Ask question

Ask question

All categories

2 years ago

13

Your friend says that inertia is a force that keeps things in their place, either at rest or in motion. Do you and your discussi

on partners agree? Why or why not?

1 answer:

Galina-37 [17]2 years ago

8 0

Answer:

yes i agree

Explanation:

because law of inertia state that object remain at rest or in motion unless external force apply on it

You might be interested in

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790

Nezavi [6.7K]

Answer:

$w_2=0.467rad/s$

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

$m_t=4*70kg=280kg$

The radius $r=4.2/2=2.1m$

Angular velocity $w_1=0.79rad/s$

The moment of inertia total is $I_t=1790 kg/m^2$

Momento if inertia

$I_1=m_t*r^2$

$I_1=280kg*(2.1m)^2=1234.8kg*m^2$

Angular momentum

$I_1*w_1=I_t*w_2$

Solve to w2

$w_2=\frac{I_1*w_1}{I_t}$

$w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}$

$w_2=0.467rad/s$

8 0

2 years ago

Fossil fuels like gasoline or coal are examples of

disa [49]

Answer:

Nonrenewable Resources

8 0

2 years ago

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

6 0

2 years ago

Read 2 more answers

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

2 years ago

What is the momentum of a .005kg bumble bee that is traveling at a velocity of 3.0m/s?

stiks02 [169]

p=mv

p=0.005kg×3.0m/s

p= 0.015kgm/s

4 0

2 years ago