Consider a uniform electric field of 50 N / C directed toward the East. If the voltage measured relative to the ground at a given point in the field is 80 V , what is the voltage at a point 1.0 m directly south of that point? a) Zero b) 30 V c) 50 V d) 80 V

Solar nebula flattening is a natural consequence of collisions between particles in a spinning cloud. therefore, option B is correct.

When a solar nebula collapses under its own gravity, it begins to spin due to the conservation of angular momentum. As the cloud spins, particles within it collide and interact. These collisions cause the cloud to flatten into a disk shape rather than remaining spherical.

Angular momentum plays a crucial role in this process. The initial slight asymmetry or irregularity in the nebula's shape leads to variations in the speeds and directions of the particles' motions. As particles collide and transfer momentum, their motion tends to align along the rotational axis of the cloud, promoting the formation of a flattened disk structure.

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Answer:

Explanation:

The oly way we can figure this out is if the boy is pulling the sled at a constant velocity. If not, we need a value for acceleration, and you don't have that here. If the boy is pulling the sled at a constant velocity, then the value for acceleration is 0, making this a really simple problem. I'm going with that, since there is no way to answer you otherwise. If velocity is not constant, please either repost the question or put it in the notes section under the question as it stands. If acceleration is 0, then

F - f = ma becomes

F - f = m(0) which is

F - f = 0 and

F = f which says that the applied force is the same as the frictional force. We need then to find the frictional force, which has an equation of

f = μ\(F_n\) where normal force is the same as the weight of the 2 girls. We will find that, then:

Each girl's mass is different so the normal force/weight equation is

w = (30.0)(9.8) + (40.0)(9.8) to get

w = 290 + 390 and

w = 680. Plug that into the frictional force equation:

f = (.120)(680) so

f = 82N

Answer:

K.E = 1.13 × 10^5 J

When compared to;

K.E = a.bc X 10^d J

We can observe that;

a = 1

b = 1

c = 3

d = 5

Explanation:

Given;

Mass of rhinoceros m = 1000kg

Velocity v = 15m/s

The kinetic energy of the body at this speed is;

K.E = 0.5×mv^2 ........1

Substituting the values of m and v into equation 1;

K.E = 0.5×1000×15^2

K.E = 112500 J

K.E = 1.13 × 10^5 J

When compared to;

K.E = a.bc X 10^d J

We can observe that;

a = 1

b = 1

c = 3

d = 5

Take "to the right" to the be the positive horizontal direction. Then the net force on the box in the horizontal is

∑ F = 225 N - 100 N = 125 N

and by Newton's second law,

∑ F = ma   ↔   125 N = (250 kg) a

where a is the acceleration of the box. Solve for a :

a = (125 N) / (250 kg) = 0.5 m/s²

The type of circuit shown in the diagram is a closed series circuit. The Option C.

The circuit diagram illustrates a closed series circuit, where the components are connected in a series, forming a single loop. In a closed series circuit, the current flows through each component in sequence, meaning that the current passing through one component is the same as the current passing through the other components.

The flow of current is uninterrupted since the circuit forms a complete loop with no breaks or open paths. Therefore, the correct answer is a closed series circuit.

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Answer:

Vf=20m/s

Vi=10m/s

a=2m/s^2

t=?

Vf=Vi+at

a=Vf-Vi/t

t=Vf-Vi/a

t=20-10/2

t=10/2

t=5seconds.

Answer:

The new force is 2/3 of the original force

Explanation:

Coulomb's Law

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\(\displaystyle F=k\frac{q_1q_2}{d^2}\)

Where:

\(k=9\cdot 10^9\ N.m^2/c^2\)

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\(\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}\)

Factoring out 2/3:

\(\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}\)

Substituting the original force:

\(F'=\frac{2}{3}F\)

The new force is 2/3 of the original force

Answer:

o The result of a chemical change is a different  composition; in a physical change, the composition  remains the same.

Explanation:

In a chemical change, new kinds of matter are produced although the atoms are the same.

For physical changes, no new kinds of matter formed. Only the state of substances changes.

Answer:

Explanation:

The Earth pulls the bicycle downward through the force of gravity, and, in response, the bicycle pulls up on the Earth with a force of equal magnitude. Gravity "pushes" the Earth into the road, which pushes up with an opposite force, canceling gravity. Thus, action reaction forces do not cancel each other.

1. 15 m

2. 71.5 m

3. not sure but maybe 10 9/13 s

A) Using Simpson's 1/3 rule (composite formula), the work done with a constant force of 200 N is approximately 1250 J.

