Answer:
A:1.94
Explanation:
cause that the only one on there
B. Hand-eye Coordination because it can be used in multiple activities
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
Answer:
The final velocity of the object is 330 m/s.
Explanation:
To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:
F = ma
If we plug in the values that we are given in the problem, we get:
42 = 7 (a)
To solve for a, we simply divide both sides of the equation by 7.
42/7 = 7a/7
a = 6 m/s^2
Next, we should write out all of the information we have and what we are looking for.
a = 6 m/s^2
v1 = 0 m/s
t = 55 s
v2 = ?
We can use a kinematic equation to solve this problem. We should use:
v2 = v1 + at
If we plug in the values listed above, we should get:
v2 = 0 + (6)(55)
Next, we should solve the problem by performing the multiplication on the right side of the equation.
v2 = 330 m/s
Therefore, the final velocity reached by the object is 330 m/s.
Hope this helps!
Weight is equivalent to the product of the mass of an object and the strength of the gravitational field.
Using:
F = ma
a = 8.2 / 5
a = 1.64 N/kg
The gravitational field strength is equivalent to 1.64 N/kg.
