Answer:
171.5 m
Explanation:
To find the distance, speed x time
342 x 0.5
171.5 m
Hope this helped!
A rotating fan completes 1150 revolutions every minute. Consider the tip of a blade, at a radius of 20.0 cm. (a) Through what distance does the tip move in one revolution
Answer:
125.6 cm
Explanation:
Applying,
S = 2πr................... Equation 1
Where S = distance moved by the tip in one revolution, r = radius of the rotating fan, π = pie
From the question,
Given: r = 20 cm,
Constant: π = 3.14
Substitute these values into equation 1
S = 2(3.14)(20)
S = 125.6 cm
Hence the distance moved by the tip in one revolution is 125.6 cm
Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless.
Identify all the action/reaction pairs of forces between the three skaters. Draw a free-body diagram for skater 2, in the middle.
Answer:
Following are the responses to the given question:
Explanation:
The mass (mg) is downwards, the typical [tex]N_2[/tex] pressure is upward. This pushing force of [tex]\vec{F}_{3 \to 2}[/tex] is pushed forward by skater 3 on skater 2. (considered as rightward direction). The strength of the slater 2 in reply is the slater [tex]\vec{F}_{2\to 3}[/tex] Skater three moves to the left.
Its push force [tex]\vec{F}_{2\to1}[/tex] imparted to the skateboarder 2 in relation those above forces The force[tex]\vec{F}_ {1\to 2}[/tex] on skater 2 by skater 1 is as a response, to a left. Skater 2's free body system is as follows:
List and briefly explain the incidents leading to the occurrence of any five nuclear accidents that have taken place in different parts of the world.
Answer:
Chernobyl Nuclear Disaster Nuclear Disaster. Japan 2011 Kyshtym Nuclear Disaster. Russia 1957 Windscale Fire Nuclear Disaster. Sellafield, UK 1957 Three Mile Island Nuclear Accident. Pennsylvania, USA 1979
Explanation:
Hope this helps... pls vote as brainliest
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of 1.1 m/s, while Ender recoils with a velocity of -4.3 m/s. Determine the ratio Bonzo/mEnder of the masses of these two enemies.
Answer:
the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
Explanation:
Given the data in the question;
Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s
Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s
the ratio Bonzo/mEnder of the masses of these two enemies = ?
Now, using the law of conservation of momentum.
momentum of both Bonzo and Ender are conserved
so
Initial momentum = final momentum
we have
0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]
[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex] ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]
we substitute
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -4.3 m/s / 1.1 m/s ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -3.9090 ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = 3.91
Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
The diagram shows the molecular structure of ethane. What is the chemical
formula for ethane?
Ethane
H H
H-C-C-H
| |
H H
PLS HELP ME 100 POINTS PLS I NEED HELP QUICK PLS
For this project, you are expected to submit the following:
1. Your Student Guide with completed Student Worksheet
2. Your scale model of the solar system
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Conduct research on the actual sizes of the planets.
a) Do research to find the actual sizes of the Sun and the planets. This information is typically represented as diameter in kilometers (km). Recall that diameter is the length of the imaginary straight line from one side of a figure, such as a sphere, to the opposite side of the figure. This line passes through the center of the figure.
b) Record the actual diameters of the Sun and the planets in the first column of the table in the Student Worksheet.
c) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 3: Determine the scaled sizes of the planets.
a) Go to a reliable website to find a solar system model calculator.
b) Decide how big you want the Sun in your model to be. For example, you could assign your Sun to be 300 mm. Input this figure in the calculator, and the calculator will determine the diameters of the eight planets for you. You want to make sure that the Sun is big enough so that the smallest planet will still be big enough to draw.
c) Record information from the calculator in the second column of the table in the Student Worksheet.
d) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 4: Create a scale model of the solar system.
a) Draw and cut construction paper models of the Sun and the planets using the scaled measurements from the table.
b) Glue the models on the poster board. You can glue or tape poster boards together if necessary. Be sure to put the Sun in the center and to put the planets and a drawing of their orbits in order from nearest to farthest from the Sun.
Note: Remember that in this model, the diameter of the planets is scaled but the distance of the planets from the Sun is not. That means your model does not accurately represent the distances of the planets from the Sun so you need not worry about these measurements.
c) Label the Sun and the planets.
d) Put an attention-catching title above or below your model.
e) Write your name on the back of your poster board.
