A 8.2 kg ball moving with a velocity of 14.6 m/s collides with a 3.3 kg ball moving with a velocity of 4.3 m/s in the opposite direction. Find the common velocity of the balls (in m/s) if they coalesce on impact. (Enter the magnitude only)

Answer:

Option C. is correct

Explanation:

The magnetic field is the area around a magnet in which there is magnetic force. When an electric current flows through a wire a  magnetic field is created. A single wire does not produce a strong magnetic field. So, to increase the power  of the magnetic field, increase the number of coils in the wire.

Each time gravity pulls the man down 0.5 m, it does work of -367.5 J. This work is converted into kinetic energy as the man gains velocity while moving downwards.

The work performed by gravity pulling a man down 0.5 m after each step can be determined utilizing the recipe: work = force x distance x cos(theta), where power is the heaviness of the man (mass x speed increase because of gravity), distance is 0.5 m, and theta is the point between the course of power and the heading of movement (which is 180 degrees since the man is moving downwards).

Expecting the man has a mass of 75 kg, the weight can be determined as 75 kg x 9.8 m/s^2 = 735 N. In this way, the work done by gravity after each step is:

work = 735 N x 0.5 m x cos(180 degrees) = - 367.5 J

The negative sign demonstrates that the work done by gravity is the other way to the relocation of the man. In this way, each time gravity pulls the man down 0.5 m, it takes care of business of - 367.5 J. This work is changed over into active energy as the man acquires speed while moving downwards.

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Answer:

Because the truck is moving to the east faster than the ball is moving west.

Explanation:

In this problem the text reads:

Ball                                          Truck

Rolling west at 0.25 m/s Moving east at 1.0 m/s

You get asked if the observer in the truck would see the ball moving to the west, the answer is yes, cause the observer in the truck would be moving to the east along with the truck, but someone looking at the ball from the outside will see the ball moving east because the speed of the ball is less than that of the truck.

Answer:

Explanation:

First, let's find A x (B+C):

A x (B+C) = A x B + A x C (distributive property of cross product over vector addition)

Now, we need to find A x B and A x C separately:

A x B = (3i + 3j - 2k) x (i - 4j + 2k) (using the cross product formula)

= (31 + 32) i - (34 - 21) j + (3*(-2) - 3*1) k

= 9i + 10j - 9k

Similarly,

A x C = (3i + 3j - 2k) x (4i - j + 2k)

= (34 - 2(-1)) i - (32 + 31) j + (31 - 34) k

= 14i - 9j - 9k

Now, we can substitute these values into the original equation:

A x (B+C) = A x B + A x C

(3i + 3j - 2k) x [(i - 4j + 2k) + (4i - j + 2k)] = (9i + 10j - 9k) + (14i - 9j - 9k)

Simplifying the right side:

(9i + 10j - 9k) + (14i - 9j - 9k) = 23i + 1j - 18k

Now, let's calculate the left side:

(3i + 3j - 2k) x [(i - 4j + 2k) + (4i - j + 2k)]

= (3i + 3j - 2k) x (5i - 5j + 4k)

= (35 + 24) i - (3*(-5) + 34) j + (35 - 2*(-5)) k

= 23i + 1j + 15k

Therefore, we have:

23i + 1j - 18k = 23i + 1j + 15k

Since the left and right sides are equal, we have shown that:

A x (B+C) = A x B + A x C.

The difference is that Hall effect sensor detects the strength of a magnetic field perpendicular to it, while a permanent magnetic sensor detects the angle of a parallel magnetic field.

The permanent magnetic sensor tends to be more accurate as it has a bigger detectable area that has detects layout error.

What is a magnet?

A magnet is any material that produces a magnetic field. A magnet has north and south poles at opposite ends.

The difference between a hall effect magnet and a permanent magnet sensor is that hall effect detects  magnetics fields perpendicular to it while a permanent magnet detects magnetic fields parallel to it.

In conclusion, the permanent magnet is more accurate as it has a wider detectable area.

