Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Answer:
4 m/
Explanation:
From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.
The Net force = Mass * Acceleration.
Since Net Force = 20 Newton, Mass = 5kg, therefore;
20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;
Acceleration = 
which gives 4.
Note: RHS means Right Hand Side.
LHS means Left Hand Side.
Answer:
Gamma rays
Explanation:
Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.
Answer: b) pointed toward and parallel to the member.
Explanation:
It is shown in the picture attached
Answer:
48.6°
Explanation:
The forward force, F equals the component of the weight along the slope.
So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.
So a = gsinθ
Since we are given that a = 75%g = 0.75g,
0.75g = gsinθ
sinθ = 0.75
θ = sin⁻¹(0.75)
= 48.6°
