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3 years ago

A graduated cylinder should be read at the top of the meniscus. True False

2 answers:

8 0

-- If it's Mercury in the graduated cylinder, then the statement is true.

-- If it's water in the graduated cylinder, then the statement is false.

-BARSIC- [3]3 years ago

7 0

Answer:

Explained

Explanation:

A graduated cylinder should be read at the top of the meniscus or the down of the meniscus purely depends upon the kind of liquid used. Water gives a concave meniscus whereas mercury gives a convex meniscus. The measurement should account for the meniscus. Measurement should be taken such that line we are reading must be even with center of the meniscus. From the above explanation we can

If it's Mercury in the graduated cylinder, then the statement is true.

If it's water in the graduated cylinder, then the statement is false.

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A ball collides with a vertical, unmovable wall. There is no friction between the wall and the ball (the only force acting on th

user100 [1]

Answer: the same direction I.e to the left.

Explanation:

The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface will have the same speed (magnitude of velocity) as before the collision.

There will also be parallel force caused by friction, but it has to be treated separately for two reasons:

The perpendicular force is limited to coefficient of friction times the normal force. If that is not enough to stop the ball, it will skid on the surface.The perpendicular force, and this depends on the specific geometry, does not pass through the centre of mass of the ball. Therefore it imparts a moment on the ball that causes it to start rotating. And once the ball is rotating so that the point of contact is stationary, there is no momentum to cause any friction force anymore and the friction force disappears and stops decelerating the ball.

So what happens is that the vertical component of the velocity will be reversed, while the horizontal component will be somewhat reduced with the corresponding amount of kinetic energy transferred to energy of rotation. The rotation will always eliminate the friction force before the horizontal component of velocity is zeroed, so the ball will always continue in the same direction, just a bit slower.

If you instead threw an elastic box (which could not start rotating freely) it could actually bounce back.

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3 years ago

if a girl is standing in front of a smooth surface from which a sound is reflected, the girl may hear

Kamila [148]

The girl will hear an ECHO of the sounds that reach her,
(if she is standing more than about 50 feet from the wall).

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3 years ago

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Hii can someone please doo this! 50 pointss.

andrey2020 [161]

Answer:

The periodic table illustrate some of the elements from Hydrogen to Calcium

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3 years ago

Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

olga2289 [7]

<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

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3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$