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3 years ago

Which example describes constant acceleration due ONLY to a change in direction?

Physics

2 answers:

Yakvenalex [24]3 years ago

6 0

Answer:

c. traveling around a circular track

Explanation:

Acceleration is defined as the rate of change of velocity per unit time:

$a=\frac{\Delta v}{\Delta t}$

However, velocity is a vector quantity. This means that acceleration can be due to:

- A change in the speed of an object, or

- A change in the direction of the object

Let's analyze each choice:

a. increasing speed while traveling around a curve --> in this case, both speed and direction are changing (the direction is changing since the object is moving around a curve), so this is not the correct choice

b. an object at rest --> here neither the speed nor the direction are changing (in fact, the object is not accelerating), so this is not the correct choice

c. traveling around a circular track --> here the direction is changing (because the track is circular), while we don't know anything about the speed. If we assume the speed to be constant, then this is the correct choice

d. an object in free fall --> here the speed is changing, while the direction not, so this is not the correct choice

Brut [27]3 years ago

5 0

'Traveling around a circular track' can be a description of constant
acceleration due only to changes in direction. But if it is, then the
progress around the circular track must be at constant speed.

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The half-life of the reaction gets shorter as the initial concentration is increased. True or False

Art [367]

True

The half-life isn’t applicable to a first order reaction because it does not rely on the concentration of reactant present. However the 2nd order reaction is dependent on the concentration of the reactant present.

The relationship between the half life and the reactant is an inverse one.

The half life is usually reduced or shortened with an increase in the concentration and vice versa.

3 0

3 years ago

An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00

SSSSS [86.1K]

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

7 0

3 years ago

Physics question, answer completely with work

Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

$I=F\Delta t = \Delta p$

Where

I is the impulse

F is the force

$Delta t$ is the time interval during which the force is applied

$\Delta p$ is the change in momentum

In this problem,

$\Delta t = 3.0 s$ is the time interval

$I=6.0 N\cdot s$ (east) is the impulse

Therefore, the magnitude of the force is

$F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N$

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0

2 years ago

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0

3 years ago

A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago