Answer: 1. walking across a carpet and touching a metal door handle 2. pulling your hat off and having your hair stand on end.
Explanation
:)
3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in the middle of the square.The 4 electrons are held in place by some mechanism. The 4 electrons are released by the mechanism at the same time. They move and reach the corners of a square of side 0.8 meters, and keep on moving . Find the velocity of each electron at the corners of the square of side 0.8 meters.
Explanation:
3
i believe that they are all going at 3.2 meters each, I did 4 times 0.8
The velocity of each electron at the corners of the square is 15.92 m/s.
The given parameters;
charge of electron, q = 1.6 x 10⁻¹⁹ Clength of the square, L = 0.8 mThe diagonal length of the square is calculated as;
[tex]d^2 = 0.8^2 + 0.8^2\\\\d = \sqrt{0.8^2 + 0.8^2} \\\\d = 1.13 \ m[/tex]
The distance of each corner charge and the middle charge is calculated as;
[tex]r = \frac{1.13}{2} \\\\r = 0.565 \ m[/tex]
The force between each corner charge and the middle charge is calculated as;
[tex]F= \frac{kq^2}{r^2}[/tex]
The centripetal force on each charge moving around the square is calculated as;
[tex]F = \frac{mv^2}{r}[/tex]
solve the forces together;
[tex]\frac{mv^2}{r} = \frac{kq^2}{r} \\\\v^2 = \frac{kq^2}{m} \\\\v = \sqrt{ \frac{kq^2}{m} } \\\\v = \sqrt{ \frac{(9\times 10^9) \times (1.602\times 10^{-19})^2}{9.11 \times 10^{-31}} } \\\\v = 15.92 \ m/s[/tex]
Thus, the velocity of each electron at the corners of the square is 15.92 m/s.
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1. Although mercury is a metal, it is a liquid at room temperature. Mercury melts at about -39°C. If
the temperature of a block of mercury starts at -54°C and increases by 22°C, does the mercury melt?
Explain your answer.
Answer:
I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.
Answer: Mercury
Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.
A moving man is lifting a box up an inclined plane. Halfway up he sets the box down and rests. Which of the following explains why the box does not slide back down the inclined plane?
a. The force of friction balances the force of gravity.
b. The force of gravity does not affect inclined planes.
c. The force of friction does not affect inclined planes.
d. The force of friction is less than the force of gravity.
Answer: A or B
Explanation: I’m guessing that they even each other out depending on the incline, gravity will help keeping it in place
A copper wire has a mass of 29.33 mg/cm and has a length of 2.5 cm.
Find the weight of the copper wire.
What is characteristic of a good insulator?
A. Electrons are usually not moving at all.
B. Electrons are free to move around.
C. Electrons are semi-free to move around.
D. Electrons are tightly bound to the nuclei.
Answer:
D. Electrons are tightly bound to the nuclei.
Explanation:
In an insulator, the electrons of the outer most shell are bound with a very high electrostatic forces coming from the nucleus of each atom so electrons cannot flow around all atoms making up the material as in a conductor.
The characteristic of a good insulator is Electrons are tightly bound to the nuclei. (option d)
In a good insulator, electrons are tightly bound to the nuclei of their atoms. This means that they are not free to move around within the material, unlike conductors where electrons are relatively loosely bound and can move freely. Due to this strong binding, electrons in insulating materials cannot carry an electric charge or energy easily from one atom to another.
When an electric field is applied to an insulator, the electrons may experience a small displacement within their respective atoms, but they generally do not move from one atom to another or flow through the material like they would in a conductor. As a result, insulators prevent the flow of electric current and are used to isolate or protect conductive elements from accidental contact.
So, the correct answer is D. Electrons are tightly bound to the nuclei.
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A 10.0 kg gun applies a force of 250. N left on a 0.0200 kg bullet. What is the force on the gun? A. 5.00N B. 25.0N C. 12,500N D. 250.N
Answer:
The answer is D.250N.
Explanation:
ʕ•ᴥ•ʔhi how are you ？
Answer:
I’m not okay . Me and my bf are bickering once again
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
Weight = 8.162 Newton.
Explanation:
Given the following data;
Mass = 2.2 kg
Acceleration due to gravity = 3.71 N/kg
To find the weight of the textbook;
Weight = mass * acceleration due to gravity
Weight = 2.2 * 3.71
Weight = 8.162 N
Therefore, the weight of the science textbook in mars is 8.162 Newton.
A wave has a frequency of 2 Hz. Find its period
What does a step-up transformer do?
