Answer:
The leaves have a charge in each experiment, but the sign of the charge cannot be determined.
Explanation:
In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.
Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.
Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.
Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.
Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.
At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa
(kilopascals) and in Denver, 1609.344 meters above sea level, the pressure of the atmosphere
is about 83.437 kPa. Using this data, find a linear equation for pressure P in terms of
altitude h. (Hint: write the pressure and altitude in each location as a point (h, P). Then
use point-slope form to find the equation of the line.)
Answer:
[tex]P = -\frac{17978}{1609344}(h)+101.325[/tex]
Explanation:
Given
[tex]h = height[/tex]
[tex]P = Pressure[/tex]
[tex](h_1,P_1) = (0,101.325)[/tex]
[tex](h_2,P_2) = (1609.344 ,83.437 )[/tex]
Required
Determine the linear equation for P in terms of h
First, we calculate the slope/rate (m);
The following formula is used:
[tex]m = \frac{P_2 - P_1}{h_2 - h_1}[/tex]
Substitute values for P's and h's
[tex]m = \frac{83.347 - 101.325}{1609.344- 0}[/tex]
[tex]m = \frac{-17.978}{1609.344}[/tex]
[tex]m = -\frac{17.978}{1609.344}[/tex]
Multiply by 1000/1000
[tex]m = -\frac{17.978 * 1000}{1609.344*1000}[/tex]
[tex]m = -\frac{17978}{1609344}[/tex]
The equation is then calculated using:
[tex]P - P_1 = m(h - h_1)[/tex]
Substitute values for m, h1 and P1
[tex]P - P_1 = m(h - h_1)[/tex]
[tex]P - 101.325 = -\frac{17978}{1609344}(h - 0)[/tex]
[tex]P - 101.325 = -\frac{17978}{1609344}(h)[/tex]
Make P the subject
[tex]P = -\frac{17978}{1609344}(h)+101.325[/tex]
The above is the required linear equation
A tennis ball is hit with a vertical speed of 10 m/s and a horizontal speed of 30 m/s. How far will the ball travel horizontally before landing?
a. 10 m
b. 20 m
c. 40 m
d. 60 m
e. 80 m
Answer:
D) 60 m
Explanation:
We can use the constant acceleration equation that contains displacement, initial velocity, acceleration, and time. We want to solve for the time that the ball was in the air first.
Δx = v_i * t + 1/2at²Let's use this equation in terms of the y-direction.
Δx_y = (v_i)y * t + 1/2a_y * t²The vertical displacement will be 0 meters since the ball will be on the floor. The initial vertical velocity is 10 m/s, the vertical acceleration is g = 10 m/s², and we are going to solve for time t.
Let's set the upwards direction to be positive and the downwards direction to be negative. We must use -g to be consistent with our other values.
Plug the known values into the equation.
0 m = 10 m/s * t + 1/2(-10 m/s²) * t²Simplify the equation.
0 = -10t + 5t² 0 = 5t² - 10tFactor the equation.
0 = 5t(t - 2)Solve for t by setting both factors to 0.
5t = 0t - 2 = 0We get t = 0, t = 2. We must use t = 2 seconds because it is the only value for t that makes sense in the problem.
Now that we have the time that the ball was in the air, we can use the same constant acceleration equation to determine the horizontal displacement of the tennis ball. We will use this equation in terms of the x-direction:
Δx = v_i * t + 1/2at² Δx_x = (v_i)x * t + 1/2a_x * t²Plug the known values into the equation.
Δx_x = 30 m/s * 2 sec + 1/2(0 m/s²) * (2 sec)²We can eliminate the right side of the equation since anything multiplied by 0 outputs 0.
Δx_x = 30 * 2 Δx_x = 60The horizontal displacement of the ball is 60 meters. Therefore, the answer is D) 60 m.
Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slide. If they both start at the same height above the ground, which of the following is true about their kinetic energies as they reach the ground?
A) Bob's kinetic energy is greater than Esther's.B) Esther's kinetic energy is greater than Bob's.C) They have the same kinetic energy.D) The answer depends on the shape of the slide.
Answer:
Explanation:
They have the same kinetic energy
give me two reasons that may render a metalochromic indicator efficient
Answer:
A complexometric indicator is an ionochromic dye that undergoes a definite color change in presence of specific metal ions.[1] It forms a weak complex with the ions present in the solution, which has a significantly different color from the form existing outside the complex. Complexometric indicators are also known as pM indicators.[2]
What is the magnitude of the electric field on a +2 C charge if it experiences an electric force of 6 N?
Answer:
call me 7164013838 hurry bye
solve at least one question (30 points)
Answer:
(a) the particle position = 135 m
(b) the velocity of the particle = 44 m/s
(c) the acceleration of the particle = 50 m/s²
Explanation:
Solution to Question 2.
