The Force need to accelerate the object is by 3 m/s² is 15 N. Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N
<h3>Force:</h3>
This can be defined as the product of the mass and the acceleration of a body. The S.I unit of force is kgm/s or Newton(N)
To calculate the force needed to accelerate a mass of 5 kg object at 3 m/s² we use the formula below.
Formula:
- F = ma........ equation 1
Where:
- F = Net force needed to accelerate the object
- m = mass of the object
- a = acceleration of the object
From the question,
Given:
Substitute these values into equation 1
Suppose a frictional force of 5 N acts on the motion, The force applied is
- F = F'+ma............ Equation 2
Where:
- F = Frictional force = 5N
Substitute into equation 2
- F = 5(3)+5
- F' = 15+5
- F = 20 N.
Hence, The Force need to accelerate the object is by 3 m/s² is 15 N. Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N
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Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
- 
L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
the 4 resistor wired in parallel is equal to 23.946ohm
725 Kg equals 725x9.8= 7105 N
Answer:
Difference in experimental data.
Explanation:
There is difference of experimental value between the experiment that is performed on the earth and on the international space station because presence of gravity. The result of the experiment on the earth is different due to the presence of gravity that contributes in the result of the experiment as compared to international space station where no gravity is present so there is high difference of the numerical value of the result of both experiments of earth and international space station.
