Answer:
BC and DE
Explanation:
In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.
Area under AB, 
Area under BC, 
Area under CD, 
Area under DE, 
Area under EF, 
So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".
The harmonic frequency of a musical instrument is the minimum frequency at which a string that is fixed at both ends in the instrument may vibrate. The harmonic frequency is known as the first harmonic. Each subsequent harmonic has a frequency equal to:
n*f, where n is the number of the harmonic and f is the harmonic frequency. Therefore, the harmonic frequency may be calculated using:
f = 100 / 2
f = 50 Hz
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:
Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
Acceleration=force/mass=28/(10+4)=2m/s^2
force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block
vf=vi+at
-10=20*28/14 * t
t=30/2=15sec
i hope this can help you.
