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Elza [17]
3 years ago
8

Identical marbles are released from the same height on each of the following four frictionless ramps . Compare the speed of the

marbles at the end of each ramp. Explain your reasoning
Physics
1 answer:
Montano1993 [528]3 years ago
3 0

Since there is no friction on the ramp so there is no loss of energy in terms of frictional loss

So here we can use mechanical energy conservation law in this case

so here we will have

U_i = KE_f

here we know that

U_i = mgh

also we know that

K_f = \frac{1}{2}mv^2

now we will have

mgh = \frac{1}{2}mv^2

now by solving above equation we have

v = \sqrt{2gh}

so here as we know all identical balls are released from same height so all balls must have same speed at the end

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PV = nRT, is the a. equation of state of an ideal gas
romanna [79]

Ideal gas law:

PV = nRT

P = pressure, V = volume, n = # of moles, R = gas constant, T = temperature

Equipartition theorem:

Each degree of freedom that a molecule has adds 0.5kT to its total internal energy where k = Boltzmann's constant and T = temperature

2nd law of thermodynamics:

A set of governing principles that restrict the direction of net heat flow (always hot to cold, heat engines are never 100% efficient, entropy always tends to increase, etc)

Clearly the answer is Choice A

3 0
3 years ago
If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 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
Determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E].
sergeinik [125]

Answer:

3.44 metres

Explanation:

To determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E], resolve the given parameters into x - component and y - component.

Resolving into x - component

- 2.4cos32 + 4.9cos27 = 2.3306

Resolving into y - component

- 2.4sin32 - 4.9sin27 - 1.6 = - 2.553

The vector sum of the displacement will be

Sqrt( 2.3^2 + 2.6^2) =

Sqrt ( 11.81)

3.44 m

Therefore, the vector sum of the displacements is 3.44 metres

6 0
3 years ago
An infrared wave has a wavelength of 6.5 × 10–4 cm. what is this distance in angstroms, å?
Citrus2011 [14]
1 angstrom = 10^-8 cm 
6.5 x 10-4 cm  = 65 000 x 10-8 cm = 65 000 angstroms 
answer 65 000 angstroms
7 0
4 years ago
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