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Answer:

Answer: 354311 N/C, 275095 N/C, 458528 N/CExplanation:E = kQ/[r * √(r² + L/2)²], where L = 10 cm = 0.1 mE = kQ/[r * √(r² + 0.05²)]For the glass rod,E at 1 cm = E = k * 15*10^-9 / [0.01 * √…

Explanation:

Answer:

I hope this helps

We have that the angle is

\(\theta=32.53\)

From the Question we are told that

E⃗ =5.00×105ı^N/C

Generally the equation for Tension   is mathematically given

\(W=Tcos\theta\)

Where

\(tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}\)

\(\theta=32.53\)

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The time it takes the stone to fall from a height of 7.11 m is 1.2 seconds.

Time can be defined as an ongoing and continuous sequence of events that occur in succession, from past through the present, and to the future.

To calculate the time it takes the stone to drop from an height of 7.11 m, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1 and solve for t.

Hence, the time it takes the stone to fall is 1.2 seconds.

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Answer:

A.

Explanation:

this would make sense but it seems to be more like they both sound different

Answer:

deductive reasoning usually follows steps .

The total resistance of the circuit is 49 Ω. The current strength in the circuit, when connected to a voltage source of 56 V, is approximately 1.14 A.

To calculate the total resistance of the circuit, we need to determine the equivalent resistance of the resistors connected in a series.

Given:

R1 = 4 Ω

R2 = 30 Ω

R3 = 10 Ω

R4 = 5 Ω

Calculate the equivalent resistance (RT) of R1 and R2, as they are connected in series:

RT1-2 = R1 + R2

RT1-2 = 4 Ω + 30 Ω

RT1-2 = 34 Ω

Calculate the equivalent resistance (RTotal) of RT1-2 and R3, as they are connected in parallel:

1/RTotal = 1/RT1-2 + 1/R3

1/RTotal = 1/34 Ω + 1/10 Ω

1/RTotal = (10 + 34) / (34 * 10) Ω

1/RTotal = 44 / 340 Ω

1/RTotal ≈ 0.1294 Ω

RTotal ≈ 1 / 0.1294 Ω

RTotal ≈ 7.74 Ω

Calculate the equivalent resistance (RTotalCircuit) of RTotal and R4, as they are connected in series:

RTotalCircuit = RTotal + R4

RTotalCircuit = 7.74 Ω + 5 Ω

RTotalCircuit ≈ 12.74 Ω

Therefore, the total resistance of the circuit is approximately 12.74 Ω.

To determine the current strength (I) when connected to a voltage source of 56 V, we can use Ohm's Law:

I = V / RTotalCircuit

I = 56 V / 12.74 Ω

I ≈ 4.39 A

Therefore, the current strength in the circuit, when connected to a voltage source of 56 V, is approximately 4.39 A (or 1.14 A, considering significant figures).

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Answer:

Explanation:

The acceleration of an electron in an electric field is given by the equation:

a = qE/m

where a is the acceleration, q is the charge of the electron, E is the electric field, and m is the mass of the electron.

Given that the acceleration of the electron is 4.3 m/s^2, and the mass of the electron is 9.11 × 10^-31 kg, and the charge of the electron is -1.6 × 10^-19 C, we can solve for the electric field E:

E = ma/q

E = (4.3 m/s^2) × (9.11 × 10^-31 kg) / (-1.6 × 10^-19 C)

E = -2.44 × 10^4 N/C

The negative sign indicates that the direction of the electric field is opposite to the direction of the electron's motion. Therefore, the magnitude of the electric field required to accelerate an electron with an acceleration of 4.3 m/s^2 is 2.44 × 10^4 N/C and the direction is opposite to the direction of motion of the electron.

The acceleration of the falling obeject of mass 2.55 kg is  -8.2 m/s². And the right option is d) -8.2 m/s².

This can be defined as the rate of change of velocity.

To calculate the acceleration of the object, we use the formula below.

Formula:

Where:

Note: Taking forces acting upward to be positve and downward force to be negatove,

From the question,

Given:

Substitute these values into equation 2

Hence, The acceleration of the object is -8.2 m/s².

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RESPUESTA

a) C = 3.54 x 10⁻⁹ F = 3.54 nF

b) Q = 35.4 x 10⁻⁶ C = 35.4 μC

c) E = 2,000,000 V/m

EXPLICACIÓN

a) La capacitancia de un capacitor de placas paralelas se define como la razón entre la magnitud de la cantidad de carga en cada placa y la diferencia de potencial aplicado. A su vez, ésta depende de la geometría del capacitor. La ecuación que la define es:

donde C es la capacitancia, A es el área de las placas y d es la distancia que separa las placas. Además ε0 es la permitividad del vacío. En general la fórmula también incluye la permitividad relativa del material dielétrico (el material entre las placas del capacitor) pero en este problema no nos interesa, dado que el material entre las placas de este capacitor es el vacío.