Step 5: Complete the Student Worksheet.
a) Make sure the table in the Student Worksheet is complete.
b) Answer the questions in the Student Worksheet.
c) Check to make sure you added the sources you used for this project in the Student Worksheet.
Step 6: Evaluate your project using this checklist.
If you can check each of the following boxes, you are ready to submit your project.
Did you conduct research to find the actual size of the Sun and the planets? Did you record this information in the table in the Student Worksheet?
Did you use a solar system model calculator to determine the scaled size of the Sun and planets? Did you record this information in the Student Worksheet?
Did you add the links of the websites you used for this project to the Student Worksheet?
Did you use the scaled sizes to create models of the Sun and the planets?
Did you put your model together in a way that represents the solar system (Sun in the center and planets in order from nearest to farthest from the Sun)?
Did you label each component of your model?
Did you add an attention-catching title above or below your model?
Did you write your name on the back of your poster board?
Did you complete the Student Worksheet at the end of this guide?
Step 7: Revise and submit your project.
a) If you were unable to check off all the requirements on the checklist, go back and make sure that your project is complete. Save your project before submitting it.
b) Turn in your scale model of the solar system to your teacher. Be sure that your name is on it.
c) Submit your Student Guide through the virtual classroom.
d) Congratulations! You have completed your project.
Answer
I hope this help....
Explanation:
Answer:
Hope this helps
Explanation:
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.
Answer:
Explanation:
From the given information:
The initial PE [tex](PE)_i[/tex] = m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = [tex]P.E_f -P.E_i[/tex]
ΔP.E = 0 - [tex](PE)_i[/tex]
ΔP.E = [tex]-P.E_i[/tex]
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]
[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]
this can be re-written as:
[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]
[tex]\Delta U =(0.70) (490.5)[/tex]
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
A 50mf capacitor, a 0.3H inductor and an 80 ohm resistor is connected in series with a 120v, 60Hz power source
What is the impedance of a circuit?
Answer:
Z = 138.5 Ω
Explanation:
In a series RLC circuit the impedance is
Z = [tex]\sqrt{R^2 + ( X_L - X_C)^2 }[/tex]
the capacitive impedance is
X_C = 1 / wC
the inductive impedance is
X_L = wL
in this exercise indicate that C = 50 10⁻³ F, L = 0.3 H and the frequency is f=60 Hz
angular velocity and frequency are related
w = 2π f
w = 2π 60
w = 376.99 rad / s
let's calculate
Z = [tex]\sqrt{80^2 + ( 376.99 \ 0.3 - \frac{1}{376.99 \ 50 \ 10^{-3}} )^2 }[/tex]
Z = [tex]\sqrt{6400 + ( 113.1 - 0.053)^2}[/tex]
Z = √19179.6
Z = 138.5 Ω
A car starts from rest and accelerates uniformly in a straight line in the positive x direction. After 25 seconds, its speed is 90 km/h.
a) Determine the acceleration of the object. [5]
b) How far does the object travel during the first 25 seconds? [3]
c) What is the average velocity of the object during the first 25 seconds?
Answer:
A. 1 m/s²
B. 312.5 m
C. 12.5 m/s
Explanation:
We'll begin by converting the velocity i.e 90 Km/h to m/s. This can be obtained as follow:
Velocity (Km/h) = 90 Km/h
Velocity (m/s) =?
Velocity (m/s) = Velocity (Km/h) × 1000 / 3600
Velocity (m/s) = 90 × 1000 / 3600
Velocity (m/s) = 90000 / 3600
Velocity (m/s) = 25 m/s
A. Determination of the acceleration.
Initial velocity (u) = 0 m/s
Final velocity (v) = 25 m/s
Time (t) = 25 s
Acceleration (a) =?
v = u + at
25 = 0 + (a × 25)
25 = 0 + 25a
25 = 25a
Divide both side by 25
a = 25/25
a = 1 m/s²
B. Determination of the distance travelled.
Initial velocity (u) = 0 m/s
Final velocity (v) = 25 m/s
Acceleration (a) = 1 m/s²
Distance travelled (s) =?
v² = u² + 2as
25² = 0 + (2 × 1 × s)
625 = 0 + 2s
625 = 2s
Divide both side by 2
s = 625 / 2
s = 312.5 m
C. Determination of the average velocity.
Total distance travelled = 312.5 m
Total time = 25 s
Average velocity =?
