the pendulum consists of a uniform disk with radius r = 10.0 cm and mass 500 g attached to a uniform rod with length L _ 500 mm and mass 270 g. (a) Calculate the rotational inertia of the pendulum about the pivot point. (b) What is the distance between the pivot point and the center of mass of the pendulum? (c) Calculate the period of oscillation.

Answer:

The reason why times, 0.23 s, and 1.80 s makes sense is that a projectile passes through a given height level, which is lesser than the maximum height reached by the projectile, twice in its trajectory

Explanation:

The formula for the time of motion of a projectile is given as follows;

\(t = \dfrac{-V_{oy} \pm \sqrt{V_{oy} ^2 - 2 \cdot g \cdot \Delta y} }{-g}\)

Therefore, when \(V_{oy} ^2 = 2 \cdot g \cdot \Delta y}\), we have only one time value

When \(V_{oy} ^2 > 2 \cdot g \cdot \Delta y}\), two time values can be obtained and both will be positive when we have;

\(V_{oy} > \sqrt{V_{oy} ^2 - 2 \cdot g \cdot \Delta y} }\)

When \(V_{oy} < \sqrt{V_{oy} ^2 - 2 \cdot g \cdot \Delta y} }\), one of the time values will be negative and can be rejected

Therefore, given that the times obtained above are 1.80 s, and 0.23 s, and both make sense due to the following reason;

In the path of the projectile motion of the basketball, there are two points in time at which the height of the basketball above the starting point is exactly 2 meters, given that the maximum height reached is more than 2 meters

The first time the basketball is 2 meters above the point it is shot is lesser of the two calculated time values, which is during the upward motion of the basketball before it reaches the maximum height, while the second time is, which is the larger calculated time, is the time that the basketball reaches the hoop, after flying past the highest point

The coordinate on the x axis where the net electric field is zero is 45.7 cm.

The electric field produced by a point charge is given by the equation:

E = k * q / r^2

where:

E is the electric field strength

k is Coulomb's constant (8.988 × 10^9 N m^2 C^-2)

q is the charge of the point particle

r is the distance from the point particle

The net electric field at a point is the vector sum of the electric fields produced by all the point charges at that point.

In this case, we have two point charges, q1 and q2, with charges of 2.60 × 10^-8 C and -5.29q1, respectively. The charges are located at x = 23.0 cm and x = 73.0 cm, respectively.

We want to find the coordinate on the x axis where the net electric field is zero. This means that the electric field produced by q1 must be equal and opposite to the electric field produced by q2.

We can set up the following equation to solve for the x coordinate:

(k * q1 / (x - 23.0 cm)^2) = (k * (-5.29q1) / ((x - 73.0 cm)^2)

Simplifying the equation, we get:

(x - 23.0 cm)^2 = 28.1 * ((x - 73.0 cm)^2)

Solving for x, we get:

x = 45.7 cm

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The pressure acting on the top and bottom faces is\(0.1635 N/cm^2\), the pressure acting on the side faces is\(0.4905 N/cm^2,\) and the pressure acting on the front and back faces is \(0.981 N/cm^2.\)

To compute the pressure acting on each face of the rectangular block, we need to know the weight of the block and the area of each face.

The weight of the block can be calculated as follows:

Weight = Mass x Gravity

Weight = 0.3 kg x 9.81 \(m/s^2\)

Weight = 2.943 N

The area of each face can be calculated as follows:

Top and bottom face: length x width = 6 cm x 3 cm = 18 \(cm^2\)

Side faces: length x height = 6 cm x 1 cm = 6 \(cm^2\)

Front and back faces: width x height = 3 cm x 1 cm = 3\(cm^2\)

Now we can calculate the pressure acting on each face:

Top and bottom face: Pressure = Weight / Area = 2.943 N / \(18 cm^2\) = \(0.1635 N/cm^2\)

Side faces: Pressure = Weight / Area = 2.943 N / \(6 cm^2\) = 0.4905 \(N/cm^2\)

Front and back faces: Pressure = Weight / Area = 2.943 N / 3 cm^2 = 0.981 N/cm^2

Therefore, the pressure acting on the top and bottom faces is\(0.1635 N/cm^2\), the pressure acting on the side faces is\(0.4905 N/cm^2,\) and the pressure acting on the front and back faces is \(0.981 N/cm^2.\)

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Hello!

