PHYSICS COLLEGE

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.

Answers

Answer 1

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

Related Questions

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If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change?a. 4
b. 2/pi
c. 2
d. (2)^1/2

Answers

Answer:

the answers the correct one is a 4

Explanation:

The centripetal acceleration is by

a = v² / R

angular and linear velocities are related

v = w R

let's substitute

a = w² R

for initial condition

a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

a = (2w₀)² R

a = 4 (w₀² R)

a = 4 a₀

when reviewing the answers the correct one is a

If you drew magnetic field lines for this bar magnet, which statement would be true

Answers

Arrows point away from north and toward south.

and

Field lines loop around the magnet starting at the north pole and ending at the south pole.

field lines correct for future weiins

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer: $\theta =45.73^(\circ)$

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

$T\cos\theta =mg$ -----1

$T\sin \theta =(mv^2)/(r)$ ------2

Divide 2 & 1

$tan\theta =(v^2)/(rg)$

$tan\theta =(4.65^2)/(2.15* 9.8)$

$tan\theta =1.026$

$\theta =45.73^(\circ)$

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Answers

Answer:

93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

41°: 93.6 N

63°: 155.6 N

90°: 200 N

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

Force = 200 Newton.

Angle 1 = 41°

Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

$\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N$

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

$\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2$ ....equation 1.

For the vertical component:

$\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200$ ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton$

For the first tension:

$T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton$

Read more on tension here: brainly.com/question/4080400

What type of circuit measurement is made by placing a meters test leads in parallel with a deenergized component

Answers

If the component is DE-energized, meaning the whole device is
powered down, then the only thing you can measure with the meter-
probes on both ends of the component is its resistance.

If you have a fancy, expensive meter, then maybe you could measure
the component's capacitance or inductance. I never had one of those.

The normal meter measures volts, amps, and ohms. If you touch
the probes to both ends of the component and the circuit is energized,
then you measure the voltage across the component. If the circuit is
DE-energized, then you're measuring the component's resistance.

(Note: You have to know which one you're measuring, and set up the
meter properly. For example, if the circuit is energized and you try to
measure resistance, it's possible that you could fry your meter.)

a. A nucleus is made up of protons and neutrons. Protons have positive charges and neutrons have no charge. The strong nuclear force holds the nucleus together because it acts against another force inside the nucleus. What force is the strong nuclear force counteracting?

Answers

Answer:

The electromagnetic force tends to push the protons apart. (Like forces repel).

Explanation:

Answer:its the electromagnetic force

Explanation:

Protons are positive so repel themselves

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