<u>Answer:</u>
Option A is the correct answer.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
First walking 1.2 km north, displacement = 1.2 j km
Secondly 1.6 km east, displacement = 1.6 i km
Total displacement = (1.6 i + 1.2 j) km
Magnitude = 
Angle of resultant with positive X - axis = 
= 36.87⁰ east of north.
Explanation:
Unclear question. The clear rendering reads;
"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?
Answer:
Plzzzzzzzzzzzzzzzz brainliest
Explanation:
In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.
Answer:
The height at point of release is 10.20 m
Explanation:
Given:
Spring constant : K= 5 x 10 to the 3rd power n/m
compression x = 0.10 m
Mass of block m= 0.250 kg
Here spring potential energy converted into potential energy,
mgh = 1/2 kx to the 2 power
For finding at what height it rise,
0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power
h= 10.20
Therefore, the height at point of release is 10.20 m
Answer:
424.26 m/s
Explanation:
Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q
The relative speed will be calculated by using pythagorean theorem
Relative speed = sqrt(300^2 + 300^2)
Relative speed = sqrt( 180000 )
Relative speed = 424.26 m/s
Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s
