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a_sh-v [17]
2 years ago
15

Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the

argument for the sine is in radiaus. determine the largest speed of point a during the motion and the magnitude of the acceleration of point d at this instant.
Physics
1 answer:
storchak [24]2 years ago
8 0
<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s
 
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

ad= square root of adt^2+adn^2 = 288mm/s^2</span>
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A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha
Gelneren [198K]

Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

Explanation:

Given data

velocity v₀=20 cm/s at time t=3s

velocity vf=0 at time t=8 s

To find

Average Acceleration at time=3s to 8s

Solution

As we know that acceleration is first derivative of velocity with respect to time

a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\  a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\  a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

8 0
2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
2 years ago
If you have an effusive personality, then it __________ of you. motor oil will __________ across the gravel if it spills out of
erica [24]
The first answer in the blank is pours out of you, the second answer in the space provided is diffuse. It is because a person who has an effusive personality will likely pour out his or her emotions while the motor oil will likely diffuse if this has been spilled out of the can.
6 0
2 years ago
Weddell seals make holes in sea ice so that they can swim down to forage on the ocean floor below. Measurements for one seal sho
Mariulka [41]

Answer:

B. 1 m/s

Explanation:

Metric unit conversions:

0.3 km = 300m

5 minutes = 5*60 = 300 seconds

So if a seal can reach a depth of 300m in a time of 300 seconds, its diving speed is the distance divided by time duration

v = s/t = 300/300 = 1m/s

So B is the correct answer

3 0
1 year ago
Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note
love history [14]

Answer:

The energy is 9.4\times10^{-21}\ J

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

Q=\Delta U+W

Q=\Delta U+nRT

Put the value into the formula

4480=\Delta U+\dfrac{20}{28}\times8.314\times77

\Delta U=4480-\dfrac{20}{28}\times8.314\times77

\Delta U=4022.73\ J

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

N=n\times \text{Avogadro number}

Put the value into the formula

N=\dfrac{20}{28}\times6.02\times10^{23}

N=4.3\times10^{23}

We need to calculate the energy

Using formula of energy

E=\dfrac{\Delta U}{N}

Put the value into the formula

E=\dfrac{4022.73}{4.3\times10^{23}}

E=9.4\times10^{-21}\ J

Hence, The energy is 9.4\times10^{-21}\ J

4 0
1 year ago
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