-1- was created in the 1600 by william gilbert
-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
3-
An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.
Answer:
A. Sulfur _________ group 16 chalcogens
B. Sodium _________ group 1 alkali metals
C. Argon _________ group 18 noble gases
D. Silicon _________ group 14 carbon family
E. Chlorine _________ group 17 halogens
F. Phosphorus_________ group 15 pnicogens
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)

a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s


The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
![[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%2822%20-%2011.78%29%20%2A%20%287.30%29%5D%20%20%2B%20%5B%2811.78%20-%200%29%20%2A%20%287.30%29%5D)
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
The distance by the ball clear the crossbar is 1.15 m
Explanation:
Given that,
Distance = 44 m
Speed = 24 m/s
Angle = 31°
Height = 3.05 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity

Put the value into the formula


We need to calculate the vertical velocity
Using formula of vertical velocity

Put the value into the formula


We need to calculate the time
Using formula of time

Put the value into the formula


We need to calculate the vertical height
Using equation of motion

Put the value into the formula


We need to calculate the distance by the ball clear the crossbar
Using formula for vertical distance

Put the value of h


Hence, The distance by the ball clear the crossbar is 1.15 m