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EleoNora [17]
3 years ago
15

Select three different examples of accelerated motion. a body traveling in a straight line and increasing in speed a body travel

ing in a straight line and decreasing in speed a body traveling in a straight line at constant speed a body standing still a body traveling at a constant speed and changing direction
Physics
1 answer:
IrinaVladis [17]3 years ago
6 0
This is the same question that I just answered.

Have present the definition of acceleration:

         a = Δv / Δt, this is change in velocity per unit of time.

a and v are in bold to mean that they are vectors.

1) a body traveling in a straight line and increasing in speed: CORRECT:

Acceleration is the change in velocity, either magnitude or direction or both. So, a body increasing in speed is accelerated.

2) a body traveling in a straight line and decreasing in speed: CORRECT

A decrease in speed is a change in velocity, so it means acceleration.

3) a body traveling in a straight line at constant speed: FALSE.

That body is not changing either direction or speed so its motion is not accelerated but uniform.

4) a body standing still : FALSE.

That body is not changind either direction or speed.

5) a body traveling at a constant speed and changing direction: CORRECT.

The change in direction means that the body is accelerated. The acceleration due to change in direction is named centripetal acceleration.
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A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid
Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm=\frac{6}{100}=0.06 m

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

dF=IBdr

dF=IBdr

\tau=rdF=IBrdr

\tau=\int_{0}^{R}IBr dr

\tau=IB(\frac{R^2}{2}

Torque due to friction

\tau=R\times F

Where friction force=F

R\times F=\frac{IBR^2}{2}

F=\frac{IBR}{2}

Substitute the values

F=\frac{3\times 1\times 0.06}{2}

F=0.09 N

7 0
3 years ago
A roast turkey is taken from an oven when its temperature has reached 185°F and is placed on a table in a room where the tempera
skelet666 [1.2K]
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3 years ago
Read 2 more answers
The chart shows data for a moving object.
Ipatiy [6.2K]

Answer:

number 2

Explanation:

8 0
3 years ago
3. What exerts a greater force on the table of 2 kg book lying flat or a 2 kg book on its
Darina [25.2K]

Answer:

A book on its side exerts a greater force.

Explanation:

Pressure = Force / Area

Assuming that 1kg = 10N

2kg = 20N

Area of book lying flat = 0.3m × 0.2m

                                     = 0.6m²

Pressure of book lying flat = 20N / 0.6m²

                                            = 30Pa (1 s.f.)

Area of book on its side = 0.2m × 0.05m

                                        = 0.01m²

Pressure of book on its side = 20N / 0.01m²

                                               = 2000Pa (1 s.f.)

Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.

5 0
3 years ago
Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
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