If m=120kg and a=15m/s2, what is the force

Answer:

Explanation:

The pendulum is at rest ( or balance due to the equal action of opposing forces).

Answer:

Equilimbru or maximum displacement is question.

Answer:

6.39×10^23 kg is the weight on mass

Answer:

5.08cm³

Explanation:

Given parameters

The larger cube has twice the mass of the smaller cube

Mass of smaller cube = 20g

Mass of the larger cube = 2 x 20  = 40g

Density of the smaller cube  = 7.87g/cm³

Unknown:

Volume of the larger cube = ?

Solution:

Density is the mass per unit volume of substance. For all samples of a substance, the density value is the same.

So, the density of the small and large iron is the same

   Density  = \(\frac{mass}{volume}\)  

        Volume  = \(\frac{mass}{density}\)  

 So;

  Volume of larger cube  = \(\frac{40}{7.87}\)  = 5.08cm³

Answer:

5.08cm^3

Explanation:

Answer:

The mass of the plane is 5,096.8 meters

Explanation:

The formula for gravitational potential energy is

\(GPE=mgh\)

Note:

\(GPE\) is the gravitational potential energy in joules

\(m\) is the mass in kilograms

\(g\) is the acceleration due to gravity

\(h\) is the height in meters

In this example we are given the height and the gravitational potential energy. Knowing that we can evaluate the mass.

\(GPE=60000000\\g=9.81\\h=1200\\\)

\(60000000=m*9.81*1200\)

\(\frac{60000000}{11772} =m\)

\(m=5096.83996\)

\(m=5096.8\)

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Answer:

the formula v = f×lambda.

v= 4.25× 5.6

therefore speed is

23.8 meters per second

Answer:

The answer is A because it is a resource that takes many many years to make more of and we cannot make it ourselves

The displacement of the string at the position x=0, assuming that the standing wave ys(x,t) is present, is zero. This means that the particles of the string located at x = 0 do not move when the standing wave is present.

When a standing wave is present, the displacement of a string can be calculated using the equation y_s(x,t) = 2Asin(kx)sin(ωt); where y_s(x,t) is the displacement of the string, A is the amplitude of the wave, k is the wave number, x is the position of a particle on the string, ω is the angular frequency of the wave and t is the time. The wave number is given by k =\frca{ nπ}{L}; where n is the harmonic number and L is the length of the string. In the case of x = 0, the displacement of the string can be calculated asy_s(0,t) = 2Asin(kx)sin(ωt)y_s(0,t) = 2Asin(0)sin(ωt)Since sin(0) = 0, the displacement of the string at x = 0 is zero. In other words, the particles of the string located at x = 0 do not move when the standing wave is present. In conclusion, the displacement of the string at the position x=0, assuming that the standing wave ys(x,t) is present, is zero.

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Velocity and displacement take direction of motion into account while distance does not take direction of motion into account.

Vector quantities take into account both magnitude and direction of motion while scalar quantities only take the magnitude of motion into account. In describing the motion of an object, we use terms such as; velocity, distance, displacement etc.

Looking at the options provided in the question, displacement and velocity are vectors hence they take the direction of motion into account. Distance is a scalar quantity hence it does not take direction of motion into account.

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Answer:

50N

Explanation:

Impulse = Force × Time

25 = Force × 0.5

Force = 25/0.5

= 50N

Answer:

\(4.5\ \text{N}\)

Explanation:

\(F_1\) = Gravitational force between the objects = \(18\ \text{N}\)

\(r_1\) = Initial distance between the two objects

\(r_2\) = Final distance between the two objects = \(2r_1\)

Gravitational force between two objects is given by

\(F=\dfrac{Gm_1m_2}{r^2}\)

So

\(F\propto \dfrac{1}{r^2}\)

\(\dfrac{F_2}{F_1}=\dfrac{r_1^2}{r_2^2}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{r_1^2}{(2r_1)^2}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{1}{4}\\\Rightarrow F_2=\dfrac{F_1}{4}\\\Rightarrow F_2=\dfrac{18}{4}\\\Rightarrow F_2=4.5\ \text{N}\)

The new force of attraction between the objects is \(4.5\ \text{N}\).

Answer:

Explanation is in the picture and the answer is 16

7.11 meters

Explanation

to solve this we need to use this formula:

Step 1

Let

now,replace and calculate

therefore, the answer is

7.11 meters

I hope this helps you

We can see that the electric potential energy will be a combination of positive and negative values, as there are both positive and negative charges in the array. The final will depend on the distances between the charges.