When the object is hung from the spring, the force of the spring balances out the force due to gravity. In terms of equations:

\(\Sigma F = F_g - F_s\\\\0 = F_g - F_s\\\\F_s = F_g\)

The force of a spring is equivalent to:
\(F_s = -kx \\\)

k = Spring Constant (3000 N/m)
x = Extension of spring (0.05 m)

**Negative sign means the force is in the opposite direction of the extension. For example, if you pull a spring, the spring force tries to collapse the spring, aka it works against you

The force due to gravity is equivalent to:
\(F_g = mg\)

m = mass (? kg)
g = acceleration due to gravity (9.81 m/s²)

Solving for mass:
\(kx = mg\\\\m = \frac{kx}{g}\\\\m = \frac{3000(0.05)}{9.81} = \boxed{15.29 kg}\)

The final temperature of the two cups of water being mixed together, given that one cup has a temperature of 30 °C and the other has 75 °C, is  58.13 °C

To obtain the final temperature of the two cups of water, we must obtain the equilibrium temperature of the two cups of water mixture. Details below:

Heat loss by cup 2 = Heat gain by cup 1

MᵥᵥC(Tᵥᵥ - Tₑ) = MC(Tₑ - T)

250 × 4186 (75 - Tₑ) = 150 × 4186(Tₑ - 30)

1046500(75 - Tₑ) = 627900(Tₑ - 30)

Clear bracket

78487500 - 1046500Tₑ = 627900Tₑ - 18837000

Collect like terms

78487500 + 18837000 = 627900Tₑ + 1046500Tₑ

97324500 = 1674400Tₑ

Divide both side by 1674400

Tₑ = 97324500 / 1674400

Tₑ = 58.13 °C

From the above calculation, the equilibrium temperature is 58.13 °C.

Thus, we can conclude that the final temperature of the two cups of water is 58.13 °C

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Answer:

general

Explanation:

The general environment is very narrow and includes the forces that directly impact a firm and its competitors

Answer:

Chemical reactions occur when chemical bonds between atoms are formed or broken. The substances that go into a chemical reaction are called the reactants, and the substances produced at the end of the reaction are known as the products.

Explanation:

Answer:

In a chemical reaction, the products are the substance present after the reaction.

Explanation:

The best describes the image of a concave mirror when the object is at a distance greater than twice the focal point distance from the mirror is (c) virtual, upright, and magnification less than one.

When the object is placed beyond twice the focal length of a concave mirror, the image formed is virtual, meaning it cannot be projected onto a screen. The image is also upright, having the same orientation as the object.

In addition, the magnification of the image is less than one, indicating that the image is smaller in size compared to the object. This is because the concave mirror forms a reduced and virtual image when the object is located beyond the focal point.

The distance between the object and the mirror affects the size and position of the virtual image formed.

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The given problem involves calculating the definite integral of a function f(x) over a specific range. The function f(x) is defined differently for different values of x, and the final result of the definite integral \(1^2\)₁ f(x) dx, where x ≤ n, is -cos(n) - (-cos(1)) + 3cos(T) - 3cos(n) + infinity.

To calculate the definite integral 1²₁ f(x) dx, where x ≤ n, we need to evaluate the integral of the given function f(x) over the specified range. The function f(x) has different definitions depending on the value of x. For x ≤ n, the function is sin(x), and for x > n, the function is -3sin(x). Additionally, the function is defined as 2x for values of x greater than a certain threshold T.

To solve this problem, we need to consider the different intervals of the range separately. First, we integrate sin(x) over the interval 1 to n. The integral of sin(x) is -cos(x), so the value of this part of the integral becomes -cos(n) - (-cos(1)).

Next, we need to integrate -3sin(x) over the interval n to T. The integral of -3sin(x) is 3cos(x), so this part of the integral becomes 3cos(T) - 3cos(n).

Lastly, we integrate 2x over the interval T to infinity. The integral of 2x is \(x^2\), so this part of the integral becomes infinity.

Combining these three parts, the final result of the definite integral \(1^2\)₁ f(x) dx, where x ≤ n, is -cos(n) - (-cos(1)) + 3cos(T) - 3cos(n) + infinity.

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The reason at the time when particles in a solid or liquid move slower and closer together should be explained below.

The particles should be in the form of small pieces of matters. Also, the scientist believed that each and everything in the universe should be made of the particles. here the range of the particles should be in the range via the large  subatomic particles such as electrons or it has lower  microscopic particles.

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the mass of 15.9 moles of sulfuric acid (H2SO4) is approximately 1558.2 grams.

