PHYSICS COLLEGE

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.(a) Find the value of the normal force on the block.

(b) Find the magnitude of force F~2 that is acting on the block

(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .

Answers

Answer 1

Answer:

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

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What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?42 degrees
48 degrees
57 degrees
61 degrees

Answers

Answer:

61 degrees, I just did the test.

Explanation:

Answer: 61 degrees

Explanation:

I just did the question and got it right

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answers

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_(min)=√(5gR)$

Put the value into the formula

$v_(min)=√(5*9.8*0.8950)$

$v_(min)=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+(1)/(2)mv^2=(1)/(2)m(√(5gR))^2$

$v^2=(√(5gR))^2-2gH$

$v^2=(6.62)^2-2*9.8*1.377$

$v^2=16.8352$

$v=√(16.8352)$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.

Final answer:

The minimum translational speed the solid ball must have when it is at a height H=1.377 m above the bottom of the loop to successfully complete the loop without falling off the track is approximately 7.672 m/s. This was derived using principles of energy conservation.

Explanation:

The minimum translational speed must be sufficient enough to maintain contact with the track even at the highest point of the loop. Using the principle of energy conservation, the total energy at the height H, assuming potential energy to be zero here, should be equal to the total energy at the highest point of the loop. Here, the total energy at height H will consist of both kinetic and potential energy while at the top of the loop it consists of potential energy only. Setting these equations equal to each other: 0.5 * m * v² + m * g * H = m * g * 2R Solving the above equation for v:v = √2g (2R-H). Substituting known values henceforth gives us √2*9.81*(2*0.895-1.377) = 7.672 m/s. Hence, the ball must have a minimum translational speed of approximately 7.672 m/s at height H to complete the loop without falling.

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The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answers

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=(0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m)/((\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m) \n\ny=0.88\ m$

b) the mass moment of inertia about z axis passing the rotation center O is:

$I_G=(1)/(12)*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\n0.888)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+(1)/(12)*3(1.5)(1.5)^2+\n3(1.5)(0.888-0.75)^2\n\nI_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=(1)/(12)*3(0.8)(0.8)^2+ (1)/(3)* 3(1.5)(1.5)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\n(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\n\nI_o=13.4\ kgm^2$

Final answer:

To solve this problem, calculate the mass of each element of the pendulum, use that information to determine the center of mass, and then apply the parallel axis theorem to calculate the two moments of inertia.

Explanation:

To determine the center of mass and the mass moment of inertia of the pendulum, first we calculate the individual masses of the rods: AB and OC, and the plate. Each rod has a mass of 2 kg (given mass per unit length is 3kg/m and length of each rod is 1 m from the first reference paragraph).

The center of mass ӯ can be determined using the formula for center of mass, averaging distances to each mass element weighted by their individual masses. The mass moment of inertia, also known as the angular mass, for rotation about the z axis through G is determined using the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance D away from an axis through the center of mass is the sum of the moment of inertia for rotation about the center of mass and the total mass of the body times D squared.

Finally, the moment of inertia about the z axis passing through the center of rotation O can be calculated again using the parallel axis theorem, with distance d being the distance between points G and O.

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You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by:

Vx,f = sqrt (2*g*x*sin(Q))

Enunciado: Una bola se lanza verticalmente de la parte superior de un edificio con una velocidad inicial de 25 m/s. La bola impacta al suelo en la base del edificio 7 segundos después de ser lanzada. (Marque la respuesta correcta) ¿Qué altura subió la bola (medida desde la parte superior del edificio)? a) 19.6 m b) 12.75 m c) 31.88 m d) 40 m e) 20 m

Answers

La altura vertical máxima alcanzada es de 31,88 m.

Tenemos la siguiente información de la pregunta;

Velocidad inicial = 25 m/s

Velocidad final = 0 m/s (a la altura máxima)

tiempo empleado = 3,5 minutos (el tiempo empleado para subir y bajar es igual).

Usando la ecuación;

v^2 = u^2 - 2gh

Dado que v = 0

u^2 = 2gh

h = tu^2/2g

h = (25)^2/2 *9.8

h = 31,88 m

Obtenga más información sobre las ecuaciones de movimiento: brainly.com/question/8898885

The subject of this question is kinematics. The ball reached a height of 65.1 meters.

To determine the height that the ball reached, we can use the kinematic equation for vertical motion:

Final height = Initial height + Initial vertical velocity * Time + (1/2) * Acceleration * Time^2

In this case, the initial height is the height of the building, the initial vertical velocity is 25 m/s, the time is 7 seconds, and the acceleration is -9.8 m/s^2. Plugging in these values, we get:

Final height = 0 + 25 * 7 + (1/2) * (-9.8) * 7^2 = 0 + 175 - 240.1 = -65.1.

Since the ball is at ground level, the height it reached is the negative of the calculated value, so the correct answer is 65.1 m.

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Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)

Answers

The formula for calculating the horizontal displacement of a horizontally launched projectile is $x=v*t$

A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.

The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.

Thus, the horizontal displacement of the projectile is given by the expression.

$x=v*t$

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