy = -5.2 m
Explanation:
Vector C is in the third quadrangle then Cx and Cy are negatives. The answer were both components are negatives is letter D. But let's do the operations to prove it.
cos Ф = os/hyp clear os
os = hyp x cosФ
os = 6 x cos 60
os = 6 x 0.5 = 3 but is negative os or Cx = -3 m
sen Ф = as / hyp clear as
as = hyp x sen Ф but is negative as or Cy = -5.2 m
as = 6 x sen 60
as = 6 x 0.87
as = 5.2 m
A vector consists of two components: magnitude and direction. In this case, for vector C, the components are Cx and Cy, where Cx represents the component in the x-direction, and Cy in the y-direction. In the context of this question, the components of vector C are Cx = -5.20m and Cy = 3.00m.
Explanation:In the context of physics, vectors consist of both magnitude and direction, represented in components. The components of a vector are the projections of the vector along the axes. Here, Cx is the component of the vector C in the x-direction, and Cy is the component in the y-direction. Considering the options given, vector C is defined by the components: Cx = -5.20m, Cy = 3.00m if we consider the first option (A). Please note the choice would depend upon the context or given system of coordinates.
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Identify the situation where work is being done.
A.Carrying a box of crayons across the room.
B.Lifting a backpack off the floor
C.Sitting on a stool
D.Holding a football
Answer:
A. carrying a box of crayons across the room.
Explanation:
work is said to be done when a force moves something over a distance.
Answer:
A.
Explanation:
Carrying a box of crayons across the room.
You have two beakers, one filled to the 100-mL mark with sugar (the sugar has a mass of 180.0 g) and the other filled to the 100-mL mark with water (the water has a mass of 100.0 g). You pour all the sugar and all the water together in a bigger beaker and stir until the sugar is completely dissolved.
b. Which of the following is true about the volume of the solution? Explain.
i. It is much greater than 200.0 mL.
ii. It is somewhat greater than 200.0 mL.
iii. It is exactly 200.0 mL.
iv. It is somewhat less than 200.0 mL.
v. It is much less than 200.0 mL.
Answer:
V. It is much less than 200 ml.
Explanation:
The final volume of the sugar-water mixture is gonna be something very close to the volume of water itself. The reason to this contraintuitive answer is that sugar molecules can dissolve and "find spaces inside the water molecular structure". In other words in a sugar beaker or cup the volume is mostly free air, because its a crystalline net structure.
The big change is going to be in the density of the solution and the mass is going always to be preserved. So there will be
180 g from sugar + 100g from water = 280 g of total volume
Density=mass/volume
Density of water=100/100=1
Density of sugar=180/100=1.8
Density of solution aprox=280/100=2.8
An object is dropped from the top of a cliff 640 meters high. Its height above the ground t seconds after it is dropped is 640−4.9t^2. Determine its speed 44 seconds after it is dropped.
Answer:
v = -431.2 m/s
Explanation:
Given that,
Initial position of the object, [tex]x=640-4.9t^2[/tex]
Let v is its speed 44 second after it is dropped. The relation between the speed and the position is given by :
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(640-4.9t^2)}{dt}[/tex]
[tex]v=-9.8t[/tex]
Put t = 44 seconds in above equation. So,
[tex]v=-9.8\times 44[/tex]
v = -431.2 m/s
So, the speed of the ball 44 seconds after it is dropped is 431.2 m/s and it is in moving downwards.
Final answer:
The speed of the object 44 seconds after being dropped from a 640-meter-high cliff is 431.2 m/s. This is found by differentiating the height function to get the velocity function and substituting the time into the velocity equation.
Explanation:
To determine the speed of an object 44 seconds after it is dropped from the top of a 640-meter-high cliff, you can differentiate the height function to find the velocity function.
Given the height formula h(t) = 640 - 4.9t², we get the velocity function v(t) = h'(t) = -9.8t by differentiation. At t = 44 seconds, the speed is simply the absolute value of the velocity. So, we calculate: v(44) = -9.8 * 44
This yields a velocity of -431.2 m/s, and since speed is the magnitude of velocity, the speed is 431.2 m/s.
In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector C→ has a magnitude of 13.8 m and is angled 20.7° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of B→? State your angle as a positive number.
