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2 years ago

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The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the

acceleration of the particle A at point O?

1 answer:

lara [203]2 years ago

7 0

As velocities are tangent, the value of both Particle A and Particle B would be same for that point O (Intersecting point)

a = v / t
Here, v = 7, t = 6
So, a = 7/6
a = 1.17
As the graph is decreasing, value of acceleration would be negative.
So, a = -1.17 m/s²

In short, Your Answer would be Option C

Hope this helps!

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Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment

riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a look at the free bodily diagrammatic representation.

The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ = $\frac{360+240}{450}$

tan θ = $\frac{600}{450}$

tan θ = 1.333

θ = tan⁻¹ (1.333)

θ = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

$M__B}=(Fcos \theta *240)-(Fsin \theta *450)$

where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:

$M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)$

$M__B}= 194400.463-485999.348$

$M__B}=-291598.885 N-mm\\$

$M__B}=-291.598 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B $(M__B})$ ,

we have:

$M__B}=-Fcos \theta *360 - Fsin \theta * 0$

$M__B}=-Fcos \theta *360 - 0$

$M__B}=-Fcos \theta *360$

where Force(F) = 1350 N and θ = 53.13° ; we have:

$M__B}= -1350*cos53.13^0*360$

$M__B}= -291600 N-mm$

$M__B}= -291.6 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

6 0

2 years ago

A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0

2 years ago

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

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

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An

Svetradugi [14.3K]

efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

=55%

5 0

2 years ago

Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo

LenaWriter [7]

Answer:

$6.33\times 10^8\ kg\cdot m/s$

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = $33\times \frac{5}{18}=9.17\ m/s$

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

= $(6.9\times 10^7\ kg)(9.17\ m/s)$

= $6.33\times 10^8\ kg\cdot m/s$

Therefore, the magnitude of ship's momentum is $6.33\times 10^8\ kg\cdot m/s$ .

6 0

2 years ago