A student pushes a 0.5 kg lab cart with a force of 3 Newtons. Determine the acceleration of the cart in
m/s2.

After plugging all the data into the equation, the result of the relative centrifugal force (RCF)  is measured in terms of g.

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

\(RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2\)

where;

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

\(RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g\)

Thus, after plugging all the data into the equation, the result is measured in terms of g.

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The temperature gradient in the flow of direction is 294525 W.

A temperature gradient is the gradual variance in temperature with distance. The slope of the gradient is consistent within a material. A gradient is established anytime two materials at different temperatures are in physical contact with each other.

Q= T/( L/ KA)

Q= ( 1500 − 450) / 0.15 / 9.35v * 4.35)

   = 294525 W

Units of measure of temperature gradients are degrees per unit distance, such as °F per inch or °C per meter.

Many temperature gradients exist naturally, while others are created. The largest temperature gradient on Earth is the Earth itself. Q= T/Ka.

Therefore, The temperature gradient in the flow of direction is 294525 W.

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Answer:

Explanation:

From the given information;

Let assume that the distance travelled by the speeder prior to the time the trooper catches with it to be = d

the time interval to be = t

Then, the speeder speed = distance/time

Making distance the subject; then:

distance (d) = speed × time

d = (50 m/s)t

d = 50 t --- (1)

Now, for the trooper; Using the equation of motion:

\(d = ut + \dfrac{1}{2}at^2\)

\(d = (20)t+\dfrac{1}{2}(2.5)t^2\)

d = 20t + 1.25t²

Replace the value of d in (1) to the above equation, we have:

50 t = 20 t + 1.25t²

50t - 20t = 1.25t²

30t = 1.25t²

30 = 1.25t

\(t = \dfrac{30}{1.25}\)

t = 24 seconds

From (1), the distance far down the highway the trooper will travel prior to the time it catches up with the speeder is:

= 50t

= 50(24)

= 1200 seconds

Explanation:

A concave lens is also known as a diverging lens because it is shaped round inwards at the centre and bulges outwards through the edges, making the light diverge. They are used to treat myopia as they make faraway objects look smaller than they are.

Answer:

Motor cortex

Both hemispheres have a motor cortex, with each side controlling muscles on the opposite side of the body (i.e, the left hemisphere controls muscles on the right side of the body).

Explanation:

Mirrors are a common tool used by painters to depict a variety of viewpoints and nuanced interactions between the observer and the artist.

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Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point =  mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

Answer:

9.23m

Explanation:

Max height = u²sin²theta/2g

u is the speed = 34.7m/s

theta is the angle of elevation = 33°

g is the acceleration due to gravity = 9.8m/s²

Substitute into the formula

Max height = 24.7²sin²33/2(9.8)

Max height = 610.09sin²33/2(9.8)

Max height = 610.09(0.29663)/19.6

Max height = 180.97/19.6

Max Height = 9.23m

Hence the max height the football reaches during flight is 9.23m

Based on the diagram, the correct statement is: Government and consumers make economic decisions in a mixed economy, while consumers make economic decisions in a market economy.

In a mixed economy, economic decisions are made by both the government and consumers.

The government plays a significant role in regulating and influencing economic activities through policies, regulations, and interventions.

In market economy, economic decisions are primarily made by consumers. The market forces of supply and demand dictate the allocation of resources, production levels, and pricing.

The freedom to buy and sell whatever they choose is what ultimately determines how commodities and services are produced and distributed.

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The uniform line of charge and \(\rho\) for the uniformly charged sphere. Using slope the value of \(\lambda\) is  \(\\\frac{4\pi\epsilon_0\times slope}{2}$\).

For the uniform line of charge, the electric field $E$ is inversely proportional to the distance $r$, so $Er$ should be constant. For the uniformly charged sphere, the electric field $E$ is inversely proportional to the square of the distance $r$, so $Er^2$ should be constant. Therefore, we can use the graphs of $Er^2$ versus $r$ and $Er$ versus $r$ to determine which data set is for the uniform line of charge and which set is for the uniformly charged sphere.

For data set A, the graph of $Er^2$ versus $r$ is a straight line, which means that $Er^2$ is constant. Therefore, data set A is for the uniformly charged sphere. For data set B, the graph of $Er$ versus $r$ is a straight line, which means that $Er$ is constant. Therefore, data set B is for the uniform line of charge.

For the uniform line of charge, the electric field $E$ is given by

E =\(\frac{2\lambda}{4\pi\epsilon_0r}$,\)

where $\lambda$ is the charge per unit length and $\epsilon_0$ is the permittivity of free space. Since $Er$ is constant, we can write $Er = \frac{2\lambda}{4\pi\epsilon_0}$. From the graph of $Er$ versus $r$ for data set B, we can find the slope of the straight line, which is equal to $Er$. Therefore, we can use the slope to calculate $\lambda$:

\($\lambda = \frac{4\pi\epsilon_0Er}{2} = \frac{4\pi\epsilon_0\times slope}{2}$\)

For the uniformly charged sphere, the electric field $E$ is given by $E = \frac{\rho r}{3\epsilon_0}$, where $\rho$ is the volume charge density. Since $Er^2$ is constant, we can write $Er^2 = \frac{\rho r^3}{3\epsilon_0}$. From the graph of $Er^2$ versus $r$ for data set A, we can find the slope of the straight line, which is equal to $Er^2$. Therefore, we can use the slope to calculate \($\rho$\)

\($\rho = \frac{3\epsilon_0Er^2}{r^3} = \frac{3\epsilon_0\times slope}{r^3}$\)

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Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

a) The velocity of the 3 kg ball after the collision will be 19.33 m/sec.

b) The kinetic energy before and after the collision IS 294 and 560 J.

c)The collision is elastic in nature.

