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2 years ago

4

A cylindrical container is filled with oil to a depth of 32 cm if the pressure exerted at the base of the container is known to

be. 2560pa then how much is the density if the oil?

1 answer:

ikadub [295]2 years ago

6 0

Answer:

Ro = 815 [kg/m³]

Explanation:

The static pressure of any fluid can be calculated by means of the following expression.

$P=Ro*g*h\\$

where:

Ro = density [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = depth [m]

P = pressure [Pa]

Now we can calculate the depth

$2560=Ro*9.81*(32/100)\\Ro=815 [kg/m^{3} ]$

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All of the following are types of perceptual constancies except __________ constancy. A. direction B. size C. shape D. color

Oksi-84 [34.3K]

A direction can be a perceptual constancy, it's the same reason how blind folded, or people suffering from any kind of visual problem or psychological problem are able to make Thier right decision
size, shape, color are perceptual constancy
so the most correct option would DIRECTION

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2 years ago

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Why is the transverse spatial extent of a photon proportional to its wavelength which is a longitudinal quantity?

otez555 [7]

In quantum mechanics, particularly the wave-particle theory, it states that light behaves like a wave or a particle. For the wave behavior, its movement is measured in wavelengths while the time for each wavelength is the frequency. For the particle behavior, according to Planck, the energy of the photon (light particle) is determined as

E = hc/wavelength, where h is the Planck's constant (<span>6.626 x 10-34 J-s per particle) and c is the speed of light ( 3 x 10^8m/s)

As you can see, the energy of the photon is INVERSELY PROPORTIONAL to the wavelength with the Planck's constant as the constant of proportionality.</span>

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2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

$B(t)=5t\ T$ , t is time in seconds

$\dfrac{dB}{dt}=5\ T$

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$ , r is the radius of the solenoid

$r^2=\dfrac{2xE}{(dE/dt)}$

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So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

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2 years ago

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A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

$(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}$

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

$Sf=So+(a*t)$

$15m/s=So+1.67m/s^2*6s$

$15m/s-(1.67m/s^2*6s)=So$

$4.98m/s=So$

For part c:

We are going to use:

$Sf^2=So^2+(2*a*x)$

$(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)$

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$

5 0

2 years ago