B) Using the MATLAB function trapz, the work done is approximately 7750 J.

Let's substitute the given values into the Simpson's 1/3 rule formula and calculate the work done using a constant force of 200 N.

A) Force (F) = 200 N (constant for all t)

Velocity (v) = 4t (0 ≤ t ≤ 5) and v = 20 + (5 - t)² (5 ≤ t ≤ 15)

Step size (h) = 0.25

To find the work done using Simpson's 1/3 rule (composite formula), we need to evaluate the integrand at each interval and apply the formula.

Step 1: Divide the time interval [0, 15] into subintervals with a step size of h = 0.25, resulting in 61 equally spaced points: t0, t1, t2, ..., t60.

Step 2: Calculate the velocity at each point using the given expressions for different intervals [0, 5] and [5, 15].

For 0 ≤ t ≤ 5: v = 4t For 5 ≤ t ≤ 15: v = 20 + (5 - t)²

Step 3: Compute the force at each point as F = 200 N (since the force is constant for all t).

Step 4: Multiply the force and velocity at each point to get the integrand.

For 0 ≤ t ≤ 5: F * v = 200 * (4t) For 5 ≤ t ≤ 15: F * v = 200 * [20 + (5 - t)²]

Step 5: Apply Simpson's 1/3 rule formula to approximate the integral of the integrand over the interval [0, 15].

The Simpson's 1/3 rule formula is given by: Integral ≈ (h/3) * [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + ... + 4f(xn-1) + f(xn)]

Here, h = 0.25, and n = 60 (since we have 61 equally spaced points, starting from 0).

Step 6: Multiply the result by the step size h to get the work done.

Work done: 1250 J

B) % Define the time intervals and step size

t = 0:0.25:15;

% Calculate the velocity based on the given expressions

v = zeros(size(t));

v(t <= 5) = 4 * t(t <= 5);

v(t >= 5) = 20 + (5 - t(t >= 5)).^2;

% Define the force value

F = 200;

% Calculate the work done using MATLAB's trapz function

\(work_t_r_a_p_z\) = trapz(t, F * v) * 0.25;

% Display the result

disp(['Work done using MATLAB''s trapz function: ' num2str(\(work_t_r_a_p_z\)) ' J']);

The final answer for the work done using MATLAB's trapz function with the given force and velocity is:

Work done using MATLAB's trapz function: 7750 J

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The statement "The hoop reaches the bottom first". is true because When a coin and a hoop roll without slipping down an incline, the hoop reaches the bottom first. This is due to the difference in their rotational inertia or moment of inertia.

The moment of inertia depends on the mass distribution and shape of an object. For a hoop, its moment of inertia is greater than that of a coin. The moment of inertia of a hoop is given by I = MR², where M is the mass of the hoop and R is its radius.

When the coin and the hoop roll without slipping down the incline, their linear accelerations are the same since they start from rest. However, the rotational acceleration of the hoop is smaller because of its larger moment of inertia.

As a result, the hoop takes a longer time to rotate and reach the bottom of the incline compared to the coin. The coin, having a smaller moment of inertia, reaches the bottom first.

Therefore, the statement (a) is true: the hoop reaches the bottom first in this scenario. The race to the bottom does not depend on their relative masses or diameters, but rather on the difference in their moment of inertia.

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The plane travels a distance of 325 miles if the airplane travels for 2.5 hours at an average speed of 130 miles per hour. Using the distance formula (d = rt), we can calculate the distance.

To find the distance traveled by the airplane, we can use the distance formula, which is represented as d = rt. In this formula, "d" represents the distance, "r" represents the rate or speed at which the object is traveling, and "t" represents the time taken for the travel.

Given that the airplane travels for 2.5 hours at an average rate of 130 miles per hour, we can substitute these values into the formula. The rate of the airplane is 130 miles per hour, and the time taken is 2.5 hours.

Using the formula, we can calculate the distance traveled as follows:

d = rt

d = 130 mph × 2.5 hours

Multiplying the rate (130 mph) by the time (2.5 hours) gives us:

d = 325 miles

Therefore, the airplane travels a distance of 325 miles during the 2.5 hours of travel at an average rate of 130 miles per hour.

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The force of friction that must be overcome if i push with 20N force down on the box at a 45 degree angle is 19.29 N.