The mass of 15.9 moles of sulfuric acid (H2SO4) can be calculated using the molar mass of the compound. To find the molar mass, we need to sum the individual atomic masses of all the elements in the molecule:
1. Hydrogen (H): There are 2 hydrogen atoms in H2SO4, each with an atomic mass of approximately 1 g/mol. So, the total mass for hydrogen is 2 x 1 = 2 g/mol.
2. Sulfur (S): There is 1 sulfur atom in H2SO4 with an atomic mass of approximately 32 g/mol.
3. Oxygen (O): There are 4 oxygen atoms in H2SO4, each with an atomic mass of approximately 16 g/mol. So, the total mass for oxygen is 4 x 16 = 64 g/mol.
Now, add the masses of all elements to get the molar mass of H2SO4: 2 g/mol (H) + 32 g/mol (S) + 64 g/mol (O) = 98 g/mol.
Finally, multiply the molar mass by the given number of moles to find the mass of 15.9 moles of sulfuric acid:
Mass = (Moles) x (Molar Mass) = 15.9 moles x 98 g/mol = 1558.2 g.
Therefore, the mass of 15.9 moles of sulfuric acid (H2SO4) is approximately 1558.2 grams.

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When a hockey puck is struck with a hockey stick, a force acts on the puck and the puck exerts an equal and opposite force on the stick.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. When the hockey stick strikes the puck, it applies a force to the puck in one direction. As a result, the puck exerts an equal magnitude of force but in the opposite direction on the stick.

This interaction between the stick and the puck is what allows the puck to be propelled forward. The force applied to the puck by the stick causes it to accelerate and move in the direction of the applied force. The reaction force exerted by the puck on the stick also affects the motion and stability of the stick in the opposite direction.

The combination of these forces and reactions contributes to the transfer of momentum and energy from the stick to the puck, enabling the puck to move with speed and travel in the desired direction on the ice.

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Answer:

1754.3 m

Explanation:

331*5.3=1754.3

Answer:

0.120m

Explanation:

Answer:

0.8s

Explanation:

Given parameters:

Height of shelf  = 3m

Unknown:

Time it will take to hit the ground  = ?

Solution:

To solve this problem, we use the expression below;

         x  = ut + \(\frac{1}{2}\) gt²  

x is the height

u is the initial velocity  = 0m/s

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

t is the time taken  = ?

Now insert the parameters and solve for t;

    3  = (0 x t) +( \(\frac{1}{2}\) x 9.8 x t²)

    3 = 4.9t²

       t² = 0.6

       t = 0.8s

Answer:

W₂ / W₁ = 4

Explanation:

Given:

m

V₁ = V

V₂ = 2·V

_________

W₂ / W₁ - ?

Kinetic energy of the car:

W₁ = m·V₁² / 2 = m·V² /2

W₂ = m·V₂² / 2 = m·(2·V)² /2 = 4·m·V² / 2 = 4·W₁

W₂ / W₁ = 4·W₁/ W₁ = 4

Kinetic energy increased 4 times

Answer:

Distance = 70 meters

Explanation:

Given the following data;

Speed = 90 km/h

Time = 2.8 seconds

Conversion:

90 km/h to meters per seconds = 90 * 1000/3600 = 90000/3600 = 25 m/s

To find the distance covered during this inattentive period;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

\(Speed = \frac{distance}{time}\)

Making distance the subject of formula, we have;

\( Distance = speed * time \)

Substituting into the above formula, we have;

\( Distance = 25 * 2.8 \)

Distance = 70 meters

Answer:

50 km/h

Explanation:

\((40 + 60) / 2\)

-> \(50\) (km/h)

The answer is B.

We use the spring constant in Hooke's Law, which is : F = -kx

It can be rearranged so it is equal to the constant :

k = F ÷ x

Common units of Force (F) : N, kg m/s²

Common units of displacement (x) : cm, m, km

The dimensionally correct unit from the given options is :

⇒ N/km

(a) The angle between the direction of and the positive direction of x is 60⁰.

(b) The magnitude of the vector is 47.2 m.

The angle between the two vectors is calculated by applying the following formula as shown below.

θ = arc tan ( Vy / Vx )

where;

θ = arc tan ( 40.0 / 25.0 )

θ = arc tan ( 1.6 )

θ = 60⁰

The magnitude of the vector is calculated as;

V = √ ( 25² + 40² )

V = 47.2 m

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Given data:Current (I) in the long straight wire = 6.0 A Current (i) in the circular loop = 1.0 A Radius (r) of the circular loop = 10 cm = 0.1 mDistance (d) of the point P from the z-axis = 25 cm = 0.25 mVelocity of the electron (v) = 1.0 x 106 m/sNow, the magnetic field due to the circular loop at the center (P) of the loop is given by the formula:B = μ0 i / 2 R where,μ0 = Permeability of free space = 4π × 10⁻⁷ Tm/ir = Radius of the circular loopi = Current flowing in the circular loopSubstitute the given values,μ0 = 4π × 10⁻⁷ Tm/i = 1.0 AR = 0.1 mB = (4π × 10⁻⁷ × 1.0) / (2 × 0.1)B = 6.28 × 10⁻⁵ T.

