Help me again...(science)

- A preliminary stage of a spacecraft was to place it in a parking orbit 190km above the

solniwko [45]

The correct answer is C the time to complete one orbit

8 0

3 years ago

What is one example of the use of rolling kinetic friction?

RideAnS [48]

Railroad steel wheels. Bowling ball.

3 0

4 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago

Does earths surface heat up the same everywhere

melomori [17]

the correct answer is no ;)

4 0

3 years ago

Read 2 more answers

Two asteroids collide and stick together. The first asteroid has mass of 1.50 × 104 kg andis initially moving at 0.77 × 103 m/s.

KengaRu [80]

Answer:

Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.

Explanation:

This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction (in the plane of the two velocity vectors) as y-axis. So, we have that:

$p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}$

And, since $p=mv$ , we get:

$m_1v_{1o}+m_2v_{2o}\cos\theta=(m_1+m_2)v_{fx}\\\\m_2v_{2o}\sin\theta=(m_1+m_2)v_{fy}$

Solving for v_fx and v_fy, and calculating their values, we get:

$v_{fx}=\frac{m_1v_{1o}+m_2v_{2o}\cos\theta}{m_1+m_2}\\\\\implies v_{fx}=\frac{(1.50*10^{4}kg)(0.77*10^{3}m/s)+(2.00*10^{4}kg)(1.02*10^{3}m/s)\cos20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=878m/s\\\\\\v_{fy}=\frac{m_2v_{2o}\sin\theta}{m_1+m_2}\\\\\implies v_{fy}=\frac{(2.00*10^{4}kg)(1.02*10^{3}m/s)\sin20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=199m/s$

Now, the final speed can be calculated using the Pythagorean Theorem:

$v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s$

And the direction $\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°$ can be obtained using trigonometry:

$\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°$

That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.

7 0

3 years ago