A 0.70 kg teddy bear is nudged off a window sill and falls 3.35 m to the ground. what is its kinetic energy at the instant it hits the ground?

OSHA was established to ensure the safety and health of workers in their workplaces. It enforces standards, conducts inspections, and provides resources. Its implementation has significantly improved the health and safety conditions in American workplaces.

The Occupational Safety and Health Administration (OSHA) was established as a necessary agency to ensure the health and safety of workers within their work environments. Prior to OSHA's creation, workplaces suffered from numerous safety issues, including hazardous materials, dangerous machinery, and poor working conditions that could lead to injury or even death. Therefore, the U.S. government saw the need to establish a set of standards for businesses to follow in order to prioritize their employees' safety and wellbeing.

OSHA was necessary because it took responsibility for enforcing these standards, conducting inspections to ensure compliance, and providing training and resources for both employers and workers. Without OSHA, workplaces could be highly dangerous and there might be no legal basis for holding employers accountable for harm caused to their employees in the workplace. The implementation of OSHA helped reduce workplace accidents, injuries and illnesses, significantly improving the health and safety conditions in American workplaces.

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Having a short calf and long thigh would result in better torque for leg rotation during running, as a longer lever arm (thigh) from the pivot point (knee) allows for greater torque. However, proportions should be balanced for optimal running biomechanics.

In terms of the torque needed to rotate your leg as you run, it would be better to have a short calf and long thigh. This is because torque is the rotational equivalent of force and is calculated by multiplying the force by the distance from the pivot point. In this case, the pivot point is the knee.

Therefore, a longer thigh would result in a greater torque because the force (muscle contraction) is applied further from the pivot point (knee). Conversely, a short calf means less mass is being rotated around the pivot point, reducing the torque needed for movement.

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In terms of the torque needed to rotate your leg as you run, it be better to have b) Short calf, long thigh

In terms of the torque needed to rotate your leg as you run, it would be better to have a short calf and long thigh.

This is because torque is the rotational equivalent of force, and it depends on both the magnitude of the force and the distance from the pivot point where the force is applied. In running, the longer the distance from the pivot point (knee), the more torque is generated by the same muscle force. A long thigh and short calf combination helps in maximizing this distance, thereby reducing the effort required to achieve the same angular acceleration of the leg.

Let's break this down step-by-step:

The distance between the charge and the source of the electric field is 2.2 m.

The potential energy U of a charge q placed in an electric field created by a source charge Q, at a distance r from the source charge is given by,

[tex] U=\frac{kQq}{r} [/tex] ...... (1)

Here, k is the Coulomb constant.

The electric field E at a distance r from the source charge is given by,

[tex] E =\frac{kQ}{r^2} [/tex] ......(2)

From equations (1) and (2)

[tex] U=E*q*r [/tex]

Rewrite the expression for ri.

[tex] r=\frac{U}{Eq} [/tex]

Substitute 75 J for U, [tex] 4.5*10^5 V/m [/tex] for E and [tex] 7.5*10^{-5} C [/tex] for q.

[tex] r=\frac{U}{Eq} \\ =\frac{75 J}{(4.8*10^{5} V/m)(7.2*10^{-5}C)} \\ =2.17 m [/tex]

Rounding off to the nearest tenth, the the distance between the charge and the source charge is 2.2 m

In a circuit, a voltage drop occurs over a resistor as the electric current flows through it, transforming electrical energy into heat due to resistance, not storing charge or electrical energy. the correct answer is voltage is dropped.

Among the choices provided for what occurs over a resistor in a circuit, the correct answer is voltage is dropped. When electric current flows through a resistor, it encounters resistance which impedes the flow of charge. This results in a voltage drop across the resistor. The energy that the charges lose as they pass through the resistor is dissipated mainly in the form of heat. This concept is reflected in the equation for electric power dissipation, P = IV, where P represents power, I is current, and V is voltage. This can also be expressed as P = I²R or P = V²/R using Ohm's law, where R is the resistance. Contrary to some of the other choices, resistors do not store charge or electrical energy; that function is typically carried out by capacitors in a circuit.

Answer:

Q = m x C x T

C = Q / m x T

C = 3.25 x 10^3 /0.5kg  x  50K = 130 J/kgK

The substance is gold.

Explanation:

The minimum amount of energy required to move the satellite from its orbit to a location very far away from Earth is approximately [tex]\( 6.245 \times 10^{11} \, \text{J} \)[/tex]

To move a satellite from a stable orbit around the Earth to a location very far away, such as into deep space, we need to provide enough energy to overcome the gravitational pull of the Earth and to accelerate the satellite to a speed sufficient to escape Earth's gravitational field entirely. This energy required to escape Earth's gravitational field is called the escape velocity.