The electric potential energy for the array of three charges can be determined using the formula:

U = k(q1q2/r12 + q1q3/r13 + q2q3/r23)

where k is the Coulomb's constant, q1, q2, and q3 are the charges, and r12, r13, and r23 are the distances between the charges.

Given that q1=8.48 μC, q2=19.2 μC, and q3=−16.1μC, we can plug these values into the formula and calculate the electric potential energy:

U = k(8.48 μC * 19.2 μC/r12 + 8.48 μC * - 16.1μC/r13 + 19.2 μC * -16.1μC/r23)

Without the values of r12, r13, and r23, we cannot determine the exact value of the electric potential energy.

However, we can see that the electric potential energy will be a combination of positive and negative values, as there are both positive and negative charges in the array. The final value will depend on the distances between the charges.

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The tangential component of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating is approximately \(1.48 m/s^2.\)

First, let's convert the initial and final speeds from revolutions per minute (rpm) to radians per second:

ω1 = 130 rpm = 130(2π/60) rad/s ≈ 13.6 rad/s

ω2 = 280 rpm = 280(2π/60) rad/s ≈ 29.3 rad/s

The angular acceleration can be calculated as:

α = (ω2 - ω1)/t = (29.3 - 13.6)/5.0 ≈ \(3.14 rad/s^2\)

At time t = 2.0 s, the angular velocity is:

ω = ω1 + αt = 13.6 + 3.14(2.0) ≈ 20.9 rad/s

The tangential component of the linear acceleration can be calculated as:

aT = rα

where r is the radius of the wheel. Substituting r = 0.47 m and α = \(3.14 rad/s^2\), we get:

aT = (0.47)(3.14) ≈ \(1.48 m/s^2\)

Therefore, the tangential component of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating is approximately \(1.48 m/s^2.\)

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The number of valence electrons in neutral atoms equals the atom's main group number. A periodic table column can be used to determine an element's main group number.

The position-time graph depicts the movement of an item across time. Time in seconds is traditionally represented on the x-axis, while object location in metres is plotted along the y-axis. The slope of the position-time graph offers vital information about the object's velocity.

Gravity causes objects in free fall to accelerate. As a result, the free fall motion position-time graph must be curved. This means that objects in free fall begin with a moderate velocity and gradually accelerate, as illustrated by the graph's steep downward curve.

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Answer:

Wavelength of light wave = 0.6 m

Explanation:

Given:

Frequency of wave = 5 x 10⁸

Speed of light = 3 x 10⁸ m/s

Find:

Wavelength of light wave

Computation:

The size of a wave form is its spatial duration, or the duration in which the wave's form repeats.

Wavelength of light wave = Speed of light / Frequency of wave

Wavelength of light wave = [3 x 10⁸] / [5 x 10⁸]

Wavelength of light wave = 0.6 m

The minimum applied force that will keep the box at rest is 55 N.

The term frictional force refers to the force that opposes the motion of a body. We have the following information'

Given that;

μ = F/R

R = mgcosθ = 26.8 kg × 9.8 ms-2 × cos13 =255.9 N

F = μR

F =  0.215 × 255.9 N

F = 55 N

The minimum applied force that will keep the box at rest is 55 N.

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The ball is on the 50-yard line. The ball travels west 5 yards. The distance ball traveled was 60 yards.

Distance can be defined as the length of any path between any two locations. Displacement is the direct distance between any two points when calculated via the shortest path between them. When computing distance, the direction is disregarded. The displacement computation takes the direction into consideration.

The 5 yard gain would bring the ball to the 45 yard line (ball was tackled  at the 50 yard line, so 50 - 5 = 45), but the 15 yards lost due to the tackle and push back would bring ball to the 60 yard line (45 + 15 = 60).

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When two point charges are 2.0 cm apart, and each charge experiences a 1.0 N electric force due to the other charge, we can use Coulomb's law to determine the electric force at a new separation of 8.0 cm.

Coulomb's law states that the electric force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Let's denote the magnitudes of the charges as q1 and q2, and the initial separation as r1. Since each charge experiences a 1.0 N electric force, we can set up the equation as follows:

\(k * (q1 * q2) / (r1^2) = 1.0 N\)

where k is the electrostatic constant.

Now, we need to find the new electric force when the separation is 8.0 cm. Let's denote the new separation as r2. Using the same equation, we can set it up as:

\(k * (q1 * q2) / (r2^2)\)= ?

To find the closest value to the electric force, we need to compare the two equations:

\(k * (q1 * q2) / (r1^2) = 1.0 N\)
\(k * (q1 * q2) / (r2^2)\) = ?