Answer:
[tex]|B|=27.00425726m[/tex]
[tex]\alpha =210.3781372[/tex]°
Explanation:
Let's use the component method of vector addition:
[tex]A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844[/tex]
Now, we know:
[tex]C_x=A_x+B_x\\\\C_y=A_y+b_y[/tex]
So:
[tex]B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774[/tex]
Now lets calculate the magnitude of the vector B:
[tex]|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m[/tex]
Finally its angle is given by:
[tex]\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344[/tex]°
Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.
Vector is quantity. The magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.
What is Vector?A Vector is a quantity in physics that has both magnitude and direction.
We know that in order to add to vector we need to divide the vector into two parts, a sine(Vertical) and a cosine(Horizontal), therefore,
The vertical addition of the vectors A and B can be written as,
[tex]\vec A_y +\vec B_y = \vec C_y[/tex]
[tex]\vec A(Sin\ \theta_A) +\vec B(Sin\ \theta_B) = \vec C(Sin\ \theta_C)[/tex]
[tex]13.6(Sin\ 40.2^o) +\vec B(Sin\ \theta_B) = 13.8(Sin\ 20.7^o)\\\\\vec B(Sin\ \theta_B ) =-3.9[/tex]
The Horizontal addition of the vectors A and B can be written as,
[tex]\vec A_x +\vec B_x = \vec C_x[/tex]
[tex]\vec A(Cos\ \theta_A) +\vec B(Cos\ \theta_B) = \vec C(Cos\ \theta_C)[/tex]
[tex]13.6(Cos\ 40.2^o) +\vec B(Cos\ \theta_B) = 13.8(Cos\ 20.7^o)\\\\\vec B(Cos\ \theta_B ) =2.5215[/tex]
As the value of Sin is negative and the value of Cos is positive, therefore, Vector B will lie in the fourth quadrant.
The angle of Vector B,
[tex]\dfrac{\vec B\ Sin\ \theta_B}{\vec B\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\dfrac{\ Sin\ \theta_B}{\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\Tan\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\theta_B = Tan^{-1}\ \dfrac{-3.9}{2.5215}\\\\\theta_B = -57.12^o[/tex]
Thus, the angle of vector B is 57.12° clockwise from the -x direction.
In order to make the angle positive, we can deduct the value from 360°,
[tex]\theta_B = -57.12^o\\\\\theta_B = 360^o -57.12^o\\\\\theta_B = 302.88^o[/tex]
The magnitude of Vector B,
We know the value of the Perpendicular component of vector B,
[tex]\vec B(Sin\ \theta_B ) =-3.9\\\\\vec B(Sin\ -57.12^o ) =-3.9\\\\\vec B= 4.644\rm\ m[/tex]
Hence, the magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.
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Which of the following describes the number of times an analog wave is measured each second during an analog-to-digital conversion? Select one: A. Converting rate B. Simplifying rate C. Conversion rate D. Sampling rate
The term that describes the number of times an analog wave is measured each second during an analog-to-digital conversion is called the 'Sampling Rate'. It refers to the number of samples per second taken from a continuous signal to make a discrete signal.
Explanation:The number of times an analog wave is measured each second during an analog-to-digital conversion is represented by option D. Sampling Rate. This term is used in digital signal processing and refers to the number of samples per second (or per other unit) taken from a continuous signal to make a discrete or digital signal. For audio, this is typically done in hertz (Hz).
For example, the standard sampling rate for audio is 44.1 kHz (kilohertz, or thousands of hertz), meaning the original wanalog signal is sampled over 44,000 times per second. This is to ensure a faithful reproduction of the sound when it's converted back into an analog signal for playback.
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The function s(t) represents the position of an object at time t moving along a line. Suppose s( 1 )=123 and s( 3 )=173. Find the average velocity of the object over the interval of time [1,3]?
Answer:
The average velocity is [tex]v_{average}=25[/tex].
Explanation:
The average velocity is calculated in the following way:
If [tex]s(t)[/tex] represents the position of an object at time [tex]t[/tex],
and [tex]s(t_{1})=a[/tex] ; [tex]s(t_{2})=b[/tex], the average velocity is defined in that interval as:
[tex]v_{average}= \frac{final.position-initial.position}{elapsed.time}=\frac{b-a}{t_{2}-t_{1}}[/tex]
Taking the data from the question:
[tex]v_{average}=\frac{173-123}{3-1}=\frac{50}{2}=25[/tex].
