3 years ago

An object that is moving must have a change in A) its speed. B) its position. C) its acceleration. D) its applied force.

1 answer:

8 0

An object that's moving doesn't necessarily change its speed or acceleration. Also, the force applied to it doesn't need to change ... in fact, a moving object doesn't need ANY force applied to it in order to keep moving.

But any moving object WILL have a change in its position ... THAT's how you know it's moving, and that's WHY you say "It's moving !". (choice-B)

You might be interested in

Lee skated 36 miles in 3 hours. What was the speed at which Lee

Zanzabum

Answer:

12mph

Explanation:

36/3=mph

8 0

3 years ago

Which natural hazards are most likely to affect Florida, due to its geography? Check all that apply. PLS HELP! I WILL GIVE BEST

zhenek [66]

Answer:

drought, floods, rip currents, tropical cyclones, wildfires

Explanation:

right on edge

6 0

3 years ago

What is the kinetic energy of a 3-kilogram bali that is rolling at 2 meters per second

alex41 [277]

6 jules

hope this helps

3 0

3 years ago

Read 2 more answers

A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

atroni [7]

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

bullet mass, Mb = 4.10 g

initial bullet velocity, Vbi = 837 m/s

wooden block mass, Mw = 820 g

initial wooden block velocity, Vwi = 0 m/s

final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

6 0

3 years ago

Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, ass

zmey [24]

Answer:

632.5 MPa

Explanation:

$\sigma_{m}$ = Matrix stress at fiber failure = 10 MPa

$V_f$ = Volume fraction of fiber = 0.25

$\sigma_f$ = Fiber fracture strength = 2.5 GPa

The longitudinal strength of a composite is given by

$\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa$

The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa

8 0

3 years ago