Answer: 10 m/s
Explanation: Velocity/Time
50/5= 10
:)
The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
Answer:
8.467 gm
Explanation:
The law governing this problem is "The Law of Constant Composition "
As per this law, all compounds irrespective of their origin and source have the same composition and properties in their purest form
It is a simple proportion and ratio related problem.
1.50 grams of carbon require 2.0 grams of oxygen
1.0 grams of carbon will require oxygen
=
6.35 grams of carbon will require oxygen
= 8.4666\\= 8.467
Answer:
7 / 1
Explanation:
The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength
Ax / Ay = λx / λy = 1 / 7
λy / λx = 7 / 1
Answer:
Explanation:
Electric field due to a point charge Q at a point at distance d is given by the relation
E = 
Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.
The charges are situated on the corners of a square in such a way that
equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.
On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero
Overall, net field due to all the four charges will be zero