A man drags a 12.0 kg bag of mulch at a constant speed applying a 39.5 N force at 41°. What is the normal force acting on the bag?

In the simple harmonic motion of a block attached to a spring, the maximum elasticity potential energy (U) occurs when displacement from equilibrium is the greatest. Given the provided spring constant (k) and displacement (x), the energy can be calculated using the formula U = (1/2)kx². The maximum elastic potential energy in this scenario is close to 1.44 joules.

In physics, the problem you're dealing with relates to simple harmonic motion associated with a block attached to a spring on a frictionless surface. When the object is displaced from equilibrium and let go, it performs simple harmonic motion. During this motion, there is a constant interconversion between the kinetic and potential energy within the system.

The 'spring constant' (k) of an ideal spring is given as 450 n/m and the displacement from the equilibrium position (x) is 0.080 m. The maximum elastic potential energy is at extremes of the motion, when the displacement is the greatest (+/- x). It is given by the formula: U = (1/2)kx², where U is the elastic potential energy.

Substituting the given spring constant and displacement values into the formula: U = (1/2) * 450 * (0.080)² = 1.44 joules. Hence, the maximum elastic potential energy of the system is closest to 1.44 joules.

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A 7 L sample of gas has a pressure of 1.1 atm at a temperature of 285 K. If the pressure decreases to 0.6 atm, causing the volume to increase to 10 L, what is the new temperature? Round your answer to the nearest tenth.

Answer: 222.1K

The angular acceleration of the automobile is 50.62 rad/s².

The total number of revolutions within the given time is 290 revolutions.

The angular acceleration of the automobile is calculated as follows;

[tex]\alpha = \frac{\omega _f - \omega _i}{t} \\\\[/tex]

ωf = 1600 rpm = 167.57 rad/s

ωi = 4500 rpm = 471.3 rad/s

[tex]\alpha = \frac{167.57 - 471.3}{6} \\\\\alpha = -50.62 \ rad/s^2[/tex]

N = (4500 rpm - 1600 rpm)

N = 2,900 rpm

N = 2,900 x (6/60)

N = 290 rev

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Answer:

the 10-kg object has higher acceleration

Explanation:

Work done by the ideal gas in the second situation is 20 Joule.

We know that amount of energy given to a ideal gas is distributed in it to increase its internal energy and to work done by increasing it volume.

Mathematically:

energy given to a ideal gas (dQ) = Increase in internal energy (dU) + work done (dW).

Now in this question:  a certain amount of an ideal gas requires 30 j when heated at constant volume. So, this energy is used to increase internal energy (as no volume change occurs).

So, dQ₁ = dU = 30 Joule.

When heated at constant pressure, the certain amount of an ideal gas requires 50 J.  So, this energy is used to increase internal energy and work done.

So, dQ₂ = dU + dW

⇒ dW = dQ₂ - dU = 50 Joule - 30 joule = 20 Joule.

Hence, work is done by an amount 20 joule by the ideal gas in the second situation.

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The fulcrum in a hammer when removing a nail is at the part where the hammer pivots.

By applying effort to the handle of a claw hammer, the output force at the nail puller end is increased due to the lever principle. Understanding these concepts aids in efficiently removing nails from wood.

Answer: Option (B) is the correct answer.

Explanation:

It is known that oxygen, sulfur and selenium are all group 16 elements.

The electronic configuration of oxygen is as follows.

     [tex]1s^{2}2s^{2}2p^{4}[/tex]

The electronic configuration of sulfur is as follows.

     [tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}[/tex]

The electronic configuration of selenium is as follows.

     [tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4p^{4}[/tex]

Hence, we can see that on moving down the group there is increase in energy levels of the atoms from 2p to 4p.

Therefore, we can conclude that when traveling from oxygen to sulfur to selenium, through this group in the periodic table, change is that the number of energy levels increases.

The x component of the center of mass velocity is calculated as the momentum-weighted average of the individual blocks' velocities, using the formula (m1*v1x + m2*v2x) / (m1 + m2).

To find the x component of the velocity of the center of mass (vcm)x, we use the formula for the center of mass velocity in one dimension, which is given by:

(vcm)x = (m1*v1x + m2*v2x) / (m1 + m2)

This equation reveals that the center of mass velocity is the momentum-weighted average of the velocities of the individual blocks. Since momentum is mass times velocity, the product m1*v1x is the momentum of block 1 in the x direction, and m2*v2x is the momentum of block 2 in the x direction. The sum of these momenta gives the total momentum in the x direction. By dividing this sum by the total mass (m1 + m2), we obtain the velocity of the center of mass in the x direction.

