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3 years ago

7

A bus is moving with an acceleration of 4 m/s^2. If it was initially at rest,

2 answers:

shusha [124]3 years ago

6 0

Probably b) 6 m/s
Not sure

AlladinOne [14]3 years ago

6 0

Answer is a because it is duh

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 4.0 rev/s in 9.0 s. At thi

iris [78.8K]

Answer:

48 rev

Explanation:

a) we can calculate the distance covered by the tube using the formula:

θ = (ω + ωo)t/2

where ω is the final angular speed, θ is the distance covered, t is the time taken, ωo is the initial angular speed.

Firstly, we calculate the distance covered from 0 to 9 s then from 9s to 24 s.

within 9s, the tub runs from rest (0) to 4 rev/s, hence:

t = 9s, wo = 0, w = 4 rev/s = (4 * 2π) rad/s = 8π rad/s. Hence:

θ = (ω + ωo)t/2 = (0 + 8π)9 / 2 = 36π rad

θ = 36π rad = (36π)/2π rev = 18 rev

Also, within 15 s, the tub runs from 4 rev/s to rest, hence:

t = 15 s, wo = 4 rev/s = 8π rad/s, w = 0 rad/s. Hence:

θ = (ω + ωo)t/2 = (8π + 0)15 / 2 = 60π rad

θ = 60π rad = (60π)/2π rev = 30 rev

Therefore the total revolutions by the tube during 24 s interval = 30 rev + 18 rev = 48 rev

4 0

2 years ago

What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?

likoan [24]

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

$Q=mL$ , here m is mass of ice and L is latent heat of fusion

So heat $Q=mL=0.4535\times 334=151.469kj$

So heat required to melt 1 lb of ice is equal to 151.469 KJ

3 0

3 years ago

Which of the following is the best example of kinetic energy being transformed into potential energy

poizon [28]

A pendulum is probably the most common showing of this example. As the pendulum swings down, it converts its potential energy (height) into kinetic energy (velocity). At the lowest point the kinetic energy is the highest and the potential is the lowest. At the highest point in its swing the velocity is zero so the kinetic energy is zero and the potential energy is at a maximum (greatest height).

3 0

3 years ago

Read 2 more answers

Why are wheels and steam engine important in the history of transport

STALIN [3.7K]

Explanation:

Because it's a power to turn the wheels of industry

5 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago