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:
(1)
(2)
Where:
is the final velocity of the bullet
is the initial velocity of the bullet
is the acceleration due gravity, always directed downwards
is the time
is the vertical position of the bullet at
Let's begin by finding from (1):
(3)
(4)
Now we have to substitute (4) in (2):
(5)
Isolating :
This is the displacement of the bullet after 6.9 s
Answer:
When she stretches her arms out,<em> B) her angular speed ω increases due to her moment of inertia decreasing</em>
Explanation:
The angular momentum of a rotating object is defined as the product of its moment of inertia and angular speed.
<em>L = I ω</em>
<em>where</em>
- <em>L is the angular momentum</em>
- <em>I is the moment of inertia</em>
- <em>ω is the angular speed</em>
<em />
According to the principle of conservation of angular momentum, if there is no external torque, angular momentum of the skater must remain conserved. If the initial and final moment of inertia is <em>I_i and I_f </em>while corresponding angular velocities are <em>ω_i and ω_f , </em>then the principle of conservation of angular momentum can be expressed as the following equation:
<em>(I_f) (ω_f) = (I_i) (ω_i)</em>
<em>ω_f / ω_i = I_i / I_f</em>
<em />
From the expression above, we can see that if the moment of inertia decreases, angular velocity would increase to conserve angular momentum of the skater.
Therefore, When she stretches her arms out,<em> her angular speed ω increases due to her moment of inertia decreasing.</em>
The statements that are true are;
- the magnitude of work done by frictional forces on the crate is 100 J
- the magnitude of work done by the applied force on the crate is 100 J
<h3>What is work done?</h3>
Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.
Hence, the following are true;
- the magnitude of work done by frictional forces on the crate is 100 J
- the magnitude of work done by the applied force on the crate is 100 J
Learn more about work: brainly.com/question/18094932?
Answer:
346819 Pa or ,347000 Pa in 3 significant figures
Explanation:
P1= 325000Pa , T1= 283K ,
P2=? T1= 302 K
as here volume and mass both are constant so using ratio method for pressure temperature law we have P1/T1 = P2/T2
THIS WE CAN ALSO OBTAIN BY RATIO METHOD FOR GENERAL GAS LAW AS
P1V1/(m1T1 ) = P2 V2/ (m2 T2)
IF V1 = V2 =V AND m1=m2=m then expression reduces to
P1/T1 = P2/T2
or P2 = (P1/T1)×T2
P2 = (325000/283) × 302
P2 = 1148.41×302
P2=346819
P2 = 347000 Pa in 3 significant figures
