Answer:
Explanation:
Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as
W = F×d
or
W =
.................................. Equation 1
Where b = upper limit, a = lower limit, Fx = expression of force.
<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>
Substituting these values into equation 1
<em>W =
</em>
W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ
Work = upper limit - lower limit
Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2
Substituting the values of a and b into equation 2
Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]
Work = [5.2 + 26.53 -3.29 + C] - C
Work = 28.44 J
Work done by the force = 28.44 J.
Answer:
Velocity
Explanation:
"The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s."
^^This explanation is from physicsclassroom.com
Hi!
The answer is <span>B. Language influences how people understand their world.
Hope this helps!
-Payshence xoxo</span>
Answer:
p = mv
m = p/v = 125000/22 = 5682 kg
Explanation:
Direct application of the momentum equation
p = mv
where,
p: momentum
m: mass
v: object velocity
steps:
-------
1) check for units consistency ( SI or Imperial)
2) separate the variable you are looking for.
3) DONE! :DD
Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis