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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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You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo

dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³ / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

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2 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

2 years ago

A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far d

AlekseyPX

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

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1 year ago

A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision

BigorU [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$f= -75 \ cm = - 0.75 \ m$

b

$P = -1.33 \ diopters$

Explanation:

From the question we are told that

The image distance is $d_i = -75 cm$

The value of the image is negative because it is on the same side with the corrective glasses

The object distance is $d_o = \infty$

The reason object distance is because the object father than it being picture by the eye

General focal length is mathematically represented as

$\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}$

substituting values

$\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}$

=> $f= -75 \ cm = - 0.75 \ m$

Generally the power of the corrective lens is mathematically represented as

$P = \frac{1}{f}$

substituting values

$P = \frac{1}{-0.75}$

$P = -1.33 \ diopters$

7 0

2 years ago

A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t

Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, $a_t$ = 12.7 m/s²

mass of the truck, $m_t$ = 2490 kg

mass of the car, $m_c$ = 890 kg

let the acceleration of the car at the moment they collided = $a_c$

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

$F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2$

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0

2 years ago