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A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released

Answers

Answer 1

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

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A 0.26-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released?
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When you blow some air above a paper strip, the paper, rises. This is because 1.) the air above the paper moves faster and the pressure is lower 2.) the air above moves faster and the pressure is higher 3.) the air above the paper moves faster and the pressure remains constant 4.) the air above the paper moves slower and the pressure is higher 5.) the air above the paper moves slower and the pressure is lower

Answers

The correct option is option (1)

The faster movement of air on the upper surface of the paper creates lower pressure above the paper.

Pressure difference:

The movement of air is always from a region of higher pressure to a region of lower pressure.

As we blow air above the paper strip a low pressure is created above the strip due to the fast movement or high speed of the air. And the pressure below the strip is higher in comparison to the pressure above since the air below is not moving.

So, due to the pressure difference, a force is generated on the paper strip by the air from the lower surface to the upper surface.

Learn more about pressure difference:

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This is happened because "the air" above "moves faster" and "the pressure" is "lower".

Option: 1

Explanation:

Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.

When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum

Answers

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

$E = hf = h(c)/(\lambda) \n\nwhere;\n\nh \ is \ Planck's \ constant = 6.626 * 10^(-34) \ Js\n\nc \ is \ speed \ of \ light = 3 * 10^(8) \ m/s\n\nE = ((6.626* 10^(-34))* (3* 10^8))/(248* 10^(-9)) \n\nE = 8.02 * 10^(-19) \ J$

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)

Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

$\phi = hf = (hc)/(\lambda_(max)) \n\n\lambda_(max) = (hc)/(\phi) \n\n\lambda_(max) = ((6.626* 10^(-34)) * (3 * 10^8) )/(6.546 * 10^(-19)) \n\n\lambda_(max) = 3.037 * 10^(-7) m\n\n\lambda_(max) = 303.7 \ nm$

A uniform, solid sphere of radius 3.75 cm and mass 4.00 kg starts with a purely translational speed of 1.75 m/s at the top of an inclined plane. The surface of the incline is 3.00 m long, and is tilted at an angle of 26.0∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed ????2 at the bottom of the ramp.

Answers

Answer:

v_2=4.53m/s $v_2=4.53m/s$

Explanation:

In order to solve the exercise it is necessary to apply the energy conservation equation,

The equation says the following,

$mgdsin(\theta)+(1)/(2)mv^2_1=(1)/(2)mv^2_2+(1)/(2)Iw^2$

Replacing the formula for I of a sphere, we have

$mgdsin(\theta)+(1)/(2)mv^2_1=(1)/(2)mv^2_2+(1)/(2)(2)/(5)mr^2((v_2)/(r))^2$

$mgdsin(\theta)+(1)/(2)mv^2_1=(1)/(2)mv^2_2+(1)/(5)mv^2_2=(7)/(10)mv^2_2$

$(10)/(7)gdsin(\theta)+(5)/(7)v^2_1=v^2_2$

In this way we get the expression

$v_2=\sqrt{(10)/(7)gdsin(\theta)+(5)/(7)v^2_1}$

We proceed to replace with the given values and obtain that

$v_2=\sqrt{(10)/(7)*9.8*3sin(26))+(5)/(7)*1.75^2}$

$v_2=4.53m/s$

v_2=4.53m/sv_2=4.53m/s

The equation says the following,

mgdsin(0) + 1/2mv2/1 = 1/2mv2/2 + 1/2Iw^2

Replacing the formula for I of a sphere,

mgdsin(0) + 1/2mv2/1 = 1/2mv2/2 + 1/2 2/5mr^2 (v2/r)^2

mgdsin(0) + 1/2mv2/1 = 1/2mv2/2 + 1/5mv2/2 = 7/10mv2/2

10/7gdsin(0) + 5/7v2/1 = v2/2

In this way, we get the expression

v2 = sqrt(10/7gdsin(0) + 5/7v2/1)

v2 = sqrt(10/7 * 9.8 * 3sin(26)) + 5/7 * 1.75^2

v2 = 4.53m/s

Further Explanation

The ball that rolls on the plane will experience two movements at once, namely the rotation of the axis of the ball and the translational field being traversed. Therefore, objects that do rolling motion have a rotational equation and a translational equation. The amount of kinetic energy possessed by the rolling body is the amount of rotational kinetic energy and translational kinetic energy. You will here learn about the ball rolling on a plane and incline.

An object can experience translational motion or rotational motion. Translational motion is the motion of objects whose direction is straight or curved. In translational motion using the concept of Newton II's law. While the rotational motion is the motion that has a rotation of a particular shaft. Rotational motion is caused by the torque, which is the tendency of a force to rotate a rigid body against a particular pivot point.

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Object Experience brainly.com/question/13696852

The ball that rolls brainly.com/question/13707126

Details

Grade: College

Subject: Physics

Keyword: object, ball, roll

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Answers

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=(mv^2)/(2)+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=(mv^2)/(2)+mg(0)+15\n\nE_f=(0.05* v^2)/(2)+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\n40=(0.05v^2)/(2)+15\n\n25=(0.05v^2)/(2)\n\nv=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

Final answer:

Using the principle of conservation of energy, the speed of the ball just before hitting the Earth's surface is found to be 79.2 m/s after accounting for the 15 J increase in thermal energy.

Explanation:

This question is concerned with the concept of conservation of energy, specifically the principles of potential and kinetic energy. When the ball is 80 meters above the Earth's surface, the total gravitational potential energy is m*g*h = 50g*9.8m/s²*80m = 39200 J (where m is mass, g is gravity, and h is height), and the kinetic energy is 0.

As the ball falls, its potential energy gets converted into kinetic energy, but we also know that the total thermal energy of the ball and the air in the system increases by 15 J. That means that not all the potential energy is converted into kinetic energy, 15 J is lost to thermal energy. So, the kinetic energy of the ball when it hits the Earth is 39200 J - 15 J = 39185 J.

Finally, we know that kinetic energy equals (1/2)*m*v², where v is the speed of the ball. Rearranging this formula to solve for v we get, v = sqrt((2*kinetic energy)/m) = sqrt((2*39185 J)/50g) = 79.2 m/s. So, just before the ball hits the surface, its speed is 79.2 m/s.

Learn more about Potential and Kinetic Energy here:

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Beverage can is thrown upward and then falls back down to the floor. As usual, a y axis extends upward (positive direction). Which of the following best describes the acceleration of the can during its free flight?a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
b) +9.8 m/s^2 throughout
c) -9.8 m/s^2 throughout
d) zero throughout
e) +9.8 m/s^2, then momentarily zero, then -9.8 m/s^2

Answers

a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s

If you travel 2 km north, then travel 5 km south, what is your displacement?

Answers

Answer:

➢This is a vector addition problem which requires magnitude and direction as the answer. First is to resolve the southbound vector and the northbound vector. Since they are opposite in directions their vector sum is their algebraic sum. 3 km north + 5 km south = 2 km south.

We then add 2 km west and 2 km south using Pythagorean theorem since west and south form a right angle. (2 km)^2 west + (2 km)^2 south gives (4 + 4) km^2 southwest = 8 (km)^2 45 degrees south of west

Extracting the square root of 8 gives us about 2.83 km 45 degrees south of west.

Explanation:

I hope it will help you...

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