Given;
velocity of a particle, v = 2 - 4t + 2t³
initial position at t = 0, s₀ = 3 m
(a) the particle position at t = 3 s
s = vt
s = (2 - 4t + 2t³)t
s = 2t - 4t² + 2t⁴
s = s₀ + s₃
s = s₀ + 2(3) - 4(3²) + 2(3⁴)
s = s₀ + 6 - 36 + 162
s = s₀ + 132 m
s = 3m + 132 m
s = 135 m
(b) the velocity of the particle at t = 3 s
v = 2 - 4t + 2t³ = 2 - 4(3) + 2(3)³
v = 44 m/s
(c) the acceleration of the particle at t = 3s
v = 2 - 4t + 2t³
[tex]a = \frac{dv}{dt} \\\\a = \frac{d}{dt} (2 -4t + 2t^3)\\\\a = -4 + 6t^2[/tex]
a = -4 + 6(3)²
a = 50 m/s²
Pls someone help me with this :(
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus , the latter of which is also radioactive and will itself decay some time later. The energy emitted in the plutonium decay is and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is , while that of the uranium is (note that the ratio of the masses is 4 to 235).
(a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest.
(b) How much kinetic energy does each nucleus carry away
Answer:
a) [tex] v_{U-235} = 2.68 \cdot 10^{5} m/s [/tex]
[tex]v_{He-4} = -1.57 \cdot 10^{7} m/s[/tex]
b) [tex] E_{He-4} = 8.23 \cdot 10^{-13} J [/tex]
[tex] E_{U-235} = 1.41 \cdot 10^{-14} J [/tex]
Explanation:
Searching the missed information we have:
E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J
m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg
m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg
a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:
Linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235} [/tex]
Since the plutonium nucleus is originally at rest, [tex]v_{Pu-239} = 0[/tex]:
[tex] 0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235} [/tex]
[tex] v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}} [/tex] (1)
Kinetic Energy:
[tex] E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2} [/tex]
[tex] 2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2} [/tex]
[tex] 1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2} [/tex] (2)
By entering equation (1) into (2) we have:
[tex] 1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2} [/tex]
[tex] 1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2} [/tex]
Solving the above equation for [tex]v_{U-235}[/tex] we have:
[tex] v_{U-235} = 2.68 \cdot 10^{5} m/s [/tex]
And by entering that value into equation (1):
[tex]v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s[/tex]
The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.
b) Now, the kinetic energy of each nucleus is:
For He-4:
[tex]E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J[/tex]
For U-235:
[tex] E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J [/tex]
I hope it helps you!
An object with a mass of 32 kg has an initial energy of 500 J. At the end of the experiment, the velocity of the object is recorded as 5.1 m/s. If the object travelled 50 m to get to this point, what was the average force of friction on object during the trip? Assume no potential energy. Show all work
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
what is the velocity of a point on the edge of the wheel?
A. 6.15 m/s
B. 5.39 m/s
C. 5.84 m/s
D. 6.20 m/s
Answer:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
[tex]V_{r}[/tex] = 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!
The force of friction between an object and the surface upon which it is sliding is 46N and the coefficient of friction between them is 0.30. What is the weight of the object?
Answer:
let the coefficient of friction be ß
Explanation:
[tex]from \: third \: law \: of \: solid \: friction \\ friction = \beta \times R \\ where \: R \: is \: the \: normal \: reaction \\ 46 = 0.30 \times R \\ R = \frac{46}{0.30} \\ R = 153.33 \: N \\ but \: R = mg \\ where \: mg \: is \: the \: weight \\ hence \\ weight, W = 153.33 \: N[/tex]
If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has been released?
Answer:
500J
Explanation:
The arrow will have an energy of 500J after it has been released from its state of rest.
This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".
The energy at rest which is the potential energy is 500J This energy will be converted to kinetic energy in total after the arrow has been released. This way, no energy is lost and we can account for the energy transformations occurring.A man does 4,465 J of work in the process of pushing his 2.90 103 kg truck from rest to a speed of v, over a distance of 25.0 m. Neglecting friction between truck and road, determine the following.(a) the speed vm/s
(b) the horizontal force exerted on the truck
N
Answer:
(a)The final speed of the truck "v" is 1.76 m/s
(b) The horizontal force exerted on the truck is 178.64 N
Explanation:
Given;
work done by the man, W = 4,465 J
mass of the truck, m = 2.9 x 10³ kg
initial speed of the truck, u = 0
distance the truck was pushed to, d = 25 m
final speed of the truck, = v
The work done in pushing the truck is calculated as;
W = F x d
where;
F is the horizontal force applied
d is the distance in which the truck was pushed to
W = ma x d
[tex]a = \frac{W}{md} \\\\a = \frac{4465}{2.9 \times 10^3 \ \times 25} \\\\a = 0.0616 \ m/s^2[/tex]
(a)The final speed of the truck "v" is calculated as;
v²= u² + 2ad
v² = 0 + 2(0.0616 x 25)
v² = 3.08
v = √3.08
v = 1.76 m/s
(b) the horizontal force exerted on the truck
F = ma
F = (2.9 x 10³) (0.0616)
F = 178.64 N
A Man Moved first a Distance of 1000 m in 25 second and 2.5 km in 50 second along a in straight line?
