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Marrrta [24]
2 years ago
14

Why are citizens obligated to respond to such documents?

Physics
1 answer:
Zina [86]2 years ago
5 0
To protect the constitutional right to confront ones accused
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If earth had no atmosphere , would a falling object ever reach terminal velocity?
Bas_tet [7]
The right answer is no.
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3 years ago
A heat engine does 9200 J of work per cycle while absorbing 22.0 kcal of heat from a hightemperature reservoir. What is the effi
Morgarella [4.7K]

Answer: 9.9%

Explanation: efficiency = (work output /work input) × 100

Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.

Work output = 9200 j and work input = 92048 j

Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%

4 0
3 years ago
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
In-s [12.5K]

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0
3 years ago
Which of these would most likely be a parts of a lab procedure?
vladimir2022 [97]
C . Record the time to complete a chemical reaction
6 0
3 years ago
Read 2 more answers
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0
2 years ago
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