The magnetic field at point P is 6.28 × 10⁻⁵ T, acting in the positive y-direction.By Lorentz's law, the magnetic force acting on a charged particle moving in a magnetic field is given by:F = q (v × B)where,q = Charge of the particlev = Velocity of the particleB = Magnetic field acting on the particle Cross product of velocity and magnetic field:v × B = | i  j  k| v1  v2  v3  B1  B2  B3| = (v2B3 - v3B2)i - (v1B3 - v3B1)j + (v1B2 - v2B1)k| = (-v × B).

The electron has a charge of -1.6 × 10⁻¹⁹ C, velocity v = - 1.0 x 106 m/s in the negative x-direction and experiences a magnetic force in the positive y-direction. Hence, only the y-component of the force will be considered.Substitute the given values in the equation:F = q (v × B)F = -1.6 × 10⁻¹⁹ [(1.0 x 106) (-6.28 × 10⁻⁵)]F = 1.005 × 10⁻¹² NThe y-component of the magnetic force on the electron is 1.005 × 10⁻¹² N.Answer: 1.005 × 10⁻¹² N.

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Answer:

allows no light to pass

Explanation:

The work done on the elevator by its cable is 459348 joules. Work is defined as the product of the force acting on an object and the distance moved by the object in the direction of the force.

Therefore, the work done on the elevator by its cable will be calculated using the formula:W = Fdwhere W is the work done, F is the force applied, and d is the distance moved by the object.

The force applied on the elevator by the cable is equal in magnitude to the weight of the elevator, which is given by:F = mgwhere F is the force applied, m is the mass of the elevator, and g is the acceleration due to gravity. Substituting the given values:F = (1200 kg)(9.81 m/s²)F = 11772 N.

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Answer:

conduction i think

Explanation:

Answer:Conduction

Explanation:

When a rope of length L and mass m is suspended from the ceiling, the tension in the rope at position y can be found using the following expression:

T(y) = mg + λy where g is the acceleration due to gravity, λ is the linear mass density of the rope, and y is the distance measured upward from the free end of the rope.

Here's how to derive the expression: Let's consider an element of length dy of the rope at a distance y from the free end of the rope. The weight of the element is dm = λdy and acts downward. The tension in the rope on the element can be resolved into two components - one acting downward and another acting upward. Let T be the tension in the rope at point y and T + dT be the tension in the rope at point (y + dy).The upward component of tension on the element is given by Tsinθ, where θ is the angle between the element and the vertical. As the rope is assumed to be in equilibrium, the horizontal components of tension balance each other and the net vertical force on the element is zero. Therefore, we have,

Tsinθ - dm g = 0 ⇒ Tsinθ = dm g ⇒ Tsinθ = λdyg

The angle θ can be found using the equation tanθ = dy/dx ≈ dy/dy = 1. Therefore, sinθ = dy/√(dy²+dx²) ≈ dy and we have,T dy = λdyg ⇒ T = λgThis expression gives the tension in the rope at the free end of the rope. The tension in the rope at position y, measured upward from the free end of the rope is given by,T(y) = mg + λy

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The researchers calculated that a typical neutron star had a radius of around 11 km. A neutron star has a density of radius 3 1017 kg/m, which is the same as nuclear matter.

With their latest research, the scientists have discovered proof that the strong nuclear force produces a repulsive force between neutrons at a neutron star's core that prevents the star from collapsing in on itself when particles are arranged in much denser configurations and separated by smaller distances.

The maximum mass of neutron stars has a new upper limit, according to astronomers: It is limited to 2.16 solar masses. We are aware of their little size: A neutron star with a mass 1.4 times that of the sun is predicted to have a radius of between 8 and 16 kilometres.

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Answer:

4 km/hr^2

Explanation:

We can use the formula for acceleration:

a = (v_f - v_i) / t

where:

a = acceleration

v_f = final velocity

v_i = initial velocity

t = time taken

Substituting the given values, we get:

a = (70 km/hr - 50 km/hr) / 5 hr

a = 20 km/hr / 5 hr

a = 4 km/hr^2

Answer:

150

Explanation:

15m/s×10s= 15m/a

and that is the correct answer

Answer:

0.404

Explanation:

The calculation of confidence interval for the mean is shown below:-

Data provided in the question

Mean = 5.5

The Standard deviation for three values  is 0.7

Since there is 95% of the confidence interval so the critical value is

\(t_{0.025.5} = 4.303\\\)

Now

Margin error is

\(t_{0.025.5} = 4.303\times \frac{s}{\sqrt{n} }\)

\(= 4.303\times \frac{0.7}{\sqrt{3} }\)

= 1.74

Now,

The Standard error is

\(= \frac{s}{\sqrt{n} }\)

\(= \frac{0.7}{\sqrt{3} }\)

= 0.404

Hence, the confidence interval for the mean based is 0.404