Since the electrostatic constant (k), the charges (q1 and q2), and the distance between the charges (r1) are all the same in both equations, we can set up a ratio:

\((r1^2) / (r2^2)\) = 1.0 N / ?

Simplifying the ratio, we have:

\((r1^2) / (r2^2) = 1.0\)

To find the value of ?, we rearrange the equation:

\(? = 1.0 N * (r2^2) / (r1^2)\)

Using the given values,\(? = 1.0 N * (8.0 cm)^2 / (2.0 cm)^2 = 16.0 N.\)

Therefore, the electric force on each charge, when they are moved to a new separation of 8.0 cm, is closest to 16.0 N.

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Answer:

4.5 v

Explanation:

The lamps are wired in parallel ....they have the same potential apllied

1.5 + 1.5 + 1.5 = 4.5 v

Total mass of the IBM initials written with 35 xenon atoms is approximately 7.63 × 10⁻⁻²¹ grams.

The atomic mass of xenon (Xe) is 131.29 atomic mass units (amu). To convert this to grams, we need to use the conversion factor between atomic mass units and grams, which is 1 amu = 1.6605 × 10⁻²⁴ gm.

Therefore, the mass of one xenon atom in grams is:

131.29 amu × 1.6605 × 10⁻²⁴ gm/amu = 2.18 × 10⁻²² gm

So, the total mass of 35 xenon atoms in grams is:

= 35 × 2.18 × 10⁻²² g

= 7.63 × 10⁻⁻²¹gm

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--The given question is incomplete, the complete question is:

"Using scanning tunneling microscopy, scientists at IBM wrote the initials of their company with 35 individual xenon atoms (as shown below). Calculate the total mass of these letters in grams, if the atomic mass of xenon (Xe) is 131.29 amu"--

A lunar eclipse occurs when the Moon moves into the Earth's shadow. This can occur only when the Sun, Earth, and Moon are exactly or very closely aligned (in syzygy) with Earth between the other two, and only on the night of a full moon.

Answer:

hydrochlorine +12÷B to the power of 4 -× y reapeated zminus 2 to the power of 9

The potential energy of a stone of mass 10 kg that is lifted to a height of 8m is -784 Joule.

Potential energy is essentially what it sounds like, though there are some nuances. An object's actual potential energy is determined by its position in relation to other objects. For example, a brick suspended from a two-story building has more potential energy than a brick resting on the ground. This is due to the brick's relative position to the Earth providing it with more energy. However, because there is no force acting on them, two bricks next to each other do not give each other more energy.

Given,

Mass of stone = 10kg

Height = 8m

Potential energy can be determined by formula,

P=mgh

Where, g= gravitational acceleration = \(-9.8 m/s^2\)

Here, gravitational acceleration is negative because stone is thrown against the gravitational pull

\(P=10*8*(-9.8)=-784 J\)

Hence, the potential energy is -784 J

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The value of m for the diffraction maximum is approximately 18.

The diffraction grating has a ruling density of 3 750 rulings/cm, which means that the distance between adjacent rulings is:

d = 1 / (3750 rulings/cm) = 2.67 × \(10^{-4}\)cm

The distance between adjacent maxima for the two sodium wavelengths is given as:

Δy = 1.87 mm = 1.87 × \(10^{-1}\)cm

d sin θ = mλ

For the two closely spaced wavelengths of sodium, we have:

d sin θ = mλ1    (1)

d sin θ = mλ2    (2)

where λ1 = 589.0 nm = 5.89 × \(10^{-5}\)cm and λ2 = 589.6 nm = 5.896 × \(10^{-5}\)cm.

Subtracting equation (2) from equation (1), we get:

d sin θ = m(λ1 - λ2)

Rearranging this equation, we have:

m = (d sin θ) / (λ1 - λ2)

sin θ ≈ tan θ ≈ y / L

where y is the separation between the maxima on the screen, and L is the distance between the grating and the screen.

Substituting the given values, we have:

sin θ ≈ (1.87 × \(10^{-1}\)cm) / (3.00 m) = 6.23 × \(10^{-6}\)

Now we can calculate the value of m:

m = (d sin θ) / (λ1 - λ2)

m = [(2.67 × \(10^{-4}\)cm) × (6.23 × \(10^{-6}\))] / [(5.89 × \(10^{-5}\) cm) - (5.896 × \(10^{-5}\)cm)]

m ≈ 18

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Answer:

60cm

Explanation:

30+30=60