Answer:
10 Mil. J
Explanation:
multiplyed balls and divided by 5
The force of static friction acting on the luggage =255 N
Explanation:
Applied force= 255 N
mass of luggage=68 kg
coefficient of friction= μ=0.7
Normal force=N = mg
N= 68(9.8)=666.4 N
now the force of friction is given by
Ff= μN
Ff=0.7 (666.4)
Ff=466.5 N which is greater than the applied force.
When the applied force is less than the force of friction, in that case
force of friction= applied force
Thus the force of station friction acting on the luggage = 255 N
Answer:
a.) 20 km east
b.) 50 km west
c.) 80 km west
Explanation:
As given,
Railway station - A B C
Distance(km) - 0 30 60
Starts from A , train reaches B be in 15 minutes
Starts from A , train reaches C be in 30 minutes
a.)
If A is origin
As train covers 30km = 15 minutes
⇒ 15 minutes = 30 km
1 minute =
km
⇒ 10 minutes = 2×10 = 20 km
∴ we get
The position of the train moving from station A to station B after 10 minutes = 20 km east
b.)
If B is origin then , the position of the train moving from station A to station B after 10 minutes = 30 + 20 = 50 km west
( Because B is the origin , so firstly train go from B to A in 15 minutes ( 30 km ) then A to B in 10 minutes (20 km) )
c.)
If C is the origin then , the position of the train moving from station A to station B after 10 minutes = 60 + 20 = 80 km west
( Because C is the origin , so firstly train go from C to A in 30 minutes ( 60 km ) then A to B in 10 minutes (20 km) )
As pressure increases, temperature must <span>increase</span> for water to remain in a gaseous state.