Answers
The height at which the ball goes for the given parameters is; 0.827 m
What is the height of the ball?
We are given;
distance between the metal plates; d = 3.1 m
mass of glass; m = 1.1g = 0.0011 kg
charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C
speed of the glass ball; v = 4.8 m/s
voltage of the ceiling; V = +3.0 × 10⁶ V
The repulsive force experienced by the ball is gotten from the formula;
F = qV/d
|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1
|F| = 4.548 × 10⁻³ N
F = -4.548 × 10⁻³ N (negative because it is repulsive force)
The net horizontal force experienced by the ball is;
F_net = F - mg
F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)
F_net = -15.328 × 10⁻³ N
To get the height of the ball, we will use the formula;
F_net * h = ¹/₂mv²
h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)
We took the absolute value of F_net, hence it is not negative
h = 0.827 m
Read more about height of ball at; brainly.com/question/12446886
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
W = K.E
Therefore, the ball traveled 0.827 m
Related Questions
A paper airplane moving 2.33 m/s has 0.161 J of KE. What is its mass? (Unit = kg)
What is matter? explain and give example
A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?Express your answer in terms of the variable v0 and appropriate constants.c) For what value of h does the collision occur at the instant when the first ball is at its highest point?Express your answer in terms of the variable v0 and appropriate constants.
An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled. What is the plate's surface charge density?
Answers
Answer:
3.65 x mass
Explanation:
Given parameters:
Time = 20s
Initial velocity = 0m/s
Final velocity = 73m/s
Unknown:
Force the ball experience = ?
Solution:
To solve this problem, we apply the equation from newton's second law of motion:
F = m
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
So;
F = m ( ) = 3.65 x mass
Final answer:
To calculate the force experienced by the ball to accelerate from rest to 73 m/s, use Newton's second law of motion.
Explanation:
To calculate the force experienced by the ball to accelerate from rest to 73 m/s, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).
Since the ball starts from rest, its initial velocity (vi) is 0 m/s. The final velocity (vf) is 73 m/s. The time (t) taken for the impact is given as 2 x 10 seconds. So, the acceleration (a) can be calculated using the formula a = (vf - vi) / t.
Substituting the given values into the equation, we have a = (73 - 0) / (2 x 10) = 3.65 m/s^2.
Now, we can find the force (F) using the formula F = m * a. If the mass of the ball is known, we can substitute it into the equation to find the force experienced by the ball.
Learn more about Force here:
#SPJ3
Answers
Answer:
1.2 s
Explanation:
Given:
v₀ = 8.0 m/s
v = -4.0 m/s
a = -10 m/s²
Find: t
v = at + v₀
(-4.0 m/s) = (-10 m/s²) t + (8.0 m/s)
t = 1.2 s
(10 Points)
increases
decreases
Answers
a speed of 50kmh than it
would at a
speed of 6okmh?
Answers
Answer:
2hr much longer
Explanation:
Given parameters
Distance = 600km
Speed 1 = 50km/h
Speed 2 = 60km/h
Unknown:
How much longer will it take to travel a distance = ?
Solution:
Speed is the distance divided by time;
Speed =
Now;
Time taken =
Time 1;
=
= 12hr
Time 2;
=
= 10hr
To find how much more time;
Time 1 will take 12hr - 10hr, 2hr much longer to travel the distance at that rate.
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.
Answers
Answer:
option C
Explanation:
the correct answer is option C
When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.
According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.
This law's application is used in machines like hydraulic jacks.
Answers
Answer: i think you should place it on the red line
Explanation:
hope this helps
and need brainliest