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

a)

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

\(\rm m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\\\\ 8 \times 4 +3 \times 14 = 8 \times 2 + 3 \times v_2 \\\\ v_2 = 19.33 \ m/sec\)

b)

The kinetic energy before the collision is:

\(\rm KE_1 = \frac{1}{2} mu_1^2 \\\\ \rm KE_1 = \frac{1}{2} \times 3 \times 14 ^2 \\\\ KE_1=294 \ J\)

The kinetic energy after the collision is:

\(\rm KE_2 = \frac{1}{2} mv_2^2 \\\\ \rm KE_1 = \frac{1}{2} \times 3 \times (19.33) ^2 \\\\ KE_2=560.47 \ J\)

c)

This is an elastic collision because the kinetic energy before the collision is equal to the kinetic energy after the impact.

Hence, the velocity of the 3 kg ball after the collision will be 19.33 m/sec and the collision is elastic in nature.

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The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Answer:

a) 20.29N

b) 19.43N

c) 15N

Explanation:

To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.

In all these calculations you can asume that one of the vectors coincides with the x-axis.

a)

\(F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N\)

b)

\(F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N\)

c)

\(F_R=(9cos(90\°)+12)\hat{i}+(9sin(90\°))\hat{j}\\\\F_R=(12N)\hat{i}+(9N)\hat{j}\\\\|F_R|=\sqrt{(12N)^2+(9N)^2}=15N\)

The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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The final temperature of the mixture is 50°C.

Temperature of the hot water, T₁ = 80°C

Temperature of the cold water, T₂ = 20°C

According to the principle of calorimetry, the heat lost by the hot body is equal to the heat gained by the cold body.

So,

Heat lost by the hot water = Heat gained by the cold water

mC(T₁ - T) = mC(T - T₂)

Since, both are water and the amount of water is the same for both,

T₁ - T = T - T₂

Applying the values of T₁ and T₂,

80 - T = T - 20

2T = 100

Therefore, the final temperature of the mixture is,

T = 100/2

T = 50°C

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The statement" The direction of the force on the charge placed in the electric field is upward" is false because the direction of the force on a negative charge (-6 C) placed in an upward-directed uniform electric field of magnitude 24 N/C would be downward.

The direction of the force on a charged particle placed in an electric field is determined by the charge of the particle and the direction of the electric field. In this case, a charge of -6 C is placed in an electric field directed upward with a magnitude of 24 N/C.

The force on a charged particle in an electric field can be calculated using the formula:

F = q * E

Where F is the force, q is the charge of the particle, and E is the electric field.

Since the charge q in this case is negative (-6 C) and the electric field E is directed upward, we can substitute the values into the formula:

F = (-6 C) * (24 N/C)

F = -144 N

The negative sign in the force value indicates that the force is in the opposite direction to the electric field. Therefore, the force on the charge placed in the electric field is downward, not upward.

The force on a negative charge is always opposite to the direction of the electric field. This is because negative charges experience an attractive force towards positive charges, and electric fields are directed from positive charges to negative charges.

Therefore, the statement "The direction of the force on the charge placed in the electric field is upward." is false.

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The net force on particle q₁ is approximately +25.6 N.

The electrostatic forces between particle q1 and the other two particles, q2 and q3, must be taken into account in order to determine the net force on particle q1. Coulomb's Law describes the electrostatic force between two charged particles:

F = k * |q₁ * q₂| / r²

F is the force, k is the electrostatic constant (9 x 109 N m2/C2), q1 and q2 are the charges' magnitudes, and r is the distance separating them.

Let's first determine the force between q1 and q2:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9 x 10^9 N m²/C²) * |(-29.6 μC) * (+37.7 μC)| / (0.630 m)²

F₁₂ = (9 x 10^9 N m²/C²) * (29.6 x 10^-6 C) * (37.7 x 10^-6 C) / (0.630 m)²

F₁₂ ≈ -7.45 N

The absence of a positive sign suggests an attractive force between q1 and q2.

Let's next determine the force between q2 and q3:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9 x 10^9 N m²/C²) * |(+37.7 μC) * (-10.8 μC)| / (0.315 m)²

F₂₃ = (9 x 10^9 N m²/C²) * (37.7 x 10^-6 C) * (10.8 x 10^-6 C) / (0.315 m)²

F₂₃ ≈ +33.05 N

The presence of a positive sign suggests a repulsive force between q2 and q3.