The escape velocity, [tex]\( v_{\text{escape}} \)[/tex], is given by the formula:

[tex]\[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}} \][/tex]

Where:

- [tex]\( G \)[/tex] is the gravitational constant [tex](\( 6.67430 \times 10^{-11} \, \text{m}^3/\text{kg/s}^2 \))[/tex],

- [tex]\( M \)[/tex] is the mass of the Earth [tex](\( 5.972 \times 10^{24} \, \text{kg} \))[/tex],

- [tex]\( R \)[/tex] is the distance from the center of the Earth to the satellite's initial orbit.

The minimum amount of energy required to move the satellite from its orbit to a location very far away would be the kinetic energy required to achieve this escape velocity.

The kinetic energy [tex]\( KE \)[/tex] required to achieve a velocity [tex]\( v \)[/tex] for an object of mass [tex]\( m \)[/tex] is given by the formula:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

Therefore, to calculate the minimum energy required, we need to find the escape velocity and then calculate the kinetic energy corresponding to that velocity.

Keep in mind that in real-world scenarios, additional energy may be required to maneuver the satellite and account for factors such as atmospheric drag and gravitational influences from other celestial bodies. However, for the sake of simplicity, we will focus on the minimum energy required to achieve escape velocity from Earth's gravity.

Let's proceed with the calculations using the known values for [tex]\( G \)[/tex], [tex]\( M \)[/tex], and [tex]\( R \)[/tex].

Let's calculate the escape velocity [tex]\( v_{\text{escape}} \)[/tex] first:

Given:

- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3/\text{kg/s}^2 \)[/tex],

- [tex]\( M = 5.972 \times 10^{24} \, \text{kg} \)[/tex],

- [tex]\( R \)[/tex] (distance from the center of the Earth to the satellite's initial orbit).

Assuming the satellite is in a low Earth orbit, where [tex]\( R \)[/tex] is approximately the radius of the Earth, [tex]\( R \approx 6.371 \times 10^6 \, \text{m} \)[/tex].

Let's calculate [tex]\( v_{\text{escape}} \)[/tex]:

[tex]\[ v_{\text{escape}} = \sqrt{\frac{2 \times 6.67430 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}} \][/tex]

[tex]\[ v_{\text{escape}} = \sqrt{\frac{2 \times 6.67430 \times 5.972}{6.371}} \times 10^{11} \][/tex]

[tex]\[ v_{\text{escape}} = \sqrt{\frac{2 \times 39.83416}{6.371}} \times 10^{11} \][/tex]

[tex]\[ v_{\text{escape}} = \sqrt{\frac{79.66832}{6.371}} \times 10^{11} \][/tex]

[tex]\[ v_{\text{escape}} \approx \sqrt{12.513} \times 10^{11} \][/tex]

[tex]\[ v_{\text{escape}} \approx 3.537 \times 10^4 \, \text{m/s} \][/tex]

Now, we'll calculate the kinetic energy [tex]\( KE \)[/tex] required to achieve this velocity using the formula:

[tex]\[ KE = \frac{1}{2} m v_{\text{escape}}^2 \][/tex]

Given that the mass m, of the satellite is not specified, we'll assume a typical satellite mass of [tex]\( 1000 \, \text{kg} \)[/tex] for illustrative purposes.

[tex]\[ KE = \frac{1}{2} \times 1000 \times (3.537 \times 10^4)^2 \][/tex]

[tex]\[ KE = \frac{1}{2} \times 1000 \times 1.249 \times 10^9 \][/tex]

[tex]\[ KE \approx 6.245 \times 10^{11} \, \text{J} \][/tex]

So, the minimum amount of energy required to move the satellite from its orbit to a location very far away from Earth is approximately [tex]\( 6.245 \times 10^{11} \, \text{J} \)[/tex].

The minimum amount of energy required to move the satellite very far away from earth is approximately 3.124 x 10¹³ Joules.

To find the minimum amount of energy required to move a satellite from its orbit to a location very far away from Earth (essentially to infinity), we need to calculate the difference in gravitational potential energy between its current orbit and a point infinitely far away.

The satellite has a mass m, the Earth has a mass M, and the radius of the Earth is R. The distance of the satellite from the center of the Earth is 2R. The gravitational potential energy U of the satellite in its current orbit is given by:

[tex]U = -\frac{G \cdot M \cdot m}{2R}[/tex]

where G is the gravitational constant (6.67 × 10⁻¹¹ Nm²/kg²).