An emf is induced in a conducting loop of wire 1.07 m long as its shape is changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 4.36 s and the local 0.115-T magnetic field is perpendicular to the plane of the loop.
Answer:
0.517 mV
Explanation:
Length of wire = 1.07 m
For square:
Perimeter = 1.07 m
Let a be the side of square
So, 4a = 1.07
a = 0.2675 m
Area of square, A 1 = side x side = 0.2675 x 0.2675 = 0.07156 m^2
For circle:
Circumference = 1.07 m
Let r be the radius of circle
So, 2 π r = 1.07
2 x 3.14 x r = 1.07
r = 0.1704 m
Area of circle, A 2 = π r^2 = 3.14 x 0.1704 x 0.1704 = 0.09115 m^2
Change in area, dA = A2 - A1 = 0.09115 - 0.07156 = 0.0196 m^2
Time taken in changing the area, dt = 4.36 s
Magnetic field, B = 0.115 T
According to the Farady's law of electromagnetic induction
[tex]e = \frac{d\phi }{dt}=\frac{dBA }{dt}=B\frac{dA }{dt}[/tex]
[tex]e = 0.115\times \frac{0.0196}{4.36}[/tex]
e = 5.17 x 10^-4 V
e = 0.517 mV
Thus, the induced emf is 0.517 mV.
Two beetles run across flat sand,starting at the same point. beetle 1 runs 0.50 m due east,then 0.80 m at 30° north of due east. beetle 2 also makes two runs; the first is 1.6 m at 40° east of due north.what must be (a) the magnitude and (b) the direction of its second run if it is to end up at the new location of beetle 1?
Final answer:
To determine the magnitude and direction of the second run of beetle 2 in order to end up at the new location of beetle 1, vector addition can be used. The magnitude can be found using the Pythagorean theorem, and the direction can be calculated using trigonometry.
Explanation:
To determine the magnitude and direction of the second run of beetle 2 in order to end up at the new location of beetle 1, we can break down the given runs and use vector addition. Beetle 1 runs 0.50 m due east and then 0.80 m at 30° north of due east. Beetle 2 runs 1.6 m at 40° east of due north. Adding these vectors, we can find the resultant vector, which represents the displacement from the starting point to the new location of beetle 1. This resultant vector has both magnitude and direction.
To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The sum of the squares of the magnitudes of the individual vectors is equal to the square of the magnitude of the resultant vector. Using trigonometry, we can calculate the angle that the resultant vector makes with the east direction. This angle represents the direction of the second run of beetle 2.
Which of the following are examples of 1-D motion?
a) A person pacing back and forth down a hallway in a straight line
b) An airplane traveling from Boise to Seattle
c) A student walking from Boone to KAIC
d) A car driving on a straight road
Answer: a, d
Explanation:
A- eaven if the direction changes it's still 1D
B- an airplane always needs more than 1D for take off and landing
C- it isn't likely to get there by an straight way without any land relief
D- a car in a straight road its 1D if the load has no land relief.
Which of the following statements is/are true?
Check all that apply.
- A conservative force permits a two-way conversion between kinetic and potential energies.
- The work done by a conservative force depends on the path taken.
- A potential energy function can be specified for a nonconservative force.
- A potential energy function can be specified for a conservative force.
- The work done by a nonconservative force depends on the path taken.
- A nonconservative force permits a two-way conversion between kinetic and potential energies.
- A conservative force permits a two-way conversion between kinetic and potential energies.
- A potential energy function can be specified for a conservative force.
- The work done by a nonconservative force depends on the path taken.
Answer:
- A conservative force permits a two-way conversion between kinetic and potential energies. TRUE
- The work done by a conservative force depends on the path taken. FALSE
- A potential energy function can be specified for a nonconservative force.
