Consulting your data table, choose the date that the Moon is furthest from the Ecliptic.

How much force must you apply to give the go cart an acceleration of 0.9m/s^2

qaws [65]

It's dependent on the mass. You can fimd the force needed using the formula F = ma. Where F is force, m is mass of the cart and a is the acceleration (0.9m/s^2). The heavier it is the more force you are going to need. Remember unit of force is N

5 0

3 years ago

HELPPP MEEE PLEADEEEE How could a cow have more momentum than a tiger ? Explain

marusya05 [52]

Answer:

If a cow has a greater mass than the tiger, it has greater momentum, but only if they are moving at the same speed, or the tiger is going slower (i think)

Explanation:

the quantity of motion of a moving body, measured as a product of its mass and velocity.

4 0

3 years ago

A bullet is fired straight up from a gun with a

melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

$V=V_{o}+gt$ (1)

$V^{2}=V_{o}^{2}+2gy$ (2)

Where:

$V$ is the final velocity of the bullet

$V_{o}=152 m/s$ is the initial velocity of the bullet

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

$t=6.9 s$ is the time

$y$ is the vertical position of the bullet at $t=6.9 s$

Let's begin by finding $V$ from (1):

$V=152 m/s-9.8 m/s^{2}(6.9 s)$ (3)

$V=84.38 m/s$ (4)

Now we have to substitute (4) in (2):

$(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y$ (5)

Isolating $y$ :

$y=815.511 m$ This is the displacement of the bullet after 6.9 s

8 0

3 years ago

When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequenc

Arlecino [84]

Answer:

$m_2/m_1=9.745$

The ratio m_2/m_1 of the masses is 9.745

Explanation:

Formula for frequency when mass m_1 is hung on the spring:

$f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}$

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

$f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}$

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

$\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}$

On Solving the above equation:

$\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745$

The ratio m_2/m_1 of the masses is 9.745

8 0

4 years ago

What three sprites are needed to create the Space Invaders game that was discussed in the unit? spaceship, laser, enemy gnome, a

Lilit [14]

Answer: anlien, enemy gnome, spaceship

Explanation:

3 0

2 years ago