Answer:
<h3>1.43m/s²</h3>
Explanation:
According to newtons second law.
F = mass * acceleration
If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg
Force applied = 1N
acceleration = Force/mass
Substitute the values and get acceleration
acceleration = 1/0.7
acceleration = 1.43m/s²
Hence the magnitude of the acceleration of the robot is 1.43m/s²
D,f,g,h,i,a,e,c,j. I’m sure that it
Answer:
The items here are describing either a condition in a later interacton or a protogalactic cloud. The results matching with spiral and elliptical galaxy are:
For spiral galaxy are options 6,3,2 and 5.
and for elliptical galaxy are options 4 and 1.
Explanation:
Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.
so, both types of galaxies are matched with their respective items given:
A. Spiral galaxy:
2. A galaxy collision results tostripping of gas.
3. The protogalactic cloud rotates in a very slow motion.
5. The density of protogalactic cloud is very high.
6. when the protogalactic cloud shrinks cloud forms very rapidly.
B. Elliptical galaxy:
1. The protogalactic cloud has high angular momentum.
4. Most of the protogalactic gases settles down into a disk.
Answer:
(a) Negative Q
(b) Positive Q
Explanation:
Charge is the inherent property of matter due to the transference of electrons.
There are three methods of charging a body.
(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.
(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.
(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.
(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.
(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values
So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
