Answer:
Option A = 1.
Explanation:
So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;
(1). First, Both objects A and D represent fixed.
(2). Both objects A and D are negatively-charged particles of equal magnitude.
(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."
(4). "Object C shows a moving, positively-charged particle."
So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.
We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.
The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).
Answer:
hope helps
Explanation:
In a weightless environment a force of 5 Newtons is applied horizontally to the right on a rock with a mass of 1 kg and to a pebble with a mass of 0.1 kg.
Answer:
because
Explanation:
streasm dont flow horizontal because if it did then that would be breaking all laws of physics and we know that what goes up must com down but water cant flow upstream only down if it does flow horizontally then it would either be between two hills or in a plains
Answer:
The duration of the movie is longer than 2 hrs.
Explanation:
Given:
The duration of the movie observed by the crew on the spacecraft is 2 hrs.
According to time-dilation formula:

Here,
is the required time,
is the original time,
is the velocity of the spacecraft and
is the velocity of light.
Since
, so
.
So the time required will be large.
Answer:
392 N
Explanation:
Draw a free body diagram of the rod. There are four forces acting on the rod:
At the wall, you have horizontal and vertical reaction forces, Rx and Ry.
At the other end of the rod (point X), you have the weight of the sign pointing down, mg.
Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.
Sum of the moments at the wall:
∑τ = Iα
(T sin θ) L − (mg) L = 0
T sin θ − mg = 0
T = mg / sin θ
Given m = 20 kg and θ = 30.0°:
T = (20 kg) (9.8 m/s²) / (sin 30.0°)
T = 392 N