Ask question

Ask question

All categories

3 years ago

15

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

inate y reaches 150mm, find the v and a vectors for P. Hint: for a circle with origin shown x2+y2=r2. The radius of the slot is 300 mm.

1 answer:

SVETLANKA909090 [29]3 years ago

4 0

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

$P_x=rcos\theta$

$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

Velocity at y=150 mm

$150=300sin\theta$

thus $\theta =30^{\circ}$

position of $P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}$

$P_x=259.80 mm$

$P=259.80\hat{i}+150\hat{j}$

Velocity at this instant

$u_x=-rsin\theta =300\times sin30=-150 mm/s$

$u_y=rcos\theta =300\times cos30=259.80 mm/s$

You might be interested in

Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i

max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0

3 years ago

Potential energy becomes kinetic energy when:

Sedbober [7]

Answer:

An object has potential energy (stored energy) when it is not in motion. Once a force has been applied or it begins to move the potential energy changes to kinetic energy (energy of motion).

EXAMPLE: A rock sitting on the edge of a cliff. If the rock falls, the potential energy will be converted to kinetic energy, as the rock will be moving. A stretched elastic string in a longbow.

5 0

2 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

$\theta=65^{\circ}$

Substituting,

$d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

A mover brings a box up the stairs in 10 seconds. If he applied a force of 20 N over a distance 10 m on the box, calculate the p

Softa [21]

Answer:

20 Watts

Explanation:

Work = force × distance

W = (20 N) (10 m)

W = 200 Joules

Power = work / time

P = 200 J / 10 s

P = 20 Watts

8 0

3 years ago