Answer:
Average speed = 46.67 m/s
Explanation:
Given that the time taken in covering first 1000 m = 25 seconds.
The time taken in covering next 2.5 km = 50 seconds.
Total distance covered = 1000 m + 2500 m = 3500 m
Total time taken = 25+50=75 seconds
Average speed = Total distance covered / total time taken
= 3500/75 = 46.67 m/s
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer:
There are 4 liquids in this experiment and red is the least dense of all of them so it should float on top, which it is doing.
The red that you see at the bottom is neither liquid nor is it a part of the experiment.
It is simply the color of the bottom of the container that the experiment was conducted in.
The bottom thing doesn’t matter. That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
A hiker walks 200 m west and then walks 100 m north. In what direction is her resulting displacement?a. northb. westc. northwestd. None of the answers is correct.
Answer:
Option (C) is correct.
Explanation:
Assuming that the hiker starts walking from the origin O as shown in the figure.
First, he walks 200 m west to point A (say), then he walks 100 m north to the final point B (say) as shown in the figure.
The final point B is in the north-west direction, therefore, the resulting point is in the north-west direction.
Hence, option (C) is correct.
A slithering snake travels once around a circle of radius 3.20m. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 80.0N. How much work does the snake do against friction?
Answer:
The magnitude of the work done by the snake against friction is 402 J.
Explanation:
The work is given by:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the distance
The distance traveled by the snake is given by the perimeter of the circle:
[tex] P = 2\pi r = 2\pi 3.20 m = 20.1 m [/tex]
Now, the force in the direction of the movement is:
[tex] |\Sigma F| = F_{\mu} = \mu N = \mu W = 0.25*80.0 N = 20 N [/tex]
Finally, the work is:
[tex] W = F*d = 20 N*20.1 m = 402 J [/tex]
Therefore, the magnitude of the work done by the snake against friction is 402 J.
I hope it helps you!
*PLEASE HELP*
Light passing through a double slit
with separation d=3.33x10^-5 m
creates its first minimum (m = 1) at an
angle of 0.551 deg. What is the
wavelength of the light, IN
NANOMETERS?
(Hint: The answer will be between 400 and
700.) (Unit = nm)
Answer:
640.5
Explanation: i got it right on acellus
The wavelength of the light is 320.23 nm.
Light passing through a double slit with separation is given as
d = 3.33 × 10⁻⁵ m.
Line number m = 1
Angle θ = 0.551°
How wavelength is calculated and what is the wavelength of the light?The distance between two sound waves i.e., crest and trough is called the wavelength.
Human eye can detect the wavelength of about 380 nm to 700 nm.
To find the wavelength, here we use the formula as,
sin θ = Line number × λ ÷ ( distance )
By substituting all the values,
Sin ( 0.551 ) = 1 × λ ÷ ( 3.33 × 10⁻⁵)
Rearranging the equation,
λ= (Sin ( 0.551 ) × ( 3.33 × 10⁻⁵) ) ÷ (1)
λ = 320.23 nm.
Thus, wavelength λ of the light is 320.23 nm.
Learn more about wavelength,
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Jaden has a mass of 45 kilograms on Earth. Jupiter has more gravity than the Earth. On Jupiter, Jaden's mass will be O more than 45 kilograms o less than 45 kilograms 45 kilograms
Answer:
m = 45 kg
Explanation:
Given that,
Mass of Jadan, m = 45 kg on Earth
Jupiter has more gravity than the Earth.
Mass of an object is the amount of matter contained inside an object. We need to tell about the mass of Jaden on Jupiter. The mass of the object remains same everywhere.It does not change in any of the location.
Hence, Jaden's mass will be 45 kg on Jupiter.
If the net work done on a particle is zero, which of the following statements must be true?
A. The speed is unchanged.
B. The velocity is zero.
C. The velocity is unchanged.
D. More information is needed.
E. The velocity is decreased.
Answer:
A. The speed is unchanged.
Explanation:
In the case when the work is to be done on a particle i.e. zero so the change made in KE of the particle would be zero. This represent the work energy theroem. But when the KE remains same or does not change so it should be the same and the particle speed would also the same
Therefore as per the given statement, the first option is correct
And rest of the options are wrong
Early mines used to be ventilated as illustrated below. Explain why this works in terms of convection.
