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3 years ago

What is a possible state for an object in the absence of a net force?

Physics

2 answers:

igomit [66]3 years ago

8 0

Answer:

at rest
zero acceleration
constant speed

Hope this helps

insens350 [35]3 years ago

5 0

There is only one possible state: constant uniform motion. That means constant speed in a straight line.

(If the constant speed happens to be zero, this description also covers the case where the object isn't moving. That special case is called "at rest".)

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You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc

ahrayia [7]

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

3 0

3 years ago

Ariel's puppy weighs 15 lb. Heather's puppy weighs 50 N. Whose puppy weighs more?

yarga [219]

15 lb is nearly 6 kg amd 50 n is nearly 5 kg,so first puppy weighs more

5 0

3 years ago

Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien

otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s

The height at which the block stops rising is approximately 1.1415 m

The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

$Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m} = \mathbf{ 1000\, N/m}$

Coefficient of kinetic friction, $\mu_k$ = 0.39

Therefore, we have;

Friction force = $\mathbf{\mu_k}$ ·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, W ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

$\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d$

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, W ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

The velocity of the block after the Spring extends 7 cm, v₂ ≈ 1.73 m/s

The height at which the block will stop moving, h, is given as follows;

$At \ the \ maximum \ height, \ h, \ we \ have ; \ \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x$

Which gives;

$Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{ 7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2 } \approx 1.536$

The distance up the inclined, the block rises, at maximum height is therefore;

$x_{max}$ ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

The height at which the block stops rising, h ≈ 1.1415 m

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

Length of the incline, $x_{max}$ = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0

2 years ago

State two places where thermal conduction takes place.

Crank

Answer:

two places where thermal conduction takes place are gases and liquids, conduction is due to collisions of molecules during their random motion. Hence, the correct option is (C). Note: Though, the particle distances between gases are much more in comparison to solids and liquids, conduction slowly occurs in gases also

Explanation:

i hope it will help you

7 0

3 years ago

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

Mila [183]

Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.

5 0

3 years ago