At a distance infinitely far away, the gravitational potential energy U∞ is zero because the gravitational influence of the Earth becomes negligible:

U∞ = 0

The minimum energy ΔE required to move the satellite from its orbit to infinity is the difference in gravitational potential energy:

[tex]\Delta E = U_\infty - U = 0 - \left(-\frac{G \cdot M \cdot m}{2R}\right) = \frac{G \cdot M \cdot m}{2R}[/tex]

In numerical terms, for a 1000 kg satellite, Earth's mass M = 5.97 × 10²⁴kg, and Earth’s radius R = 6.371 × 10⁶ m, the energy required is:

[tex]\Delta E = \frac{(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \times (5.97 \times 10^{24} \, \text{kg}) \times (1000 \, \text{kg})}{2 \times 6.371 \times 10^{6} \, \text{m}}[/tex]

[tex]&= \frac{(6.67 \times 5.97 \times 1000) \times 10^{-11 + 24 + 3}}{2 \times 6.371 \times 10^{6}} \\[/tex]

[tex]&= \frac{39819.9 \times 10^{16}}{12.742 \times 10^6} \\[/tex]

[tex]&= \frac{39819.9}{12.742} \times 10^{16 - 6} \\[/tex]

[tex]&= 3.124 \times 10^{13} \, \text{Nm}[/tex]

After calculation, the minimum energy required is approximately 3.124 x 10¹³ Joules.

Answer:

Work = 940800 J

Explanation:

As we know that work done is defined as

Work = (Force)(displacement in the direction of force)

here elevator motor lift a mass of 1200 kg

so in order to lift it up motor must have to apply the force same as the weight so that it will move up with constant speed.

so here we have

[tex]F = mg[/tex]

[tex]F = (1200 kg)(9.8 m/s^2)[/tex]

[tex]F = 11760 N[/tex]

now it is displaced upwards by distance d = 80 m

so here we have

[tex]W = (11760)\times (80)[/tex]

[tex]W = 940800 J[/tex]

so above is the work done by the elevator to lift it upwards

Whale songs in an ocean in sound waves frequencies between 30 Hertz (Hz) and around 8,000 Hz (8 kHZ).

The sound is described in physics as: a pressure wave of vibration that travels through a gas, liquid, or solid medium and is audible.

Moans, groans, grunts, blasts, and shrieks are made by humpback whales. Sound waves make up each section of their song. These sound waves include some high frequency ones. These noises would resemble tall, sharp mountains if you could see them. Low frequency sound waves are also emitted by whales. These waves resemble far-flung hills in their spacing. Without losing energy, these sound waves can travel a great distance through water. Various of these low frequency sounds, according to researchers, can travel more than 10,000 miles in some ocean depths.

Hertz units are used to measure sound frequency. Whales use frequencies between 30 Hertz (Hz) and around 8,000 Hz (8 kHZ). The whales' songs are only partially audible to humans.

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Final answer:

Whale songs in the ocean are sound waves, specifically low-frequency aural signals that can travel vast distances underwater, facilitated by the SOFAR channel. Unfortunately, noise pollution seriously impacts these marine communications.

Explanation:

The whale songs that are transmitted in the ocean are a form of sound waves. These low-frequency sound waves are capable of traveling long distances underwater, allowing whales to communicate over hundreds of kilometers. This is possible due to the unique properties of water, especially in a layer known as the SOFAR channel, where sound speed is the slowest and allows for minimal attenuation of these aural signals.

However, contemporary issues such as noise pollution from ships pose a threat to this natural communication system, substantially reducing the ability of cetaceans to transmit their songs over long distances.

Answer:

F = v / w

Air: 343m/s  /  2.0 m = 171.5/s

Helium: 1007m/s / 2.0 m = 503.5/s

Carbon Dioxide: 267m/s  /  2.0 m = 133.5 /s

Explanation:

Answer:

Energy

Explanation:

Eng 2021

Final answer:

Using a crowbar to remove a nail requires less force because it uses leverage to amplify the input force and increase the mechanical advantage.

Explanation:

When using a crowbar to remove a nail, you are using a lever with a large mechanical advantage. The input force you apply to the crowbar is much smaller than the force exerted by the crowbar on the nail. This is because the length of the crowbar provides a greater lever arm, which increases the mechanical advantage.

On the other hand, when you try to pull directly on the nail, you don't have the same leverage as the crowbar, so you need to exert a larger force to overcome the resistance of the nail.

In summary, the crowbar allows you to remove a nail with less force because it uses leverage to amplify the input force and increase the mechanical advantage.