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Using the principles of energy conservation and kinetic energy, the distance the spring compresses when the block comes to rest is found through the equation x = sqrt((m*v*v)/k), where m is the mass, v is the velocity, k is the spring constant, and x is the distance of compression.

To compute the distance the spring is compressed when the block comes to rest, we need to consider both kinetic energy conservation and energy conservation through potential energy of the spring. Our initial kinetic energy supplied by the block sliding down the plane (K1) will turn into a potential energy in the spring when it's compressed (U2). Hence, we have K1 = U2.

Assuming initial kinetic energy (K1) given by 0.5*m*v*v, and potential energy in the spring (U2) equals to 0.5*k*x*x where x is the distance in which spring is compressed.

From the conservation of energy principle, 0.5*m*v*v = 0.5*k*x*x. By simplifying the equation, we get x = sqrt((m*v*v)/k). This equation provides us the distance the spring is compressed when the block comes to rest.

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U=1/2QV

U=1/2CV^2

U=Q^2/2C

Answer: geometric shape tools

Explanation: plato/edmentum answer.

hope this helps! :)

Final answer:

The calculation of energy required to move an object to an altitude twice the Earth's radius involves understanding and applying principles of gravitational potential energy and the universal law of gravitation.

Explanation:

The question involves calculating the energy required to move a 1350 kg object from the Earth's surface to an altitude twice the Earth's radius. To solve this, we use the formula for gravitational potential energy (GPE), which is GPE = mgh at close distances to Earth's surface, and the universal law of gravitation for larger distances.

However, at distances far from the surface, the formula shifts to GPE = -G * (m1*m2)/r, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers. Since the altitude is twice the Earth's radius, the effective distance r would be 3 times the Earth's radius. This question requires integration of both concepts and understanding of physics to solve comprehensively.

With the known value of v (speed of sound in the medium, e.g., 343 m/s) and the desired change in frequency (25% or 0.25), you can calculate the speed you must approach the source of sound to observe a 25% change in frequency.

To observe a 25% change in frequency (Doppler effect) when approaching a source of sound, you need to know the relative velocity between you and the source of sound. The Doppler effect occurs when there is relative motion between the observer and the source of the sound.

The formula for calculating the apparent frequency (f') observed by a moving observer due to the Doppler effect is:

f' = f * (v + vo) / (v - vs)

Where:

f' = Apparent frequency observed by the moving observer

f = Actual frequency of the sound emitted by the source

v = Speed of sound in the medium (approximately 343 meters per second in air at room temperature)

vo = Velocity of the observer (positive if moving towards the source, negative if moving away from it)

vs = Velocity of the source of sound (positive if moving away from the observer, negative if moving towards it)

Since  want to observe a 25% change in frequency, the apparent frequency (f') would be 25% different from the actual frequency (f):

f' = 1.25 * f

Assuming the observer is moving towards the source (vo is positive), we can rewrite the equation as:

1.25 × f = f × (v + vo) / (v - vs)

Now, we can solve for the relative velocity vo:

(v + vo) / (v - vs) = 1.25

Cross-multiply:

v + vo = 1.25 * (v - vs)

Now, isolate vo:

vo = 1.25 × (v - vs) - v

With the known value of v (speed of sound in the medium, e.g., 343 m/s) and the desired change in frequency (25% or 0.25), you can calculate the speed you must approach the source of sound to observe a 25% change in frequency.

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Sir Isaac Newton is the primary contributor to the universal law of gravitation, with his precise mathematical formula that unified terrestrial and celestial phenomena. Johannes Kepler's laws of planetary motion were foundational for Newton's work, and Albert Einstein expanded on these ideas with his theory of general relativity.

Several prominent individuals contributed to the discovery and understanding of the universal law of gravitation. Sir Isaac Newton is the most well-known figure associated with the law, as he defined the gravitational force, proposing that it was a universal force that explained both why objects fall to Earth and the motions of celestial bodies. He was the first to provide a precise mathematical formula for the law of gravitation.