Average velocity = Total distance / total time
Average velocity = 312.5 / 25
Average velocity = 12.5 m/s
A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equilibrium point and released. Determine
a) the spring constant
b) the maximum velocity of the mass
c) the maximum acceleration of the mass
d) the total mechanical energy of the mass
e) the period and frequency of the mass and spring and
f) the equation of time-dependent vertical position of the mass
Answer:
a) [tex]k=19.6N/m[/tex]
b) [tex]V_m=0.81m/s[/tex]
c) [tex]a_m=6.561m/s^2[/tex]
d) [tex]K.E=0.096J[/tex]
e) [tex]T=0.78sec[/tex] &[tex]F=1.29sec[/tex]
f) [tex]mx'' + kx' =0[/tex]
Explanation:
From the question we are told that:
Stretch Length [tex]L=0.150m[/tex]
Mass [tex]m=0.30kg[/tex]
Total stretch length[tex]L_t=0.150+0.100=>0.25[/tex]
a)
Generally the equation for Force F on the spring is mathematically given by
[tex]F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}[/tex]
[tex]k=19.6N/m[/tex]
b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by
[tex]V_m=A\omega[/tex]
Where
A=Amplitude
[tex]A=0.100m[/tex]
And
[tex]\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s[/tex]
Therefore
[tex]V_m=A\omega\\\\V_m=8.1*0.1[/tex]
[tex]V_m=0.81m/s[/tex]
c)
Generally the equation for Max Acceleration of Mass on the spring is mathematically given by
[tex]a_m=\omega^2A[/tex]
[tex]a_m=8.1^2*0.1[/tex]
[tex]a_m=6.561m/s^2[/tex]
d)
Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]K.E=\frac{1}{2}*0.3*0.8^2[/tex]
[tex]K.E=0.096J[/tex]
e)
Generally the equation for the period T is mathematically given by
[tex]\omega=\frac{2\pi}{T}[/tex]
[tex]T=\frac{2*3.142}{8.1}[/tex]
[tex]T=0.78sec[/tex]
Generally the equation for the Frequency is mathematically given by
[tex]F=\frac{1}{T}[/tex]
[tex]F=1.29sec[/tex]
f)
Generally the Equation of time-dependent vertical position of the mass is mathematically given by
[tex]mx'' + kx' =0[/tex]
Where
'= signify order of differentiation
when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its coefficient of linear expansion is 1.7*10^5/°C, what is the change in coins diameter?
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f = [tex]\sqrt{4 A_f /\pi }[/tex]
D_f = [tex]\sqrt{4 \ 2.551 10^{-4} /\pi }[/tex]
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field component transmitted
Answer:
the rms value of the electric field component transmitted is 3.295 V/m
Explanation:
Given;
intensity of the unpolarized light, I = 0.0288 W/m²
For unpolarized light, the relationship between the amplitude electric field and intensity is given as;
[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]
The relationship between the rms value of the electric field and the amplitude electric field is given as;
[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]
Therefore, the rms value of the electric field component transmitted is 3.295 V/m
What is the SI unit for momentum?
0 kg.m
m
O kg
S
m
O kg
S
O kg •
m
A train is moving with a speed of 100 m/s. If the train is traveling south, at what position will it be 3 minutes after passing the +1,000-meter position marker ?
Remember, south is the negative direction and when you use the time, it must be in units of seconds. You will be applying one of the equations from above to solve this problem. And you must include a + sign if the final position is positive or a - sign if the final position is negative.
Answer:
The position after 3 minutes is - 800 m.
Explanation:
speed, v = 100 m/s
time, t = 3min = 180 s
initial position, x = 1000 m
let the distance traveled in 3 minutes is d
d = 100 x 180 = - 18000 m
So, the position is
= - 18000 + 1000 = - 800 m
Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up for 391 m with an acceleration of +18.9 m/s2. A parachute then opens, slowing the car down with an acceleration of -9.92 m/s2. How fast is the racer moving 332 m after the parachute opens?