De esta manera, si reemplazamos los datos de este problema:

• ε0 = 8.85 x 10⁻¹² F/m

• A = 2 m²

• d = 5 mm

Nota que tanto la permitividad como el área de las placas tienen unidades de metros, pero la distancia de separación entre placas está en milímetros, por lo tanto debemos convertirla a metros:

Ahora sí, reemplazamos en la ecuación de arriba y obtenemos la capacitancia:

Cuando escribimos una cantidad en notación científica, y el exponente del 10 es -9 podemos no escribir esta parte y usar el prefijo nano. De esta manera, decimos que la capacitancia es de 3.54 nF.

b) La carga de las placas podemos obtenerla de la otra ecuación para definir la capacitancia:

C es la capacitancia que encontramos en el punto a), V es la diferencia de potencial aplicada y Q lo que estamos buscando, la carga de las placas. Del enunciado, tenemos que la diferencia de potencial es V = 10,000V. Reemplazando y resolviendo para Q:

Si movemos el punto decimal un lugar hacia la derecha tenemos: 35.4 x 10⁻⁶ C, que puede llevar el prefijo de micro. Para el prefijo micro usamos la letra griega mu (μ). Entonces la carga de las placas es 35.4 μC

c) Finalmente, para encontrar la magnitud del campo eléctrico, utilizaremos la relación entre la diferencia de potencial y la distancia entre las placas:

Reemplazando con V = 10,000V y d = 0.005m, el campo eléctrico es:

The acceleration on a body that approached the earth and comes within 6 earth radii of the earth’s surface is 0.2 m/s².

The acceleration of the body can be calculated using the equation-

g = g0 R/(R + H)².

In the question, we have, h = 6R

g = 9.8 x R/(R + 6R)²

By putting in the values and calculating, we get,

g = 0.2 m/s².

Where g' = Gravitational Acceleration at Height, R = Earth's Radius, and g = Gravitational Acceleration at Earth's Surface. As a result, at a distance of 2R from the earth's surface, the acceleration of a body caused by the earth's attraction is equal to g/9.

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Answer:

When speed reduces, acceleration decreases

Explanation:

From first equation of motion;

\({ \tt{v = u + at}}\)

When time is constant:

\({ \tt{v = 0 + kt}} \\ { \tt{v = kt}} \\{ \tt{v \: \alpha \: \: a}}\)

When speed is decreasing, velocity and acceleration are in opposite directions.

This implies that the opening price was 21080.28.

Closing index = 21029.47

plus down movement =50.81

The Dow Jones Industrial Average measures the stock price performance of 30 large companies listed on US stock exchanges. Stock index. Others include the S&P 500 Index and NASDAQ.

The Opening Price Index represents the average opening price of the day before the trading day. During trading, the price must have experienced fluctuations, ups and downs. However, at the end of the trading day, the closing price was reported at 21029.47, down 50.81.

Adding back from this closing index report, the opening price was 50.81 or about 51 basis points higher than the closing price.

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The period of the pendulum, given that the pendulum has an angular velocity about the pivot point of 0.66 radians/sec is 9.52 seconds

First, we shall list out the given parameters from the question. This is shown below:

The period and angular velocity are related according to the following formula:

ω = 2π/ T

Inputting the given parameters, we can obtain the period of the pendulum as follow:

0.66 = (2 × 3.14) / T

0.66 = 6.28 / T

Cross multiply

0.66 × T = 6.28

Divide both sides by 0.66

T = 6.28 / 0.66

T = 9.52 seconds

Thus, we can conclude that the period of the pendulum is 9.52 seconds

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H⁣⁣⁣⁣ere's l⁣⁣⁣ink t⁣⁣⁣o t⁣⁣⁣he a⁣⁣⁣nswer:

bit.\(^{}\)ly/3a8Nt8n

Answer:

C. Gold

Explanation:

Why?

