Answer:
Mass remains constant but weight reduces
Explanation:
Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.
Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.
Therefore, for this case, since g decreases, the weight decreases but mass remains constant.
Answer:
44.6 N
Explanation:
Draw a free body diagram of the block. There are four forces on the block:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force F pulling right 30° above horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30° − mg = 0
N = mg − F sin 30°
Sum of forces in the x direction:
∑F = ma
F cos 30° − Nμ = 0
F cos 30° = Nμ
N = F cos 30° / μ
Substitute:
mg − F sin 30° = F cos 30° / μ
mg = F sin 30° + (F cos 30° / μ)
Plug in values:
mg = 20 N sin 30° + (20 N cos 30° / 0.5)
mg = 44.6 N
Answer:
An example in which liquid pressure phenomena can be used in daily life is in Water blasting
Explanation:
Water blasting refers application of pressurized water to remove materials from the surface of objects.
There are different varieties of water blasting, including;
Hydrocleaning; Cleaning enabled by the use of high pressure water
Hydrodemolition; Demolition or removal of concrete using pressurized water
Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.
Answer:
The magnitude of electric force is 
Explanation:
Coulomb's Law:
The force of attraction or repletion is
- directly proportional to the products of charges i.e
- inversely proportional to the square of distance i.e
[ k is proportional constant=9×10⁹N m²/C²]
There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C
Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C
and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C
Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.
If we draw a line from q₁ to Q .
The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.
Let hypotenuse = r
Therefore, 
we know,
Total force 
[ r=5]
N
The magnitude of electric force is
