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3 years ago

The speed at which an object falls and the acceleration at which it falls are

Physics

1 answer:

allochka39001 [22]3 years ago

6 0

No, the speed at which an object falls is not equal to the acceleration at which it falls.

Answer:

Option B

Explanation:

Speed is defined as how fast an object can cover a specific distance and in what time it covers. So it is measured as the ratio of distance covered to the time taken to cover that distance. While acceleration is the rate of change of velocity. Moreover, speed is a scalar quantity and acceleration is a vector quantity. So most of the times, the direction will play an important role in the varying values of speed and acceleration. Also, acceleration of an object will depend upon the force and mass of the object. Thus, speed and acceleration will not attain same value always.

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A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?

Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

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3 years ago

A boy got out of a boat and as he

emmainna [20.7K]

Answer:

third law of motion

Explanation:

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3 years ago

When scientists observed the light from stars and galaxies, they noticed that their color shifted toward the end of the visible

lakkis [162]

Color shifted towards the end is C red

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3 years ago

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A box of unknown mass is sliding with an initial speed vi = 4.70 m/s across a horizontal frictionless warehouse floor when it en

malfutka [58]

Here Change in Kinetic Energy = Work Done by Friction

Therefore, substituting the given values to the equation, we get

0.5 * m * (vFinal^2 - vInitial^2) = µ m g * d

Therefore

0.5*( 5.90^2 - Vfinal^2 ) = 0.100*9.8*2.10

Therefore

vfinal = 5.54 m/sec

8 0

3 years ago

refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a

Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature. )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0

3 years ago