Answer:
V = 90.51 m/s
Explanation:
From the given information:
Initial speed (u) = 0
Distance (S) = 391 m
Acceleration (a) = 18.9 m/s²
Using the relation for the equation of motion:
v² - u² = 2as
v² - 0² = 2as
v² = 2as
[tex]v = \sqrt{2as}[/tex]
[tex]v = \sqrt{2*18.9*391}[/tex]
v = 121.57 m/s
After the parachute opens:
The initial velocity = 121.57 m/ss
Distance S' = 332 m
Acceleration = -9.92 m/s²
How fast is the racer can be determined by using the relation:
[tex]V= \sqrt{v^2 + 2aS'}[/tex]
[tex]V = \sqrt{121.57^2+ 2 (-9.92)(332)}[/tex]
V = 90.51 m/s
The velocity of an object traveling in a circle is quadrupled and its radius is tripled The acceleration of this object will change by factor of?
Answer:
The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.
Sample Problem #1
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 900 kg
v = 10.0 m/s
R = 25.0 m
Requested Information:
a = ????
Fnet = ????
To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
Sample Problem #2
A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.
The solution of this problem begins with the identification of the known and requested information.
Known Information:
m = 95.0 kg
R = 12.0 m
Traveled 1/4-th of the circumference in 2.1 s
Requested Information:
v = ????
a = ????
Fnet = ????
To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:
v = d / t
v = (0.25 • 2 • pi • R) / t
v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)
v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:
a = v2 / R
a = (8.97 m/s)2 / (12.0 m)
a = (80.5 m2/s2) / (12.0 m)
a = 6.71 m/s2
To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N
In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.
Definition of Xenophobia
Answer:
dislike of or prejudice against people from other countries.
Explanation:
Answer: dislike of or prejudice against people from other countries.
Explanation:
A soccer player kicks a ball. Why does the action force exerted by the player's foot cause a different motion than the reaction force?
The reaction force is greater than the action force.
The action force and the reaction force act on different objects.
The action force is greater than the reaction force.
The action force and the reaction force act in opposite directions.
Answer:
D because of POE
Explanation:
the reaction force is the ball exerting the same newtons of force back to your leg opposite of the ball, but the reaction force and the action force is never stronger than each-other. and action is only being done on the soccer ball, so process of elimination, the answer is D
Answer:
B
Explanation:
it can only have a different motion if it is acted upon a different ball
What recommendations would you give to the global government to help Decrease the global effects of human impact on the environment mystery recommendations and how they will positively impact our planet
Answer:
We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.
Explanation:
5 ways our governments can confront climate change
PROTECT AND RESTORE KEY ECOSYSTEMS
SUPPORT SMALL AGRICULTURAL PRODUCERS
PROMOTE GREEN ENERGY
COMBAT SHORT-LIVED CLIMATE POLLUTANTS
BET ON ADAPTATION, NOT JUST MITIGATION
Which statement accurately represents the arrangement of electrons in Bohr’s atomic model?
Electrons move randomly in the relatively large space surrounding the nucleus.
Electrons move around the nucleus in fixed orbits of equal levels of energy.
Electrons move around the nucleus in fixed orbits of increasing levels of energy.
Electrons vibrate in fixed locations around the nucleus.
Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy
Explanation:
The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.
It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.
Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.
two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges
Answer:
The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.
Explanation:
Given;
charge of each particle, Q = 2 x 10⁻⁷ C
distance between the two charges, r = 15 cm = 0.15 m
distance midway between the charges = 0.075 m
The magnitude of the electric field is calculated as;
[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]
The direction of the electric field is from positive particle to negative particle.
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :
[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]
[tex]v_1t= 0.35[/tex] ...........(i)
[tex]v_0t= 1.09[/tex] ...........(ii)
Dividing the equation (ii) by (i)
[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]
∴ [tex]v_0=3.11 \ v_1[/tex]
Now loss of energy = change in the kinetic energy
[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]
[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]
[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]
If f is average friction force, then
(f)(L) = W
(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
(f) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
The Average force of friction is ( F ) = 7.307 * 10⁻³ v₀²
Given data:
Predicted range ( v₁t ) = 0.3503 m
Actual range ( v₀t ) = 1.09 m
mass = 16.3 g
First step : Determine the value of V₀
[tex]t = \sqrt{\frac{2H}{g} }[/tex] , v₁t = 0.3503 , ( v₀t ) = 1.09 m
To obtain the value of V₀
Divide ( v₀t ) by ( v₁t ) = 1.09 / 0.3503 = 3.11 v₁
∴ V₀ = 3.11 v₁
Next step : Determine the average force of friction ( f )
given that loss of energy results in a change in kinetic energy
W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]
= 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]
∴ W = 7.307 * 10⁻³ v₀²
Average force of friction = W / Actual length
= 7.307 * 10⁻³ v₀² / 1
∴ Average force of friction ( F ) = 7.307 * 10⁻³ v₀²
Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²
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Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
which object has potential energy but not kinetic energy
Answer:
A ball resting on the edge of a cliff
Explanation:
An object must be in motion or be moving to have kinetic energy. Since the ball is resting on the edge of a cliff, it is not actually moving so it does not have kinetic energy but the resting place of the ball is potential energy.
