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marshall27 [118]
2 years ago
5

A crate slides off the back of a flatbed truck that accelerated forward from rest. If the bed of the truck is frictionless, the

crate falls off the back of the truck because
A. air friction pushes it backwards.
B. it moves forward with the truck but the truck goes too fast.
C. it is at rest and it tends to stay at rest.
D. it accelerates in the opposite direction of the truck.
Physics
1 answer:
jolli1 [7]2 years ago
5 0
The right answer should be B
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O professor Hosney, levou os alunos da segunda série ao laboratório para realizar um experimento. Pegou um recipiente de capacid
givi [52]

Answer:

The temperature beyond which the substance overflows the container is 86.23°C.

Explanation:

English Translation

Professor Hosney took the second grade students to the laboratory to perform an experiment. He took a 1000ml capacity container at a temperature of 68oF and poured 980 ml of a substance at 20oC into it. While placing the set to heat, he consulted a table where he found the volumetric expansion coefficient of the substance, 4 x 10-4 ºC-1 and the linear expansion coefficient of the container material, 3 x 10-5 ºC-1. Hosney then asked students to determine the temperature from which the substance would overflow. A student then asked, what is the melting temperature of the substance, and the teacher answered promptly 290.8 K. What is the temperature from which the substance will overflow?

Solution

The change in volume of a substance is given as

ΔV = γV₀(ΔT)

where

γ = coefficient of volume expansion

V₀ = Initial volume

(ΔT) = change in temperature.

At the temperature where the substance will overflow, the volume of the substance and the container will both be the same.

Let this temperature be T.

For the substance,

γ = coefficient of volume expansion = (4 × 10⁻⁴) °C⁻¹

V₀ = Initial volume = 980 mL

(ΔT) = change in temperature = (T - 20)

We will still leave ΔT as ΔT

ΔV₁ = (4 × 10⁻⁴) × 980 × ΔT

ΔV₁ = 0.392 ΔT

New volume of the substance at that temperature = V₀ + ΔV₁ = 980 + 0.392ΔT

For the container

γ = coefficient of volume expansion = 3 × coefficient of linear expansion = 3 × (3 × 10⁻⁵) °C⁻¹ = (9 × 10⁻⁵) °C⁻¹

V₀ = Initial volume = 1000 mL

(ΔT) = change in temperature = (T - 20) (note that 68°F = 20°C)

We will still leave ΔT as ΔT

ΔV₂ = (9 × 10⁻⁵) × 1000 × ΔT

ΔV₂ = 0.09 ΔT

New volume of the container at that temperature = V₀ + ΔV₂ = 1000 + 0.09 ΔT

At the temperature where overflow occurs, the two volumes are initially first the same.

980 + 0.392ΔT = 1000 + 0.09 ΔT

0.392ΔT - 0.09ΔT = 1000 - 980

0.302ΔT = 20

ΔT = (20/0.302) = 66.23°C

T - 20° = 66.23°

T = 66.23 + 20 = 86.23°C

The temperature beyond which the substance overflows the container is 86.23°C.

Hope this Helps!!!

8 0
3 years ago
A student measures out 5.0 g of magnesium and burns it. After the reaction, the student
kondor19780726 [428]

Answer:

The answer is 2.9. g

Explanation:

7.9 - 5.0 = 2.9

I hope this helps!

4 0
2 years ago
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on
Agata [3.3K]

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

   

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

3 0
3 years ago
Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the
MariettaO [177]

Answer:

(a) 0.115 m

(b) 2.08 x 10^-5 J

Explanation:

mass of bob, m = 81 g = 0.081 kg

The equation of oscillation is given by

θ = 0.068 Cos {9.2 t + Ф}

Now by comparison

The angular velocity

ω = 9.2 rad/s

(a) \omega^{2} =\frac{g}{L}

where, L be the length of the pendulum

L =\frac{g}{\omega ^{2}}

L =\frac{9.8}{9.2 \times 9.2}

L = 0.115 m

(b) A = L Sinθ

A = 0.115 x Sin 0.068

A = 7.8 x 10^-3 m

Maximum kinetic energy

K = 0.5 x mω²A²

K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6

K = 2.08 x 10^-5 J

8 0
3 years ago
A heavy boy and a lightweight girl are balanced on a massless seesaw. if they both move forward so that they are one-half their
Novay_Z [31]
The saw will tip towards the heavy boy. they are centered on the main pivot point so weight dosnt matter but as they go out. mass and weight will determin if the saw moves or not
7 0
3 years ago
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