Because air , water, glucose is pure substance or liquid

Complete Question

A rolling ball moves from \(x_1 = 8.0 \ cm\) to \(x_2 = - 4.1 \ cm\) during the time from \(t_1 = 2.9 s\)  to  \(t_2 = 6.0s\)

What is its average velocity over this time interval?

Answer:

The velocity is  \(v = 3.903 \ m/s\)

Explanation:

From the question we are told that

    The first position of the ball is  \(x_1 = 8.0 \ cm\)

    The second position of the ball is  \(x_2 = - 4.1 \ cm\)

Generally the average velocity is mathematically represented as

       \(v = \frac{ x_1 - x_2}{t_2 - t_1}\)

=>    \(v = \frac{ 8 - -4.1 }{ 6 - 2.9 }\)

=>    \(v = 3.903 \ m/s\)

Answer:

Based on the information provided, the lightbulb lights when the loop rotates because the movement of the wire in an electric or magnetic field causes electrons in the wire to move and become either mechanical energy or an electric current. This energy causes the light to glow. The energy change involved is the conversion of electrical or mechanical energy used to rotate the loop into either light or electrical energy

Explanation:

The point through which this ray will pass, after refraction by the lens is point D.

The refraction of light refers to the bending or change in direction that occurs when light passes from one medium to another. It is a phenomenon that happens due to the difference in the speed of light in different substances.

From the ray diagram given, after the light incident from point O, it will pass the converging at point D which is the focal length of the lens after refraction.

Thus, based on the converging lens given in the ray diagram, we can conclude that, the point through which this ray will pass, after refraction by the lens is point D.

So point D is the correct answer.

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Answer:

\(39000\ \text{kg m/s}\)

West

Explanation:

m = Mass of car = \(1.3\times 10^{3}\ \text{kg}\)

t = Time = 9 seconds

u = Initial velocity = 30 m/s

v = Final velocity = 0

Impulse is given by

\(J=m(v-u)\\\Rightarrow J=1.3\times 10^3(0-30)\\\Rightarrow J=-39000\ \text{kg m/s}\)

The magnitude of the total impulse applied to the car to bring it to rest is \(39000\ \text{kg m/s}\).

The direction is towards west as the sign is negative.

the force the woman exerts to do a push-up at constant speed is 333 N.

the work the woman does is 152 J.

her useful power output is 76 W.

(a) To calculate the force the woman exerts to do a push-up, we need to use torque methods. The woman is doing a push-up at constant speed, which means that the net torque on her body is zero. The only torque acting on her body is due to her weight W, which acts at the center of mass of her body. The distance between her center of mass and her hands is 0.76 m, and the angle between her body and the horizontal is 45 degrees.

The torque due to her weight about her hands is given by:

τ = r x W = (0.76 m) x (cos 45°)(W)

where r is the distance between her hands and her center of mass and cos 45° is the component of the distance perpendicular to the weight vector. Since the woman is at constant speed, the torque she exerts about her hands must be equal and opposite to the torque due to her weight. Therefore:

τ = (0.76 m)(cos 45°)(W) = (1/2)(W)(0.76 m)

Solving for W, we get:

W = 2(τ/0.76 m) = 2[(0.5)(mg)(0.76 m)/(0.76 m cos 45°)] = 333 N

Therefore, the force the woman exerts to do a push-up at constant speed is 333 N.

(b) The work the woman does is equal to the change in her potential energy as her center of mass rises. The woman's mass is not given, so we will assume a value of 60 kg. The gravitational potential energy of the woman is given by:

U = mgh

where m is the mass of the woman, g is the acceleration due to gravity (9.81 m/s^2), and h is the height her center of mass rises (0.26 m). Therefore:

U = (60 kg)(9.81 m/s^2)(0.26 m) = 152 J

Therefore, the work the woman does is 152 J.

(c) The useful power output of the woman is the work she does per unit time, taking into account the work done in lowering her body. Each push-up involves two phases: lifting her body and lowering her body. When she lowers her body, the work done is negative, as the force she exerts is in the opposite direction to the displacement. The work done in lowering her body is equal to the work done in lifting her body, so the total work done in one push-up is zero.

The woman does 30 push-ups in 1 minute, which means she does one push-up every 2 seconds. Therefore, the useful power output of the woman is:

P = (152 J)/(2 s) = 76 W

Therefore, her useful power output is 76 W.

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