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1 year ago

6

A substance occupies one half of an open container. The atoms of the substance are closely packed but are still able to slide pa

st each other.
What is most likely the phase of the substance?

gas
liquid
solid and gas
liquid and solid

1 answer:

Law Incorporation [45]1 year ago

3 0

Answer:

D. Liquid and Solid.

Explanation:

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A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at

Allisa [31]

Answer:

x = 1,185 m , t = 4/3 s , F = - 4 N

Explanation:

For this exercise we use Newton's second law

F = m a = m dv /dt

β - α t = m dv / dt

dv = (β – α t) dt

We integrate

v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

0 = β t - ½ α t²

t = 2 β / α

t = 2 4/6

t = 4/3 s

Let's find the distance at this time

v = dx / dt

dx / dt = v₀ + β t - ½ α t2

dx = (v₀ + β t - ½ α t2) dt

We integrate

x = v₀ t + ½ β t - ½ 1/3 α t³

x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

x = 3.5556 - 2.37

x = 1,185 m

The value of force is

F = β - α t

F = 4 - 6 4/3

F = - 4 N

8 0

1 year ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit

Reika [66]

The Answer is C Both at the same time

3 0

1 year ago

A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force

VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0

2 years ago

A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with

valentinak56 [21]

Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

7 0

2 years ago