FALSE
- A potential energy function can be specified for a conservative force. TRUE
- The work done by a nonconservative force depends on the path taken. TRUE
- A nonconservative force permits a two-way conversion between kinetic and potential energies. TRUE
- A conservative force permits a two-way conversion between kinetic and potential energies. FALSE
- A potential energy function can be specified for a conservative force. TRUE
- The work done by a nonconservative force depends on the path taken. TRUE
Explanation:
A conservative force permits a two-way conversion between kinetic and potential energies. TRUE
The action of conservative force on a system can produce energy potential and kinetic. Example of this: the gravitational force. This claim that the work by extenal forces ( conservatives and non conservatives) is equal to the variation of kinetic energy of the system so the work made by conservative forces can modify both potential and kinetic energy.
- The work done by a conservative force depends on the path taken. FALSE
This kind of force can be obtained from potential function so the work made by this kind of force depend only to initial and final point of teh path made.
- A potential energy function can be specified for a nonconservative force.
FALSE
Considering that the work made by kind of force depend of the taken path they kind of forces can not be determined by a potential fuction.
- A potential energy function can be specified for a conservative force. TRUE
This is as consequence of the definition of conservative force that it can be determined from a potential function.
- The work done by a nonconservative force depends on the path taken. TRUE
This kind of force can not be obtained from potential function so the work made by this kind of force depend of the path taken to do this.
- A nonconservative force permits a two-way conversion between kinetic and potential energies. TRUE
A nonconservative force permits conversion to kinetic energy plus potential energy during it made work over the system. This statement is supported by taking into account the energy conservation for system, this claim that the work by extenal forces ( conservatives and non conservatives) is equal to the variation of kinetic energy of the system so the work made by non conservative forces can modify both potential and kinetic energy.
- A conservative force permits a two-way conversion between kinetic and potential energies. FALSE
As the conservative force is determined from a potential function it can only modify the potential energy of the system.
- A potential energy function can be specified for a conservative force. TRUE
This is as consequence of the definition of conservative force that it can be determined from a potential function.
- The work done by a nonconservative force depends on the path taken. TRUE
This kind of force can not be obtained from a potential function so the work made by this kind of force depend of the path taken.
A conservative force permits a two-way conversion between kinetic and potential energies. The work done by a conservative force depends on the path taken. A potential energy function can be specified for a conservative force.
Explanation:A conservative force permits a two-way conversion between kinetic and potential energies. A potential energy function can be specified for a conservative force. The work done by a conservative force depends on the path taken.
A nonconservative force does not permit a two-way conversion between kinetic and potential energies. The work done by a nonconservative force depends on the path taken. A potential energy function cannot be specified for a nonconservative force.
Lena is studying the properties of light in a laboratory. If she increases the amplitude of the light waves she is studying, what effect will this have on her perception of the light?
Answer:
A brighter light
Explanation:
Light waves travel through space via light particles called photons. This particles have in essence 2 properties: 1. Amplitude and 2.Frequency. The first one has to do with the intensity of light we see and the second one has to do with the energy (color). If we change only the amplitude, we will see a lighter or darker light and will keep the same color in all amplitude changes. But if we modify the frequency, the intensity will keep the same and the color changes as we move into the light spectrum.
Thus, increasing the amplitude, we will perceive a brigher light.
Car B is following Car A and has a greater speed than Car A. The two cars are moving in a straight line and in the same direction, and have the same mass. In situation one, Car A is traveling at 10 mph and Car B at 20 mph. In situation two, Car A is traveling at 30 mph and Car B at 40 mph. Assuming a perfectly inelastic collision in which the cars stick together after the collision, which of the following will be true?
a. The force of the collision in the two situations will be equal.
b. Situation two will cause the greater force of collision.
c. Situation one will cause the greater force of collision.
Answer:
we see that force of collision is the same in both the case.
Explanation:
Let the time of impact in both the situation be t and mass of each be m
Applying conservation of momentum in the first case
m₁v₁ + m₂v₂ = (m₁ +m₂ ) v
m x 20 + m x 10 = 2m x v
v = 15 mph.
So the speed of B will be reduced from 20 to 15 mph and speed of A will be increased from 10 to 15 mph.
Considering impact on B only
Impulse on B is equal to change in momentum
F X t = m ( 20 - 15 )
F is force of collision .
F = 5m / t
In the second case ,
Applying conservation of momentum in the second case
m₁v₁ + m₂v₂ = (m₁ +m₂ ) v
m x 40 + m x 30 = 2m x v
v = 35 mph.
