All categories

3 years ago

What must occur for work to be done on an object?

Physics

2 answers:

Slav-nsk [51]3 years ago

6 0

The answer is D, The Object must move.

stiv31 [10]3 years ago

4 0

The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs

You might be interested in

A particular circuit toolbox contains only 80 Ω resistors and switches, which can be open or closed. Construct a circuit, fillin

Strike441 [17]

Answer:

Here we need to make parallel connection of two 80 ohm resistors to achieve 40 ohm net resistance.

Explanation:

As we know that the resistances in series add up directly and here we are given with only the resistors of 80 Ω.

So when we connect two resistors of 80 ohm in parallel we get the resultant of 40 ohm.

Mathematically:

$\frac{1}{R_p} =\frac{1}{R} +\frac{1}{R}$

$\frac{1}{40} =\frac{1}{R} +\frac{1}{R}$

$\frac{1}{40} =\frac{2}{R}$

$R=80\Omega$ gives us the only combination of two resistors in parallel.

3 0

3 years ago

Which kind of mirror can produce real images? A)concave B)convex C)flat

Anton [14]

Answer:

Explanation:

5 0

3 years ago

A student travels 4.0m East, 8.0m South, 4.0m west, and finally 8.0m north. What is her displacement

Vedmedyk [2.9K]

Answer:

travel travelled 24 meters

3 0

2 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

5 0

3 years ago

What's the difference between the speed and velocity of an object?

Artemon [7]

Speed is just how fast the object is moving, while velocity is the speed and direction of the object.

edit* Direction would be something such as North, East, etc.

4 0

3 years ago