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Sedaia [141]
2 years ago
12

A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the

mass of the object is
Physics
2 answers:
pantera1 [17]2 years ago
8 0
Newton's second law states that Fnet = ma, where Fnet is the net force applied, m is the mass of the object, and a is the object's acceleration. You have the values for Fnet and a, so you simply use this equation to solve for m, mass.
Darina [25.2K]2 years ago
4 0

Answer:

m=5.2kg

Explanation:

A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the mass of the object is

from newton's second law of motion , which states that

the rate change in momentum of an object is directly proportional to the force applied

definition of terms  you should know.

a force is tat which tends to cane a body state of rest or uniform motion in a strait line

mass is the quantity of matter in a body

acceleration is change in velocity per time

therefore,

f∝m(V-U)/t

f=km(V-U)/t

k=1

f=m(V-U)/t

(V-U)/t=a

f=ma

f=125N

a=24m/s^2

m=5.2kg

the mass of the object

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Answer:

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Explanation:

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The width is given by

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The required width is 0.000003782 m

Minimum resolvable line separation is given by

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The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

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6 0
2 years ago
Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration
worty [1.4K]

Answer:

3433.5 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of person = 70 kg

According to the question

a = Acceleration

4g=4\times 9.81\\\Rightarrow a=39.24\ m/s^2

Balancing the forces we have

F-w=ma\\\Rightarrow F=ma+w\\\Rightarrow F=ma+mg\\\Rightarrow F=m(a+g)\\\Rightarrow F=70(39.24+9.81)\\\Rightarrow F=3433.5\ N

The required force is 3433.5 N

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QveST [7]

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While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is
erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.  

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0
1 year ago
The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is
Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

            W = ΔEm

Where the work is

             W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

           Em₀ = K = ½ m v²

Final. When the system is at rest

            Em_{f} = K_{e1}b +K_{e2} = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

            ∑ F = F - fr

Axis y

           N- W = 0

           N = W

The friction force has the equation

          fr = μ N

          fr = μ W

  The job

         W = (F – μ W) x

We substitute in the equation

            (F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

           ½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values ​​and solve

           ½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

         x² 30 + 14.4 x - 1,125 = 0

        x² + 0.48 x - 0.0375 = 0

           

We solve the second degree equation

        x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

        x = [-0.48 ± 0.617] / 2

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        x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0
2 years ago
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