If a magnifying glass has a power of 10.0 D, what is the magnification it produces when held 6.55 cm from an object?
Answer: The magnification of the magnifying glass is -2.9
Explanation:
The equation for power is given as:
[tex]P=\frac{1}{f}[/tex]
where,
P = Power = 10 D
f = focal length
Putting values in above equation, we get:
[tex]f=\frac{1}{10}=0.1m=10 cm[/tex] (Conversion factor: 1 m = 100 cm)
The equation for lens formula follows:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]
where,
v = image distance
u = Object distance = 6.55 cm
Putting values in above equation, we get:
[tex]\frac{1}{v}=\frac{1}{10}-\frac{1}{(6.55)}\\\\\frac{1}{v}=\frac{6.55-10}{10\times 6.55}\\\\v=\frac{65.5}{-3.45}=-18.98cm[/tex]
Magnification (m) can be written as:
[tex]m=\frac{-v}{u}[/tex]
Putting values in above equation, we get:
[tex]m=\frac{-18.98}{6.55}\\\\m=-2.9[/tex]
Hence, the magnification of the magnifying glass is -2.9
If you pitch a baseball with twice the kinetic energy you gave it in the
previous pitch, the magnitude of its momentum is
Answer:
the magnitude of momentum is √2≈ b
Explanation:
hope that helped
a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an angle of 45 and is 51.0 m long/ What will its kinetic energy be wehn it reaches the bottom
Answer: 81.619 kJ
Explanation:
Given
Mass of roller coaster is [tex]m=230\ kg[/tex]
It reaches the steepest hill with speed of [tex]u=6.2\ km/h\ or \ 1.72\ m/s[/tex]
Hill to bottom is 51 m long with inclination of [tex]45^{\circ}[/tex]
Height of the hill is [tex]h=51\sin 45^{\circ}=36.06\ m[/tex]
Conserving energy to get kinetic energy at bottom
Energy at top=Energy at bottom
[tex]\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ[/tex]
A force of 1.35 newtons is required to accelerate a book by 1.5 meters/second2 along a frictionless surface. What is the mass of the book?
Answer:
0.9 kg
Explanation:
We would use the equation F=m*a to solve this equation. First, we would need to get mass by itself therefore we divide out acceleration from both sides ( F/a=m*a / a ) acceleration would cancel out and the end equation should look like this ( F/a = m or m = F/a) After we do that we plug in the numbers 1.35 N / 1.5 m/s^2 we get 0.9 kg, assuming you are using kg.
Answer: 0.90 kilograms
Explanation:
During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during this trial
Answer: hello your question is poorly written attached below is the complete question
answer :
TA = 1.6*10^-24 * 60 * 2, TB = 1.6*10^-24 * ( 60 + 30 ) * 2 -- ( option 1 )
Explanation:
a = 2m/s^2
Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц
Tb - Ta = m₂ a
∴ Tb = m₂ a + Ta
= ( 30 * 1.6 * 10^-24 * 2 ) + ( 60 * 1.6 * 10^-24 * 2 )
= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц
jonatha want to put ketchup on his hamburger.he truns the ketchup bottle at an angle toward his plate and smacks the bottom of the bottle until the ketchup comes.what is unblanced force
Answer:
The force is Inertia
Explanation:
This is an example: He turns the ketchup bottle at an angle toward his plate and smacks the bottom of the bottle until the ketchup comes out. How does inertia affect the ketchup in the bottle? The Inertia keeps the ketchup in the bottle. The Jacksons are driving to the lake when a car in front of theirs slams on its brakes.
Your welcome! :)
Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons
Answer:
No of proton is 13 and nucleus is 13