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arsen [322]
3 years ago
15

Which forces are capable of affecting particles or objects from large distance

Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

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A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of
musickatia [10]
The frictional force is given by F = μmg 

<span>where μ is the coeficient of friction. </span>

<span>Work done by frictional force = Fd = μmgd </span>

<span>Kinetic energy "lost" = 1/2 mv² </span>


<span>Fd = μmgd = 1/2 mv² </span>

<span>The m's cancel μgd = v² / 2 </span>



<span>d = v² / 2μg </span>

<span>d = 8² / 2(0.41)(9.8) </span>

<span>d = 32 / (0.41)(9.8) </span>

<span>d = 7.96 </span>

<span>Player slides 8 m . </span>



<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>



<span>d = v² / 2μg </span>




<span>= 4² / 2(0.46)(9.8) </span>

<span>= 8 / (0.46)(9.8) </span>

<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
3 0
3 years ago
Read 2 more answers
Which of these common substances is a heterogeneous mixture?
LenaWriter [7]

Answer:

<em>(C) If the composition of a mixture appears uniform no matter where you sample it, is homogeneous; sand on a beach *IS HETEROGENEOUS* because when you look at it up close, you can identify different types of particles, such as sand, shells, and organic matter.</em>

Explanation:

<em>(A) Pure Water is a collection of solely H2O molecules therefore Pure Water is classified as a *Compound*.</em>

<em>(B) Table Salt is NOT a heterogeneous mixture because the particles of salt can't be separated, and it is a *Pure Substance*.</em>

<em>(D) Maple Syrup is a homogeneous mixture because the solutes are fully dissolved and not easily identified. In other words, Maple Syrup is uniform throughout.</em>

<em>-Hope this helps!</em>

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4 0
3 years ago
Can you answer this math homework? Please!
steposvetlana [31]
Using the count data and observational data you acquired, calculate the number of CFUs in the original sample
5 0
2 years ago
A tangent line drawn on a velocity-time graph has a rise of 19 m/s and a run of 4.0 m/s. How large is the acceleration? What typ
klemol [59]

Answer:

Acceleration = 4.8 m/s²

Explanation:

Given:

Change in velocity = 19 m/s

Change in time = 4 s

Find:

Acceleration

Computation:

Acceleration = Change in velocity / Change in time

Acceleration = 19/4

Acceleration = 4.8 m/s²

Positive acceleration

8 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
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