Ask question

Ask question

All categories

schepotkina [342]

3 years ago

7

the two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different di

stances from the axis of rotation. if you release the hanging blocks simultaneously from rest, and if the ropes do not slip, which block lands first?

1 answer:

svetoff [14.1K]3 years ago

7 0

Answer: the block at the right lands first

Explanation:

You might be interested in

How can you use a magnet to induce a current?

Furkat [3]

A coil of wire with a current flowing thru it becomes a magnet

3 0

3 years ago

Read 2 more answers

PLEASE HELP ME I JUST CAN'T FIGURE THIS ONE OUT

kupik [55]

Indeed the answer is c!!

4 0

3 years ago

Read 2 more answers

Match each term to its definition.

den301095 [7]

Answer: first one is electrochemical
Second one is combustion
Third one is photosynthesis
Fourth one is respiration

8 0

3 years ago

Read 2 more answers

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a

Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

$PV=nRT$

$P_{i}=\dfrac{nRT}{V}$

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

$P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}$

$P_{i}=5.75\times10^{6}\ Pa$

$P_{i}=56.827\ atm$

We need to calculate the final temperature

Using relation pressure and volume

$P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}$

$P_{2}=\dfrac{56.827\times70}{400}$

$P_{2}=9.94\ atm$

Hence, The final pressure of the gas is 9.94 atm.

3 0

3 years ago

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago