)
5
-5
1 2 3
4
5
Other than at t = 0, when is the velocity of
the object equal to zero?
1. 5.0 s
2. 4.0 s
3. 3.5 s
4. At no other time on this graph. correct
5. During the interval from 1.0 s to 3.0 s.
Explanation:
Since vt =
Z t
0
a dt, vt
is the area between
the acceleration curve and the t axis during
the time period from 0 to t. If the area is above
the horizontal axis, it is positive; otherwise, it
is negative. In order for the velocity to be zero
at any given time t, there would have to be
equal amounts of positive and negative area
between 0 and t. According to the graph, this
condition is never satisfied.
005 (part 1 of 1) 0 points
Identify all of those graphs that represent motion
at constant speed (note the axes carefully).
a) t
x
b) t
v
c) t
a
d) t
v
e) t
a
Answer:
A) Φ = 0
, B) T = 7.76 s
Explanation:
A) to find the value of the phase constant replace the value
0 = a sin (b (0- 0) + Φ)
0 = sin Φ
Φ = sin⁻¹ 0
Φ = 0
B) the period is defined by time or when the movement begins to repeat itself
So that the sine function is repeated when the angle passes 2pi
b (x- ct) = 2pi
If we are at a fixed point x = 0
b c t = 2pi
t = 2π / bc
Let's calculate
T = 2π / (33.05 245)
T = 7.76 s
Answer:
The answer is for your question is :
Explanation:
True
Answer: 321 J
Explanation:
Given
Mass of the box 
Force applied is 
Displacement of the box is 
Velocity acquired by the box is 
acceleration associated with it is 

Work done by force is 

change in kinetic energy is 

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
![\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%2BW_f%3D%5CDelta%20K%5Cquad%20%5BW_f%3D%5Ctext%7BWork%20done%20by%20friction%7D%5D%5C%5C%5C%5C%5CRightarrow%20375%2BW_f%3D54%5C%5C%5CRightarrow%20W_f%3D-321%5C%20J)
Therefore, the magnitude of work done by friction is 
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.