Ask question

Ask question

All categories

3 years ago

14

A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at

150 rpm . Part A What is the speed of the 150 g ball? Express your answer to two significant figures and include the appropriate units

1 answer:

stira [4]3 years ago

6 0

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

Angular speed ,ω = 150 rpm

The speed in rad/s

$\omega =\dfrac{2\pi N}{60}$

$\omega =\dfrac{2\pi \times 150}{60}$

ω = 15.7 rad/s

The distance of center of mass from 150 g

$r=\dfrac{150\times 0+240\times 33}{150+240}\ cm$

r= 20.30 cm

The speed of the mass 150 g

v= ω r

v= 20.30 x 15.7 cm/s

v= 318.71 cm/s

v= 3.18 m/s

You might be interested in

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

As altitude increases in the troposphere and stratosphere, the air temperature does what?

natima [27]

Answer:

1

Explanation:

3 0

3 years ago

6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750

zysi [14]

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

$v_{f}^{2}=v_{i}^{2} +2*a*x$

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

$v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]$

7 0

2 years ago

What is the density of a cube that has a mass of 3.75 g and a volume of 3 mL?

valina [46]

Answer:

$\displaystyle \rho=1.25\ g/ml$

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

The cube has a mass of m=3.75 g and a volume of V=3 ml, thus the density is:

$\displaystyle \rho=\frac{3.75\ g}{3\ ml}$

$\boxed{\displaystyle \rho=1.25\ g/ml}$

Since 1 kg=1000 mg and 1 lt = 1000 ml, the density has the same value but with different units:

$\displaystyle \rho=1.25\ kg/l$

6 0

2 years ago

A space vehicle is traveling at 5425 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent

lana66690 [7]

Answer:

$V_{cE}=1489m/s$

Explanation:

Given data

Space vehicle speed=5425 km/h relative to earth

The rocket motor speed=81 km/h and mass 4m

The command has mass m

From the conservation of momentum as the system isolated

$p_{i}=p_{f}\\$

Since the motion in on direction we can drop the unit vector direction

$MV_{i}=4mV_{mE}+mV_{CE}$

Where M is the mass of space vehicle which equals to sum of the motors mass and command mass.

The velocity of the motor relative to the earth equals the velocity of the motor relative to command plus the velocity of the command relative to earth

$V_{mE}=V_{mc}+V_{cE}$

Where Vmc is the velocity of motor relative to command

This yields

$5mV_{i}=4m(V_{mc}+V_{cE})+mV_{cE}\\5V_{i}=4V_{mc}+5V_{cE}$

Substitute the given values

$V_{cE}=\frac{5V_{i}-4V_{mc}}{5}\\ V_{cE}=5425*\frac{1000}{60*60}(m/s)-\frac{4}{5}*81\frac{1000}{60*60}(m/s)\\ V_{cE}=1489m/s$

5 0

3 years ago