A. It steps up the energy.
B. It steps up the power.
C. It steps up the voltage.
D. It steps up the current.
Which of the following is NOT true about
newspaper weather maps?
A. They report the temperature of the area in degrees
Fahrenheit.
B. They let you know how much precipitation to expect.
C. They provide more detail than weather service maps.
D. They tell you about the warm and cold air fronts.
How much mechanical work is required to catch a 14.715N ball traveling at a velocity of 37.5m/s?
To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.
What is work?Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.
In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that
Work is the product of the component of the force acting in the displacement's direction and its magnitude.
Weight of the ball = 14.715 N.
Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.
Velocity of the ball = 37.5 m/s
Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.
Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.
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(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg buddy Bubba has been riding at the front of your 60 kg, 4 m long boat. You come to a stop in the water and then switch places. A) What is the center of mass before you switch places
Answer:
Explanation:
From the given information:
Let the first weight be [tex]m_ 1[/tex] = 80 kg
The weight of the buddy be [tex]m_2[/tex] = 120 kg
The weight of Bubba be [tex]m_3[/tex] = 60 kg
Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:
[tex]x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2[/tex]
The length of the boat be [tex]x_2[/tex] = 4 m
∴
We can find the center of mass of the system by using the formula:
[tex]X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}[/tex]
A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
Answer:
a) v = 3.116 m / s, b) μ = 1.65 10⁻²
Explanation:
a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = M v₀
final instant. After the crash
p_f = (M + m) v
the moment is preserved
M v₀ = (M + m) v
v = [tex]\frac{M}{M+m} \ v_o[/tex]
let's calculate
v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]
v = 3.116 m / s
b) for this part let's use the relationship between work and kinetic energy
W = ΔK
as the body has its final kinetic energy is zero
the work of the friction forces is
W = - fr x
the negative sign is because the friction forces always oppose the movement
let's write Newton's second law
Y axis
N - W_sled -W_person = 0
N = mg + M g
N = (m + M) g
X axis
fr = ma
the friction force has the expression
fr = μ N
fr = μ g (m + M)
we substitute
- μg (m + M) x = 0- ½ (m + M) v²
μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]
let's calculate
μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]
μ = 0.0165
μ = 1.65 10⁻²
Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the order of 107K. Find the order of magnitude of the wavelength radiated with greatest intensity by each of these sources. Name the part of the EM spectrum where you would expect to radiate most strongly.
Answer:
tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region
[tex]2.898\times 10^{-10}\ \text{m}[/tex] x-ray region
Explanation:
T = Temperature
b = Constant of proportionality = [tex]2.898\times 10^{-3}\ \text{m K}[/tex]
[tex]\lambda[/tex] = Wavelength
[tex]T=10^4\ \text{K}[/tex]
From Wein's law we have
[tex]\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}[/tex]
The wavelength of the radiation will be [tex]2.898\times 10^{-7}\ \text{m}[/tex] and it is in the ultraviolet region.
[tex]T=10^7\ \text{K}[/tex]
[tex]\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}[/tex]
The wavelength of the radiation will be [tex]2.898\times 10^{-10}\ \text{m}[/tex] and it is in the x-ray region.
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.17.
What is the final speed of the crate after being pulled these 20.5 meters?
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
[tex] F = ma [/tex]
[tex] a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2} [/tex]
Now, we can calculate the final speed of the crate at the end of 10.0 m:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{1} [/tex]
[tex] v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s [/tex]
For the next 10.5 meters we have frictional force:
[tex] F - F_{\mu} = ma [/tex]
[tex] F - \mu mg = ma [/tex]
So, the acceleration is:
[tex] a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2} [/tex]
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{2} [/tex]
[tex] v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s [/tex]
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!
Which of the following is true of the deep
water layer of the ocean?
A. warmest and least dense of the ocean layers
B. experiences a rapid decrease in temperature
C. is warm in the summer and cold in the winter
D. cold all year round
question is included in the picture!!! PUT REAL ANSWERS OR I WILL REPORT YOU
Answer:
Explanation:
this is like rubbing a balloon on your head to make your hair stand up. Do that to the can. The balloon is filled , ofc, and then just rub the balloon on the can. This will charge the can with static electricity. :P
A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position
Answer:
t = 9.52 s
Explanation:
This is an oscillatory motion exercise, in which the angular velocity is
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
Let's use hooke's law to find the spring constant, let's write the equilibrium equation
F_e - W = 0
F_e = W
k x = m g
k = [tex]\frac{m g}{x}[/tex]
k = 0.545 9.8 /0.0356
k = 150 N / m
now the angular velocity is related to the period
W = 2π / T
we substitute
4π² T² = k /m
T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]
we substitute
T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]
T = 2.38 s
therefore for the spring to oscillate 4 complete periods the time is
t = 4 T
t = 4 2.38
t = 9.52 s
Explain why synovial joints have greater flexibility than fixed and cartilaginous joint.
