An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

Answers

Answer 1

Answer:

Final answer:

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

"Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

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What is the meaning of relative as a noun?

Answers

Answer:

noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Answers

Answer:

22.1 m

Explanation:

$v_(o)$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball along the vertical direction.

$v_(oy)$ = initial speed of ball = $v_(o) Sin\theta = 14.3 Sin27 = 6.5 ms^(-1)$

$a_(y)$ = acceleration due to gravity = - 9.8 ms⁻²

$t$ = time of travel

$y$ = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_(oy) t + (0.5) a_(y) t^(2) \n- 3.50 = (6.5) t + (0.5) (- 9.8) t^(2)\n- 3.50 = (6.5) t - 4.9 t^(2)\nt = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_(ox)$ = initial speed of ball = $v_(o) Cos\theta = 14.3 Cos27 = 12.7 ms^(-1)$

$X$ = Horizontal distance traveled

$t$ = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t\nX = (12.7)(1.74)\nX = 22.1 m$

3. Which object has more inertia?A. A tractor trailer rig moving at 2 m/s
B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s

Answers

Answer:

A. A tractor trailer rig moving at 2 m/s

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a tractor trailer rig, the greater will be its tendency to continuously remain at rest.

Hence, the object that has more inertia is a tractor trailer rig moving at 2 m/s because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia.

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

Answers

Answer:

The speed of the arrow immediately after it leaves the bow is 38.73 m/s

Explanation:

given information:

force, F = 150 N

x = 50 cm = 0.5 m

mass of arrow, m = 50 g = 0.05 kg

We start from the force of the spring

F = kx

k = $(F)/(x)$

= $(150)/(0.5)$

= 300 N/m

The potential energy, EP of the spring is

EP = $(1)/(2) kx^(2)$

the kinetic energy, EK of the spring

EK = $(1)/(2) mv^(2)$

According to conservative energy,

EP = EK

$(1)/(2) kx^(2)$ = $(1)/(2) mv^(2)$

$kx^(2)$ = $mv^(2)$

$v^(2)$ = $(kx^(2) )/(m)$

v = $x\sqrt{(k)/(m) }$

= $0.5\sqrt{(300)/(0.05) }$

= 38.73 m/s

Final answer:

Using Hooke's Law, we can determine the speed of the arrow. The speed of the arrow immediately after it leaves the bow is approximately 38.7 m/s.

Explanation:

In this problem, we can use Hooke's Law to determine the speed of the arrow. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring:

F = -kx

Where F is the force, x is the displacement, and k is the spring constant.

In this case, the force exerted by the bow on the arrow is acting like a spring force. The force of the bow is 150N, and the displacement is 50cm (which is equivalent to 0.5m). So we can set up the equation as:

150N = -k * 0.5m

Now we can solve for k:

k = -150N / 0.5m = -300 N/m

Now that we have the spring constant, we can use it to find the potential energy stored in the bow:

PE = 0.5kx^2 = 0.5*(-300N/m)*(0.5m)^2 = 37.5 J

Next, we can use the conservation of energy to find the kinetic energy of the arrow right after it leaves the bow. The potential energy stored in the bow is converted into kinetic energy:

KE = PE = 37.5 J

The kinetic energy is given by the equation:

KE = 0.5mv^2

Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for v:

v = sqrt(2KE/m) = sqrt(2*37.5 J / 0.05 kg) = sqrt(1500) m/s ≈ 38.7 m/s.

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25 POINTS FIRST CORRECT GET BRAINLIEST!!!!!!!!!!!!!!!!1

Answers

Answer:

carbon isnt 12

Explanation:

You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

Answers

Answer:

μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

a = 2 x / t²

a = 2 0.5 /0.5²

a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

X axis

fr - Wₓ = m a (1)

Y Axis

N- $W_(y)$ = 0

N = W_{y}

We use trigonometry to find the components of the weight

sin 45 = Wₓ / W

cos 45 = W_{y} / W

Wₓ = W sin 45

W_{y} = W cos 45

The out of touch has the expression

fr = μ N

fr = μ W_{y}

We substitute in 1

μ mg cos 45 - mg sin 45 = m a

W_{y} = (a + g sin 45) / g cos 45

μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

a = -4 m / s²

μ = 1- 4 / (9.8 cos 45)

μ = 0.423

Answer:

The μ = 0.422

Explanation:

The distance travelled by the mass is equal to:

$L=ut+(1)/(2)at^(2) \n0.5=(0*5)+(1)/(2) a(0.5^(2) )\na=4m/s^(2)$

The sum of forces in y-direction equals zero:

∑Fy = 0

N - (m * g * cosθ) = 0

N - (1 * 9.8 * cos45) = 0

N = 6.93 N

The sum of forces in x-direction is equal to:

∑Fx = ma

(m * g * sinθ) - fk = m * a

(1 * 9.8 * sin45) - fk = 1 * 4

fk = 2.93 N

fk = μ * N

2.93 = μ * 6.93

μ = 0.422

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