Calculate the average power produced when a car of mass 400 kg accelerates from rest to a speed of 20m/s in a time interval of 1 minute.
Explanation:
w=F×d 1kg=9.807N
p=w/t 400kg=9.807×400=3922.8N
p=Fd/t
=3922.8×20×1
power=78456 watt
..sorry..if the answer is not right..
1. Write a list of characteristics to describe the waves. Describe each characteristic in your own words
so that any person could understand waves. Use images to help with the descriptions.
Tips: Later during this lab, you will relate your own descriptions to scientific ones. It is impþrtant
that you have your own words to begin your leaming, so don't do any research yet. Also,
learning is best when you make your own drawings. Perhaps you can insert images of your
own drawings. if not then use images from Waves on a String that look like your drawings.
Answer:
Explanation:
Maybe start by trying to draw out some transverse waves, longitudinal waves, stationary waves etc. From there you could explain things about each type. For example, particles oscillate in the direction of energy transfer in longitudinal waves or you could include that all electromagnetic waves are examples of transverse waves. Hope this helps you get started :)
Waves are a form of energy transfer through matter or space. Waves can be characters using their amplitude, frequency, wavelength, energy etc.
What is wave?A wave can be defined as the energy transfer or disturbances in air or matter. Sound and light are two types of waves.
There are broadly two types of waves namely longitudinal waves which vibration parallel to the direction of the wave and transverse wave vibrating perpendicular to the direction of wave.
A wave is associated with an amplitude. It is the maximum distance that its molecules are displaced or it is the measure of intensity.
Frequency of a wave is the number crests or troughs passing per unit time. Wavelength is the distance between two consecutive crests or throughs.
Each wave is associated with certain amount of energy which is directly proportional to its frequency. High energy wave can easily penetrate through a body.
Hence, by the terms such as amplitude, frequency, wavelength and energy a wave can be characterized.
To find more about waves, refer the link below:
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The higher an object is raised compared to the ground
Question 5 options:
the less potential energy it has.
the more potential energy it has.
the less kinetic energy it has.
the more kinetic energy it has.
Answer:
B. The more potential energy it has.
Explanation:
I majored in Physics
Study the image. Point 1 is beneath water moving toward land. Point 2 is next to an erupting vocano. Point 3 is beneath Earth apostrophe s crust moving toward the water. Point 4 is the same layer as Point 1 moving down below Point 3. At which point is subduction occurring? 1 2 3 4
Answer:
I believe the answer is D) 4
Explanation:
Subduction is the sideways and downward movement of the edge of a plate of the earth's crust into the mantle beneath another plate.
Answer: point 4
Explanation:
A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor raises the block from the floor to a height of 8.0 meters in 5 seconds what was the motor’s power output?
Answer:
156.8 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 10 kg
Height (h) = 8 m
Time (t) = 5 s
Power (P) =?
Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:
Mass (m) = 10 kg
Height (h) = 8 m
Acceleration due to gravity (g) = 9.8 m/s²
Energy (E) =?
E = mgh
E = 10 × 9. 8 × 8
E = 784 J
Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:
Time (t) = 5 s
Energy (E) = 784 J
Power (P) =?
P = E/t
P = 784 / 5
P = 156.8 Watts
Therefore, the power output of the motor is 156.8 Watts
Vin Diesel jumps out of a plane. Gravity pulls on him with 184N of force and his parachute pushes him up with 82 N of force.
Net Force = (your answer)
Body Diagram or Explanation = (your answer)
Answer:
Fr = 102[N]
Explanation:
We must perform a free body diagram to understand the forces acting on Vin Diesel. In the attached image we see the forces related to his weight and the air resistance force.
We must analyze the forces in the Y axis, we see that we have the force of the air opposing the fall of Vin Diesel, in this way this force is negative since it opposes the movement, the fall.
∑Fy = 0
[tex]F_{weight}-F_{air}=F_{r}\\F_{r}= 184-82\\F_{r}=102[N][/tex]
What is the formula to finding your target heart rate
4. Which of the following would NOT be a major process in the formation of sedimentary
rocks?
a Erosion
b. Melting
C Deposition
d. Weathering
Answer:
b. melting
Explanation:
it is made of sediments and that is not necessary
Prove that the vectors A = 2i - 3j + k and B = i+ j+ k are
mùtually perpendicular
Answer:
theta = 90°
Explanation:
hope it's helpful
Explanation:
A • B = (2i - 3j + k) • (i + j + k)
= (2)(1) + (- 3)(1) + (1)(1)
= 0 = AB cos θ
∴ cos θ = 0 (as A ≠ 0,B ≠ 0)
or θ = 90°
or the vectors A and B are mutually perpendicular.
-TheUnknownScientist