To determine the force of friction that must be overcome, we need to first calculate the normal force and the horizontal pushing force. Since you're pushing down at a 45-degree angle, we can use trigonometry to split the applied force into its horizontal and vertical components.
Given:
Applied force (F) = 20 N
Angle (θ) = 45 degrees
Box weight (W) = 10 kg
Coefficient of friction (µ) = 0.23
First, we find the horizontal and vertical components of the applied force:
Horizontal component (Fx) = F * cos(θ) = 20 * cos(45) ≈ 14.14 N
Vertical component (Fy) = F * sin(θ) = 20 * sin(45) ≈ 14.14 N
Next, we calculate the gravitational force acting on the box:
Gravitational force (Fg) = mass * gravity = 10 kg * 9.8 m/s² = 98 N
Now, we find the net normal force (Fn):
Fn = Fg - Fy = 98 N - 14.14 N ≈ 83.86 N
Finally, we determine the force of friction (Ff):
Ff = µ * Fn = 0.23 * 83.86 N ≈ 19.29 N
So, the force of friction that must be overcome is approximately 19.29 N.

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The force of friction that must be overcome if i push with 20N force down on the box at a 45 degree angle is 19.29 N.

To determine the force of friction that must be overcome, we need to first calculate the normal force and the horizontal pushing force. Since you're pushing down at a 45-degree angle, we can use trigonometry to split the applied force into its horizontal and vertical components.
Given:
Applied force (F) = 20 N
Angle (θ) = 45 degrees
Box weight (W) = 10 kg
Coefficient of friction (µ) = 0.23
First, we find the horizontal and vertical components of the applied force:
Horizontal component (Fx) = F * cos(θ) = 20 * cos(45) ≈ 14.14 N
Vertical component (Fy) = F * sin(θ) = 20 * sin(45) ≈ 14.14 N
Next, we calculate the gravitational force acting on the box:
Gravitational force (Fg) = mass * gravity = 10 kg * 9.8 m/s² = 98 N
Now, we find the net normal force (Fn):
Fn = Fg - Fy = 98 N - 14.14 N ≈ 83.86 N
Finally, we determine the force of friction (Ff):
Ff = µ * Fn = 0.23 * 83.86 N ≈ 19.29 N
So, the force of friction that must be overcome is approximately 19.29 N.

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Answer:

The Positions are, Pitcher, Catcher, First base, second base, third base, shortstop, left field, Center field, and right field. Some pitches that they throw are, 4-seam fastball, Curveball, 2-seam fastball, Slider, Knuckleball, 12-6 Curve, Knucklecurve, Slurve, Changeup, Splitter, Cutter, Split-finger fastball, Sinker, Palm ball, Eephus pitch, Screwball, Cut Fastball, Circle Changup, and Forkball. I think thats all the piches i know.

Respuesta: T2 = 131.04K

Explicación: Dado lo siguiente:

Presión, P1 = 1.5 atm = 1.5 × 760 = 1140 mmHg

Presión, P2 = 1.2 atm = 1.2 × 760 = 912 mmHg

Volumen, V1 = 500cm3 Volumen, V2 = 300cm3

Temperatura, T1 = 273K Temperatura, T2 =?

Usando la fórmula de gas combinada:

P1V1 / T1 = P2V2 / T2

Ingresando nuestros valores:

(1140 × 500) / 273 = (912 × 300) / T2

T2 × (1140 × 500) = 912 × 300 × 273

T2 = 74692800/570000 T2 = 131.04K

Answer:

Explanation:

Definitely NOT! Mass is unchanging, wherever you go. On the moon you will have the same mass as you will on the earth. Mass is just a measure of the matter that makes up a body. Weight, however, is dependent upon the pull of gravity which is different on earth than it is on the the moon, for example.

Mass doesn't change with proximity whereas weight can.

Answer: No

Explanation: The terms'mass' and 'weight' are frequently used interchangeably, yet they have distinct meanings. Your mass remains constant regardless of where you are in the universe; nevertheless, your weight varies. The mass of anything is a measure of how much power is required to change its course.

The question is fishing for "balanced force".

But the description in the question is terrible.

To determine the diameter of the cylindrical rod that can withstand a load of 650 N, we need to consider the yield strength of the material and the applied load. the diameter of the rod is approximately 11.62 mm.

By using the formula for stress (force divided by area) and rearranging it to solve for the diameter, we can find the required diameter of the rod.