Before Newton, Johannes Kepler discovered three laws of planetary motion, which Newton found crucial for his own work, as they showed gravitation's effects on planetary orbits. Moreover, Albert Einstein expanded upon the concept of gravitation with his theory of general relativity which showed that there is more to the gravity story than Newton's law suggested. Although their contributions were indirect, scientists such as Galileo Galilei and Robert Hooke, also helped set the stage for Newton's discoveries through their work on planetary motions and gravitational investigations respectively.

The critical angle is the smallest incident angle at which light is no longer refracted into water, but instead is totally internally reflected. This can be calculated using Snell's law, where the sine of the critical angle is the ratio of the indices of refraction for water and glass.

The smallest value of the incident angle θa for which none of the ray refracts into water, and instead exhibits total internal reflection, is known as the critical angle. To find this critical angle, we can apply Snell's law (n1 × sin(θa) = n2 × sin(θb)), where θa is the incident angle and θb is the refracted angle when θb is 90°, the angle of refraction is at the maximum and therefore indicates the critical angle condition. Using the indices of refraction for glass (n1) and water (n2), we can solve for the critical angle which will indicate the threshold above which light will not refract into water but instead be totally internally reflected.

If the speed of an object increases, then its kinetic energy will increase proportionally because speed and kinetic energy have a linear relationship when graphed.

just got it right...

For an uniformly accelerated motion, the following relationship is used:

[tex]2 a S=v_f^2 -v_i ^2[/tex] (1)

where a is the acceleration, S the distance covered, and vf and vi the final and initial speeds of the motion.

In our problem we are dealing with a rotational motion. Initially, the salad spinner has constant angular speed, which is given by

[tex]\omega _i = 2 \pi f[/tex]

where f is the rotational frequency, which is the number of revolutions per second:

[tex]f=  \frac{20 rev}{5 s}=4 Hz [/tex]

so the initial angular speed is

[tex]\omega _i = 2 \pi (4 Hz)=25.2 rad/s[/tex]

Then, the salad spinner starts to decelerate with constant deceleration [tex]\alpha[/tex], and during its deceleration it spins for other 6 revolutions, so covering a total angle of

[tex]\theta = 2 \pi (6 rev)=37.7 rad[/tex]

until it stops, so until it reaches a final speed of [tex]\omega _f=0[/tex].

To find the angular acceleration, we can use the equivalent of equation (1) for angular motions:

[tex]2 \alpha \theta = \omega_f^2 - \omega_i^2[/tex]

and so, since the final speed is [tex]\omega _f=0[/tex]:

[tex]\alpha = -  \frac{\omega _i^2}{2 \theta}=- \frac{(25.2 rad/s)^2}{2\cdot 37.7 rad}=-8.4 rad/s^2  [/tex]

where the negative sign means the salad spinner is decelerating.

Answer:

[tex] \alpha = -\frac{(25.132 rad/s)^2}{2* 37.7 rad}= -8.377 rad/s^2[/tex]

And we can convert this into degrees like this:

[tex] \alpha= -8.377 rad/s^2 * (\frac{180}{\pi rad}) =-479.967 rad/s^2[/tex]

Explanation:

For this case we assume that the angular acceleration is constant and the spinner slows down and come to rest at the end

We can calculate the distance traveled each revolution with this formula:

[tex] \theta= 20 rev * \frac{2\pi rad}{1 rev}= 40 \pi rad[/tex]

And since we know that the time to reach the velocity 0 is 5 s we can find the angular velocity like this:

[tex] w_o= \frac{\theta}{t}= \frac{40 \pi rad}{5 s}= 25.132 rad/s[/tex]

We know that the spinner rotates 6 more times before come rest, so the total distance traveled is:

[tex] \theta= 6* 2\pi = 37.699 rad[/tex]

[tex] w_f = 0 rad/s[/tex]

And we have the following formula :

[tex] w^2_f = w^2_i + 2\alpha \theta[/tex]

Since we know that the final angular velocity is 0 we can solve for [tex] \alpha[/tex] the angular acceleration and we got:

[tex] \alpha = -\frac{w^2_o}{2 \theta}[/tex]

And replacing the values that we found before we have this:

[tex] \alpha = -\frac{(25.132 rad/s)^2}{2* 37.7 rad}= -8.377 rad/s^2[/tex]

And we can convert this into degrees like this:

[tex] \alpha= -8.377 rad/s^2 * (\frac{180}{\pi rad}) =-479.967 rad/s^2[/tex]