So the speed of B will be reduced and speed of A will be increased.
Considering impact on B only
Impulse on B is equal to change in momentum
F X t = m ( 40 - 35 )
F is force of collision .
F = 5m / t
So we see that force of collision is the same in both the case.
: Lithium has an atomic number of 3, which means that it has 3 protons in its nucleus and 3 orbiting electrons. If it loses its outermost electron to another element, what will be the electrical charge?
Answer:
+1
Explanation:
An atom in the neutral state has the same number of protons and electrons. Since protons carry the positive charge and electrons carry negative charge of equal magnitude as that of protons, so, in neutral state the overall charge on the atom is zero.
Atomic number of Lithium is 3. Under neutral state it has 3 protons and 3 electrons. So, its overall electric charge is 0.
If an atom of Lithium loses one of its outermost electron, it is left with 2 electrons and 3 protons. Since, number of protons is 1 more than the number of electrons, the electrical charge on Lithium atom would be positive and the magnitude of charge will be equal to the number of electrons lost, which is 1 in this case. The magnitude can also be calculated as difference in the number of protons and electrons.
Therefore, on losing one electron, the electric charge on Lithium atom would be +1.
A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no appreciable air drag. Compute the position of the sandbag at 0.250 s after its release.
Answer:
Y = 40.94m
Explanation:
The initial speed of the sandbag is the same as the balloon and so is its position, so:
[tex]Y = Yo + Vo*t-\frac{g*t^2}{2}[/tex]
Replacing these values:
Yo = 40m Vo = 5m/s g = 9.81m/s^2 t = 0.25s
We get the position of the sandbag:
[tex]Y = 40+5*(0.25)-\frac{9.81*(0.25)^2}{2}[/tex]
Y = 40.94m
Light is a form of electromagnetic wave, containing oscillating electric and magnetic fields. The wave amplitude detector mentioned above shows how the electric field oscillates in time at the location of the probe. The amplitude of the wave at the location of the probe is equal to the maximum electric field measured. How does the amplitude of the wave depend on the distance from the source?
Answer:
the decrease in intensity is due to the conservation of energy in the wavefront.
Explanation:
Electromagnetic waves are transverse waves that comply with the principle of conservation of energy, these are formed by the variation of an electric and / or magnetic field and travels spherically from their point of origin.
By the principle of conservation of energy after the wave is emitted, the energy of it is distributed throughout the space, generally in spherical form. To conserve energy the density should decrease as the radius of the sphere increases, which is the inverse of the radius squared (1 / r²)
The previous decrease is observed in the decrease of the amplitude of the wave, since the intensity is the square of the electric field.
In summary, the decrease in intensity is due to the conservation of energy in the wavefront.
Final answer:
The amplitude of light decreases as the distance from the source increases due to the inverse square law, where the wave's energy spreads out over a larger area as it propagates.
Explanation:
The amplitude of an electromagnetic wave, such as light, is the maximum field strength of the electric and magnetic fields. In terms of the dependence of amplitude on the distance from the source, as light travels away from the source, the amplitude decreases. This happens because the wave's energy spreads out as the wave propagates, and because the energy of the wave is directly related to its amplitude, a reduction in energy with distance leads to a decrease in amplitude. If we consider a point source of light, the intensity (and therefore the amplitude) of the light diminishes in proportion to the square of the distance from the source, following the inverse square law.
Which of the following is NOT part of the kinetic theory of gases?
a. There is very little empty space in a gas.
b. A gas is composed of very small particles.
c. Gas particles do not attract or repel one another.
d. Gas particles move faster when the temperature increases.
Answer:
Option a
Explanation:
The three basic and major points of the kinetic theory of gases are listed below:
1. In the collision between the gas molecules, no loss or gain of energy takes place.
2. According to the theory, the gaseous molecules exhibits constant and linear motion.
3. The molecules of the gas occupies very little space as compared to the container. Thus the empty space in a gas is not very little.
Answer:
The option that does NOT correspond to the kinetic theory of gases is the option a."There is very little empty space in a gas."
