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
Answer:
The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).
Explanation:
Hi there!
The equation for the velocity of an object thrown upward is the following:
v = v0 + g · t
Where:
v = velocity of the ball.
v0 = initial velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
t = time.
To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:
v = v0 + g · t
v = 3.0 m/s - 9.8 m/s² · 0.40 s
v = -0.92 m/s
The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).
Let
be the average acceleration over the first 2.46 seconds, and
the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let
be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have


and we're also told that

(or possibly the other way around; I'll consider that case later). We can solve for
in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:


Now we can solve for
. We find that

In the case that the ratio of accelerations is actually

we would instead have

in which case we would get a velocity of

Honed I don’t know where the question is