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Triss [41]
2 years ago
6

One of the largest barometers ever built was an oil-filled barometer constructed in Leicester, England in 1991. The oil had a he

ight of 12.2 m. Assuming a pressure of 1.013 × 105 Pa, what was the density of the oil used in the barometer?
Physics
1 answer:
ZanzabumX [31]2 years ago
5 0

Answer:

ρ = 830.32 kg/m³

Explanation:

Given that

Oil head = 12.2 m

h= 12.2 m

Pressure P = 1.013 x 10⁵ Pa

Lets take density of the liquid =ρ

The pressure due to liquid P given as

P = ρ g h

Now by putting the all values in the above equation

1.013 x 10⁵ Pa = ρ x 10 x 12.2                 ( take g =10 m/s²)

ρ = 830.32 kg/m³

Therefore the density of oil is 830.32  kg/m³

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Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.
user100 [1]
Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.
4 0
2 years ago
Read 2 more answers
A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe
galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m,  y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}

The velocity v is given by:

\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}

The acceleration a:

\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}

From the given values we get two equations:

-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4

We also know:

\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}

The magnitude of the acceleration a is:

a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7

The magnitude of position r is:

r=R=68m

Plugging in to the equation for a(t):

\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}

and solving for ω:

|\omega|=0.66

Now solve for time t:

\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83

Using the calculated values to compute v(t):

\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}

The speed of the car is:

\sqrt{38.3^2 + (-23.5)^2} = 44.9

The position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}

5 0
2 years ago
Read 2 more answers
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C is the magnitude of the charge

\Delta V = 6.0 kV = 6000 V is the potential difference between point P and infinity

Substituting into the equation, we find

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
2 years ago
A 6000 kg lorry is reversing into a parking space at a speed of 0.5 m/s but collides with a car. The crumple zone of the car sto
zysi [14]

Answer:

3000 kg.m/s

Explanation:

Momentum, p is a product of mass and velocity hence

p=mv where m is mass and v is velocity.

Change in momentum is given by m(v_f-v_i) where subscripts f and i represent final and initial respectively. Since the lorry finally comes to rest then the final velocity is zero. Substituting the given figures then

Change in momentum= 6000(0-0.5)=-3000 kg.m/s

7 0
2 years ago
A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi
Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

δ=1.18cm

6 0
2 years ago
Read 2 more answers
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