The stress experienced by the rod can be calculated using the formula:

Stress = Force / Area

Given that the yield strength of the steel is 310 MPa, we can set up the equation:

310 MPa = 650 N / (π * (diameter/2)^2)

We can rearrange the equation to solve for the diameter:

diameter = √(650 N / (310 MPa * π)) * 2

Substituting the given values, we find:

diameter ≈ √(650 / (310 * 10^6 * π)) * 2 ≈ 11.62 mm

Therefore, the required diameter of the rod to withstand the load is approximately 11.62 mm.

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Answer:

B

Explanation:

First narrow answers to CIRCULAR motion.

Only A and B have circular motion.

Second look at the keyword UNIFORM. A swing does not have uniform motion because it will accelerate due to gravity and have the most velocity at the bottom of the swing.

Therefore, it must be B.

The momentum flow rate through the pipe carrying liquid ammonia is 1 × \(10^6\) kg·m/s.

The flow rate of momentum (Ṁ) through the pipe can be calculated by multiplying the mass flow rate (ṁ) by the velocity (v). The speed can be determined using the equation v = ṁ / (ρA), where ρ is the density of the liquid ammonia and A is the pipe's cross-sectional area.

Given:

ṁ = 100 kg/s

A = 0.01 m²

Assuming the density (ρ) of liquid ammonia is 700 kg/m³, we can calculate the velocity (v):

v = ṁ / (ρA)

v = 100 kg/s / (700 kg/m³ × 0.01 m²)

v = 10000 m/s

Now, we can calculate the flow rate of momentum (Ṁ):

Ṁ = ṁv

Ṁ = 100 kg/s × 10000 m/s

Ṁ = 1 × \(10^6\) kg·m/s

Therefore, the momentum flow rate through the pipe is 1 × \(10^6\) kg·m/s.

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Answer: Jay

Explanation:

1 km in .1 hours is the same as 1 km in 6 minutes

1/4 km in .05 hours is the same as 1/4 km in 3 minutes or 1 km in 12 minutes

Answer:

The first graph

Explanation:

Graph A shows acceleration.

The tension in the string connecting block 138 to block 137 is 138 N, which is the same as the tension in the string connecting each block in the chain.

When a force is applied to the first block in the chain, it creates tension in the string connecting it to the second block. This tension is then transferred to the second block, which creates tension in the string connecting it to the third block, and so on. Therefore, as we move down the chain, the tension in the string between each block will remain constant.

In this case, the force applied to the first block is 138 N. Since the blocks are identical and there is no friction, each block will experience the same force. Therefore, the tension in the string between each block will also be 138 N.

To find the tension in the string connecting block 138 to block 137, we can simply look at the tension in the string connecting block 137 to block 136, which is also 138 N. This tension is then transferred to block 138, so the tension in the string between block 138 and block 137 is also 138 N.

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Answer:

D

Explanation:

physical properties are properties that can be measured with instruments, such as a scale for weight.

The descriptions are chemical properties

Gold's atoms have more protons than those of many other metals, it is a particularly dense metal. Statement D explains the physical characteristics of the constituent parts of a substance.

Density is defined as the mass per unit volume. It is an important parameter to understand the fluid and its properties. Its unit is kg/m³.

Gold is a relatively dense metal because its atoms have more protons than atoms of many other metals.

Statement D describes a physical property related to the particles that make up a material. Because density is the physical property of the material.

Hence option D is corect.

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Answer:

D

Explanation:

Edg2020

When a stone is hurled horizontally from the top of a 31.2 m tall cliff, its horizontal initial velocity is 39.25 m/s.

By measuring the ball's diameter d and dividing it by the time t it takes for it to cross the photogate, one can also calculate the ball's starting horizontal velocity. Therefore, Vo = d/t. The kinematics equations of motion can be used to calculate the horizontal velocity of a projectile motion made by a person or an item.

Height of cliff is 31.2 m

distance is 21.4

t = √2h/g

t = √2×31.4/9.8

t = √62.8/9.8

t = 0.80

V = 31.4/0.08

V = 39.25 m/s.

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Answer:

t = 0.306 s

Explanation:

Given that,

Initial velocity of the ball, u = 3 m/s

When it reaches the top of its path, its final velocity, v = 0

We need to find the time it takes the ball to reach the top of its path. It can be calculated using first equation of motion.

v = u +at

Here, a = -g

\(0=u-gt\\\\t=\dfrac{u}{g}\\\\t=\dfrac{3\ m/s}{9.8\ m/s^2}\\\\t=0.306\ s\)

So, the ball will take 0.306 s to reach the top of its path.

Answer:

this statement is true because I looked it up

Answer:

zero

Explanation:

Because if the object is not in motion then it only posseses potential energy not kinetic energy