Answer:Unlike fixed joints or cartilaginous joints, where the bones are connected by either connective tissue or cartilage, the bones in synovial joints are not directly joined by anything, which allows for a much greater range of motion.
Explanation: I got a 100% on my test
What is the difference between a positively and negatively charged object?
Answer:
Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.
The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.
What are Charged objects?
Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.
A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.
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assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single fixed pulley?
A. 21 N
B. 61 N
C. 121 N
D. 241 N
A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are µs = 0.50 and µk = 0.40. For P = 3.6 lb,
Determine:
(a) Whether slipping occurs between the belt and either cylinder,
(b) The angular acceleration of each cylinder.
Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.
Given the following data:
Mass A = 5 lb to kg = 2.27 kg.
Mass B = 20 lb to kg = 9.02 kg.
Force = 3.6 lb to N = 16.02 Newton.
How to calculate angular acceleration.In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.
For cylinder A:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]
For cylinder B:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]
For the belt, we have
[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]
Also, we would determine the angular acceleration of cylinder B:
[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]
Next, we would calculate the forces acting on the cylinders:
[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]
[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]
Next, we would determine the force of static friction:
[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]
From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.
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Which is true regarding a child standing up for their own rights?
Answer:
hey mate......looks like the question is incomplete
does latitude has an effect on weight?
Answer:
I think so but i could be wrong..
Explanation:
Answer: yes it does
Explanation: Yes, you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.
A rod that is 96.0 cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 19.1 cm. An object is in air on the long axis of the rod 19.9 cm from the end that has the 19.1-cm radius.
(a) Find the image distance due to refraction at the 19.1-cm radius surface.
(b) Find the position of the final image due to refraction at both surfaces.
(c) Is the final image real or virtual?
Look at this attached photo
Answer:
C) Kinetic energy
Explanation:
Since the skateboarder is ALREADY going down the hill, he is using kinetic energy.
Kinetic energy is when you're moving and in motion
An object A with mass 200 kg and an another object B with mass 1000 kg are moving with same speed. The ratio of kinetic energy of object A to B is
Answer:
Ratio of kinetic energy of object A to B = 1:5
Explanation:
Given:
Mass of object A = 200 kg
Mass of object B = 1,000 kg
Find:
Ratio of kinetic energy of object A to B
Computation:
Kinetic energy = (1/2)(m)(v²)
Kinetic energy of object A = (1/2)(200)(v²)
Kinetic energy of object A = (100)(v²)
Kinetic energy of object B = (1/2)(1,000)(v²)
Kinetic energy of object B = (500)(v²)
Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B
Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)
Ratio of kinetic energy of object A to B = 100 / 500
Ratio of kinetic energy of object A to B = 1/5
Ratio of kinetic energy of object A to B = 1:5
35 POINTSS!!! PLSSSS HELLPPP!!!
Work is the transfer of power from one object to another.
Please select the best answer from the choices provided
T
F
Answer:
T
beacuse:
Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).
Disk A, with a mass of 2.0 kg and a radius of 70 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 50 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together. After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?
Answer:
w = - 197.5 rad / s
The negative sign indicates that the rotations are clockwise
Explanation:
To solve this exercise, let's use the concept of conservation of the angular number.
We create a system formed by the two discs, in this case the forces last the shock are internal
initial instant .. just before shock
L₀ = I₀ w₀ + I₁ w₁
instnte final. Right after crash
L_f = (I₀ + I1) w
angular momentum is conserved
I₀ w₀ + I₁ w₁ = (I₀ + I₁) w
w = I₀ w₀ + I₁ w₁ / Io + I1
The moment of inertia of a disk with an axis passing through its thermometric center
I₀ = ½ m² r₀²
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ 2.0 0.70²
I₀ = 0.49 kg m
I₁ = ½ 2.0 0.5²
I₁ = 0.25
₁
let's reduce the magnitudes the SI system
w₀ = -50 rev / (2pi rad / 1rev) = -314.15 rad / s
w₁ = 70 rev (2pi rad / 1rev) = 439.82 rad / s
we will assume that the counterclockwise turns are positive
w = -0.49 314.15 + 0.25 439.82 / (0.49 + 0.25)
w = (- 4.696 + 1.0995) 102) / 0.74
w = -197.75 + 0.25
w = - 197.5 rad / s
The negative sign indicates that the rotations are clockwise