Hi can you help me with this please?

. Boxes are sitting on a conveyor belt as the conveyor is turned on, moving the boxes toward the right. The belt reaches full sp

oksian1 [2.3K]

Answer:

The acceleration of the boxes is 1.5 ft/s²

The displacement of the boxes during the speed-up period is 0.1875 ft.

Explanation:

Hi there!

Let´s convert the 45 ft/min into ft/s:

45 ft/min · 1 min/ 60 s = 0.75 ft/s

It takes the belt 0.5 s to reach this speed. Then, the acceleration of the boxes will be:

a = v/t

Where:

a = acceleration.

v = velocity.

t = time.

a = 0.75 ft/s / 0.5 s

a = 1.5 ft/s²

The acceleration of the boxes is 1.5 ft/s²

The equation of displacement is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the boxes at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the point where the boxes begin to move, x0 = 0. Since the boxes were initially at rest, v0 = 0. Then:

x = 1/2 · a · t²

x = 1/2 · 1.5 ft/s² · (0.5 s)²

x = 0. 1875 ft

The displacement of the boxes during the speed-up period is 0.1875 ft.

5 0

3 years ago

What subatomic particle has a nuetral charge?

kirill [66]

An atom is a particle with a neutral charge

6 0

4 years ago

Read no.4<br><br> which newtons law is this?<br><br> 1,2 and 3 only

horsena [70]

Explanation:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

8 0

2 years ago

Two wires are made from the same material. One wire has a resistance of 0.10 ω. The other wire is twice as long as the first wir

Anna71 [15]

Answer:

Explanation:

Given that,

First wire has a resistance of

R1 = 0.1 Ω

We are told that

Second wire is twice as long as the first wire.

Then,

Let the first wire has length

L1 = x

Then, second wire will have

L2 = 2x

Also, the radius of the second wire is half the radius of the first wire

Let the first wire has radius

r1 = y

Then, it area is

A1 = πr1² = πy²

Then, the second wire has radius

r2 = ½y

It area is also

A2 = πr2² = π(½y)² = ½πy²

Since the wire is made of the same material, then, they will have the same resistivity ρ

Then, we want to find the resistance of the second wire

Using

R = ρL/A

Where

R = resistance

ρ = sensitivity

L = length

A = area

Then, make resistivity subject of formula since it is a constant

RA = ρL

Then, ρ = RA/L

Then,

R1 • A1 / L1 = R2 • A2 / L2

Substituting each value

0.1 × πy² / x = R2 × ¼πy² / 2x

Cross multiply

0.1 × πy² × 2x = R2 ×¼π y² × x

Then,

R2 = 0.1 × πy² × 2x / ¼πy² × x

R2 = 0.1 × 2 / ¼

R2 = 0.8Ω

The resistance of the second wire is 0.8Ω

6 0

4 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago