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There are very, very few times that scientists make estimations because they don’t have enough time or data to test all the poss

Zanzabum

True, it’s limited to when they use estimations at most

6 0

3 years ago

A 112 g hockey puck glides across a frictionless ice surface with no horizontal forces acting on it. If the puck's velocity is 2

Burka [1]

22.5 m/s
This is because the puck will not slow down, and maintain a constant velocity.

7 0

4 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$

This is the time that its needed in order for the rocket to reach the required altitude.

5 0

3 years ago

the gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by u

Ilia_Sergeevich [38]

To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. 
b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. 
f = (ma), = .226 x 105.88, = 23.92N.

8 0

4 years ago

Read 2 more answers

Which of the following types of waves can transmit only through matter? Select all that apply. (Hint: there are 2)

Anarel [89]

A) seismic waves
b) sound waves

sound waves are traveling vibrations of particles in media (air, water, metal, etc) so they can’t travel through empty space (vacuums) because there are no atoms or molecules to vibrate.

seismic waves are mechanical waves and require a medium for propagation, hence, they cannot travel through a vacuum.

hope this helps :)

7 0

3 years ago