Explanation:
The kinetic theory of gases allows to deduce the properties of the ideal gas using a model in which the gas molecules are spheres that comply with the laws of classical mechanics. This theory states that heat and movement are related, that particles of all matter are moving to some extent and that heat is a sign of this movement.
The postulates of this theory are:
A gas is formed by a large number of spherical particles whose size is negligible compared to the distance between the particles. That is, there are no attractive forces between the molecules of a gas .The molecules move in a straight line at high speed and only interact when they collide. Crashes between particles and with the vessel walls are considered perfectly elastic, conserving translational kinetic energy. Expressed in other words, gas molecules possess kinetic energy. In the movement, the gas molecules collide elastically with each other and with the walls of the container that contains them in a perfectly random way, that is, in each shock the energy is delivered from one particle to another, and therefore they can continue in constant motion The frequency of collisions with the walls of the vessel explains the pressure exerted by the gases. The energy of such particles can be converted into heat or another form of energy. but the total kinetic energy of the molecules will remain constant if the volume and temperature of the gas do not vary; Therefore, the pressure of a gas is constant if the temperature and volume do not change. Then, an increase in the temperature of a gas also increases the speed at which the particles move.Taking into account the above, the option that does NOT correspond to the kinetic theory of gases is the option a."There is very little empty space in a gas."
A water rocket can reach a speed of 76 m/s in 0.060 seconds from launch.
What is its average acceleration?
Answer:
1300 m/s²
Explanation:
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (76 m/s − 0 m/s) / 0.060 s
a = 1266.67 m/s²
Rounded to two significant figures, a ≈ 1300 m/s².
you and your friend left a bus terminal at the same time and traveled in opposite directions. Your bus was in heavy traffic and had to travel 20 miles per hour slower than your freind's bus. After 3 hours, the buses were 270 miles apart. How fast was each bus going?
Answer:
The rate at which bus 1 is going is 55 mph
The rate at which bus 1 is going is 35 mph
Explanation:
As per the question:
Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h
Thus
Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.
R' = R - 20
The distance is given as the product of rate and time:
d = Rt (1)
Now, the total distance given is 270 miles:
d + d' = 270
Now, using eqn (1):
Rt + R't = 270
3(R + R - 20) = 270
6R = 270 + 60
R = 55 mph
R' = R - 20 = 55 - 20 = 35 mph
Answer:
speed of the two vehicle are 55 mph and 35 mph
Explanation:
given,
speed of friends vehicle = x mph
speed of your vehicle = (x - 20) mph
when both travel in opposite direction
distance between the two buses = 270 miles
distance = speed × time
270 = 3(x) + 3(x-20)
90 = 2 x -20
x = 55 mph
now, speed of other vehicle is (55-20) = 35 mph
hence, speed of the two vehicle are 55 mph and 35 mph
A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and neglecting air resistance and the earth’s curvature, it would have hit the ground in1. There is no way to tell from the information given.
2. less than 0.5 sec.
3. 0.5 sec.
4. more than 0.5 sec.
Answer:
3. 0.5 sec.
Explanation:
A bullet fired horizontally follows a projectile motion, which consists of two independent motions:
- A horizontal motion with constant speed
- A vertical motion with constant acceleration, g = 9.8 m/s^2, towards the ground
The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:
[tex]y(t) = h + v_{0y} t - \frac{1}{2}gt^2[/tex]
where y is the vertical position at time t, h is the initial height, and [tex]v_{0y}[/tex] is the initial vertical velocity of the bullet.
Since the bullet is fired horizontally, [tex]v_{0y}=0[/tex]. So the equation becomes
[tex]y(t) = h - \frac{1}{2}gt^2[/tex]
And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:
[tex]t=\sqrt{\frac{2h}{g}}[/tex]
As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.
We have that for the Question "A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and neglecting air resistance and the earth’s curvature, it would have hit the ground in1. There is no way to tell from the information given." it can be said that the time will remain the same
T=0.5
Option 2
From the question we are told
A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and neglecting air resistance and the earth’s curvature, it would have hit the ground in1. There is no way to tell from the information given.
1. less than 0.5 sec.
2. 0.5 sec.
3. more than 0.5 sec.
Generally
When speed is increased and the Range also increased with respect to the speed
Therefore'
The time will remain the same
T=0.5
Option 2
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