We must now add all the forces in order to determine the net force on q1:

Net force = F₁₂ + F₂₃

Net force ≈ -7.45 N + 33.05 N

Net force ≈ +25.6 N

The presence of a positive sign implies that the net force is pointing to the right, in the same direction as particle q2.

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Answer:

The period of the wave does not change looting the value that accompanies the time, the wavelength does not change since it is the constant that accompanies x.

We see that the amplitude is twice the amplitude of the incident waves. Since the wave is stationary the velocity is zero

Explanation:

In this exercise we are given the equation of two traveling waves, it is asked to find the resulting wave

    u = f + g

    u = 2 sin (x + t) + 2 sin (x-t)

we will develop double angled breasts

    u = 2 [(sin x cos t + sin t cos x) + (sin x cos t - sin t cos x)]

    u = 2 [2 sin x cos t]

    u = 4 sin x cos t

The period of the wave does not change looting the value that accompanies the time, the wavelength does not change since it is the constant that accompanies x.

We see that the amplitude is twice the amplitude of the incident waves. Since the wave is stationary the velocity is zero

With a horizontal velocity of 3.04 m/s and a horizontal displacement of 1.68 m from the end of the diving board, the swimmer enters the water 1.70 metres below the diving board.

The rate at which something moves in a certain direction is referred to as its velocity. as quickly as a car travelling north on a highway or a rocket taking flight.

We can use the kinematic equations of motion to solve this issue.

Use the swimmer's horizontal travel distance as the displacement in the x-direction. Given that the swimmer enters the water 1.68 metres from the board's end, the following is the answer:

x=1.68 m and v0x=3.04 m/s

Δx = v0x * t

calculating t:

t = 1.68 m / 3.04 m/s because x / v0x.

t = 0.5526 s

Thus, the swimmer enters the water in 0.5526 seconds.

"y" equals "v0y*t" plus "(1/2)*a*t2"

replacing the values with:

Δy = 0 + (1/2) * (-9.81 m/s²) * (0.5526 s)²

Δy = -1.70 m

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Answer:

The three main/basic parts of the cell are:

Cell Membrane (Plasma Membrane)

Cytoplasm

Nucleus

Explanation:

The frequency of f3 is approximately 716 Hz.

The frequency of a repeated event is its number of instances per unit of time. Hertz (Hz), which stands for the number of cycles per second, is a popular unit of measurement.

a. Given two frequencies, f1 and f2, the frequency ratio is as follows:

frequency ratio= \(\frac{f2}{f1}\)

Inputting the values provided yields:

frequency ratio = \(\frac{576}{320Hz}\) =1.8.

As a result, the difference in frequency between f1 = 320 Hz and f2 = 576 Hz is 1.8.

b. Since there are 12 half-steps in an octave and a fourth is a distance of 5 half-steps, going down a fourth requires dividing the frequency by \(2^{(4/12)}\). Hence, once a fourth is subtracted, the frequency ratio between f and f1 is:

frequency ratio= \(\frac{f}{ (f1 /f2 ) }\)=  \(\frac{f}{ (f1 / 1.3348) }\)

By dividing the numerator and denominator by 1.3348, we may make this more straightforward:

frequency ratio= (f × 1.33348)/f1

As a result, (f × 1.3348) / f1 is the frequency ratio between f and f1 after descending a fourth.

c. (c) To find the frequency of f3, we need to know the interval between f1 and f3. Let's assume that f3 is a fifth above f2. The frequency ratio for a fifth is given by: \(2^{(7/12)}\) = 1.49831

Therefore, the frequency of f3 is:

f3 = f1 × (\(2^{(7/12)}\)) × (\(2^{(7/12)}\)) = 320 Hz × 1.49831 ×1.49831 = 716 Hz

Therefore, the frequency of f3 is approximately 716 Hz.

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Answer:

B

Explanation:

Answer:

B

Explanation:

Answer:3 N —->

Explanation:

In the case when  the forces act on the car as shown in the picture so here the net force should be 3N.

Since

A parachute in the back of the car is released to help slow down the car.

So here the net force should be

= 10 N - 7N

= 3 N

Hence, In the case when  the forces act on the car as shown in the picture so here the net force should be 3N.

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Answer:

a) By \(v^2 = u^2 + 2as\) => a= 70291.70.

(b)By \(v = u + at\) => t= 1.58 ms.

(c)By \(v^2 = u^2 - 2gh\) => H = 46045.92 m.

Explanation:

a) By \(v^2 = u^2 + 2as\)

\((950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70\)

(b)By \(v = u + at\)

\(950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms\)

(c)By \(v^2 = u^2 - 2gh\)

\(0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m\)

Answer:

Multiply the friction factor by pipe length and divide by pipe diameter. Multiply this product by the square of velocity. Divide the answer by 2.

Explanation:

Multiply the friction factor by pipe length and divide by pipe diameter. Multiply this product with the square of velocity. Divide the answer by 2.

Hope this answer helps you

Answer:

I think its C

Answer:

D

Explanation:

If the temperature doesn't change but the mass of the object increases, the thermal energy in the object increases.

The resultant displacement of the man is 109.77 km in the direction N60°E.

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

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