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3 years ago

6

Torque can be best described as which of the following? Give an example of both a force and a torque and explain why in a couple

of sentences.
A. Rotational force.
B. Rotational velocity.
C. Rotational energy.
D. Rotational power.
E. All of the above.

1 answer:

kicyunya [14]3 years ago

4 0

Answer: rotational force

Explanation:

Torque is the twisting force which cause rotation and the axis of rotation is the point at which the object rotates.

Torque is a rotational force as it leads to the rotation of an object about an axis. Force simply means a pull or push. When an unbalanced ball acts on a force, the ball, the ball will be moved towards the linear motion.

Then, the unbalanced force that is acting in the ball produces torque which causes the ball's rotational motion.

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What kind of energy does a skier have standing still at the top of a hill?

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The skier has potential because potential energy is enery that is stored or an object that is or does not move

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3 years ago

In the future, a spaceship has traveled three percent of the distance to a space station. If the ship has traveled 2.9 x 10^7 mi

leva [86]

Answer:

$9.38 * 10^8 miles$

Explanation:

The spaceship has traveled 3% of the distance to a space station and it has traveled $2.9 * 10^7$ miles.

Let the total distance from the ship's starting point to the space station be x.

This means that:

$\frac{3}{100} * x = 2.9 * 10^7\\\\=> x = \frac{2.9 * 10^7 * 100 }{3} \\\\x = 9.67 * 10^8 miles$

The total distance to be traveled is $9.67 * 10^8 miles$ .

Therefore, the distance left to travel is:

$9.67 * 10^8 - 2.9 * 19^7\\\\= 9.38 * 10^8 miles$

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3 years ago

An object weighing 1.840 kg has a volume of 0.0015 m3. What is the density of the object in g/cm3?

olga_2 [115]

Answer:

1.22gcm³

Explanation:

D = mass/ volume

Mass=1.840kg = 1,840g

1000g = 1kg

1.840kg= x(g)

X(g) = 1.840/1000

= 1840g

Volume = 0.0015m³= 1,500cm³

1m³= 1000,000cm³.

0.0015m³= x(cm³)

X(cm³) = 1000,000×0.0015

X(cm³)= 1500.

Since density is mass/volume, now impute your data's

D=m/v

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5 0

3 years ago

Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1".The b

Gre4nikov [31]

Answer:

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

See explanation below.

Explanation:

For this case we consider the system as shown on the figure attached.

Since the system is connected the acceleration for both masses are equal, that is $a_{M_1}= a_{M_2} = a$

From the second Law of Newthon we have that the force applied for the mass $M_2$ is $F_{M_2}= M_2 a$ and we know that the force acting on the x axis for the mass 2 is $F_{M_2}= T_2$ so then we have that $T_2= M_2 a$

Now when we consider the system of $M_1 +M_2$ as a whole mass, this system have the same acceleration $a$ and on this case we will see that the only force acting on the entire system would be $T_1$ and then by the second law of Newton we have that:

$F_{M_1 +M_2} = T_1 = (M_1 +M_2) a$

And then if we compare $T_1$ and $T_2$ we see that :

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

6 0

3 years ago

Two objects each moving with speed v travel in opposite directions along a straight line passing through both their centers. The

Tpy6a [65]

Answer:

$\dfrac{1}{16}$

$\dfrac{5}{3}$

Explanation:

$m_1$ = Mass of first object

$m_2$ = Mass of second object

v = Speed of both objects

$\dfrac{v}{4}$ = Combined velocity

The ratio of final kinetic energy to initial kinetic energy will be

$\dfrac{K_f}{K_i}=\dfrac{\dfrac{1}{2}(m_1+m_2)(\dfrac{v}{4})^2}{\dfrac{1}{2}(m_1v^2+m_2v^2)} \\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{v^2}{16}}{m_1v^2+m_2v^2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{1}{16}}{m_1+m_2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{1}{16}$

The ratio is $\dfrac{1}{16}$

As the linear momentum is conserved

$m_1v-m_2v=(m_1+m_2)\dfrac{v}{4}\\\Rightarrow m_1-m_2=(m_1+m_2)\dfrac{1}{4}$

Divide by $m_2$ on both sides

$\dfrac{m_1}{m_2}-1=\dfrac{m_1}{4m_2}+\dfrac{1}{4}\\\Rightarrow \dfrac{m_1}{m_2}-\dfrac{m_1}{4m_2}=\dfrac{1}{4}+1\\\Rightarrow \dfrac{3m_1}{4m_2}=\dfrac{5}{4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5\times 4}{3\times 4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5}{3}$

The ratio of mass is $\dfrac{5}{3}$

6 0

3 years ago