The fundamental frequency of the steel piano wire, subjected to a tension of 765 N, with a length of 0.700 m, and a mass of 5.25 g, is approximately 424.6 Hz.

The frequency f1 of the string's fundamental mode of vibration can be calculated using the formula f = sqrt(T / μ) / 2L. Here:

Plugging these values into the formula, we get:
f1 = sqrt((765 N) / (0.0075 kg/m)) / (2 * 0.700 m) = 424.6 Hz

This means that the fundamental frequency of the string is roughly 424.6 Hz.

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The final volume needed for the pressure of the nitrogen gas to be 1.53 atm at a temperature of 337 K is approximately 190.8 ml. This is calculated using the combined gas law.

To find out what the final volume of nitrogen gas in the cylinder would be when the pressure is 1.53 atm and the temperature is 337 K, we can use the combined gas law, which is derived from the ideal gas law. The combined gas law states that the ratio of the product of pressure and volume to temperature remains constant for a fixed amount of gas when the temperature is measured in Kelvin. The formula for the combined gas law is (P1 * V1) / T1 = (P2 * V2) / T2, where P is pressure, V is volume, and T is temperature.

Given that:

We want to find the final volume (V2), so rearranging the equation to solve for V2 gives us:

V2 = (P1 * V1 * T2) / (P2 * T1)

Substituting the known values:

V2 = (1.23 atm * 219 ml * 337 K) / (1.53 atm * 295 K)

By performing the calculations:

V2 ≈ 190.8 ml

Therefore, the final volume needed for the pressure of the nitrogen gas to be 1.53 atm at 337 K is approximately 190.8 ml.

Answer: The waves that travel through vacuum is electromagnetic waves.

Explanation:

There are two types of waves:

Mechanical waves: These are the waves that need a medium to travel so that they can transport their energy from one location to another. For Example:  Sound waves

Electromagnetic waves: These are waves which can travel through vacuum. These have electrical and magnetic component associated with them. They travel with the speed of light. They does not require a medium to travel. For Example: Infrared waves, Microwaves

Hence, the waves that travel through vacuum is electromagnetic waves.

Answer: Option (C) is the correct answer.

Explanation:

As we known that density is the amount of mass divided by volume of the substance.

Mathematically,    Density = [tex]\frac{mass}{volume}[/tex]

So, when candy was present in the shape of a cloud then it means that it was fluffy as it has more volume.

Since, density is inversely proportional to volume therefore, with increase in volume there will occur a decrease in density.

But when the candy will become compact then there will occur a decrease in its volume. Hence, then there will occur an increase in the density of the candy.

Thus, we can conclude that the statement candy before was fluffy, and the candy after was compacted best describes the candy before and after Chanel manipulated it.

f the ray is incident on the glass-water interface at an angle to the normal greater than 52.5 ∘ then Refractive index of the glass will be  1.26 by critical angle.

The optical medium's refractive index would be a dimensionless number that indicates how well the medium bends light. The refractive index controls however much light would be refracted or twisted when it enters a substance.

Critical angle, in optics, the greatest angle at which a ray of light, travelling in one transparent medium, can strike the boundary between that medium and a second of lower refractive index without being totally reflected within the first medium.

critical angle = [tex]sin^{-1} (1/n)[/tex]

Put the value of critical angle in above equation.

52.5° =    [tex]sin^{-1}(1/n)[/tex] = 1.26  

Refractive index of the glass will be  1.26

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The stone is in the air for 6 seconds.

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

height of the cliff = h = 59.4 m

speed of the stone = u = 19.5 m/s

Asked:

total time taken = t = ?

Solution:

[tex]h = ut + \frac{1}{2}at^2[/tex]

[tex]-59.4 = 19.5t - \frac{1}{2}(9.8)t^2[/tex]

[tex]-59.4 = 19.5t - 4.9t^2[/tex]

[tex]49t^2 -195t - 594 = 0[/tex]

[tex]( t - 6 ) ( 49t + 99 ) = 0[/tex]

[tex]t - 6 = 0[/tex]

[tex]t = 6 \texttt{ s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

[tex]\texttt{ }[/tex]

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s. The stone is in the air for 6 seconds.

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Given in the question a stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s.

Acceleration is rate of change of velocity.

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

height of the cliff = h = 59.4 m

speed of the stone = u = 19.5 m/s

to find total time taken = t = ?

s = ut + 1/2 at² putting the value we get,

t = 6 sec

The stone is in the air for 6 seconds.

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With concave lenses, the characteristics of the image do not depend on the placement of the object, but with convex lenses, they do.