Answer:

Explanation:
given data:
density of water \rho = 1 gm/cm^3 = 1000 kg/m^3
height of water = 20 cm =0.2 m
Pressure p = 1.01300*10^5 Pa
pressure at bottom



= 1.01300*10^5 - 1000*0.2*9.8
= 99340 Pa
h_[fluid} = 0.307m


m = mass of the birdcage = 22.5 kg
F = net force acting on birdcage to move it = 140 N
a = acceleration produced due to the force applied
a)
Using newton's second law
a = F/m
inserting the values
a = 140/22.5
a = 6.22 m/s²
b)
t = time of travel of crate = 10.5 s
v₀ = initial velocity of the crate = 0 m/s
X = displacement of the crate
displacement of the crate is given as
X = v₀ t + (0.5) a t²
X = 0 (10.5) + (0.5) (6.22) (10.5)²
X = 342.88 m
Answer:
The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.
~Hoped this helped~
~Brainiliest?~
Answer:
2697.75N/m
Explanation:
Step one
This problem bothers on energy stored in a spring.
Step two
Given data
Compression x= 2cm
To meter = 2/100= 0.02m
Mass m= 0.01kg
Height h= 5.5m
K=?
Let us assume g= 9.81m/s²
Step three
According to the principle of conservation of energy
We know that the the energy stored in a spring is
E= 1/2kx²
1/2kx²= mgh
Making k subject of formula we have
kx²= 2mgh
k= 2mgh/x²
k= (2*0.01*9.81*5.5)/0.02²
k= 1.0791/0.0004
k= 2697.75N/m
Hence the spring constant k is 2697.75N/m
Answer:
c
Explanation:
neither the spring or hands are in the action of movemnet