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3 years ago

Which would hold more water- a teaspoon (tsp) or a milliliter?

Physics

2 answers:

ki77a [65]3 years ago

4 0

A tsp

1 milliliter equals 0.202. US teaspoons. boi did that help

salantis [7]3 years ago

3 0

Answer:

A teaspoon

Explanation:

As we know that the capacity of hold something by a standard teaspoon is related with milliliters as,

1 teaspoon is equivalent to 4.92892 milliliters

From this it can be see that the 1 teaspoon is equivalent to approximately 5 milliliters.

Therefore, the water hold by 1 teaspoon is more than the water hold by a milliliter.

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C.convection cells is the right answer

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In general, court in the United States have repeatedly struck down efforts to include creationism or intelligent design in publi

OverLord2011 [107]

Because they are not supported by the results of any legitimate investigation
that's conducted in accordance with the Scientific Method.

You may say:
"Well then, teach both lines of reasoning,
and let each student decide for himself."

This is suggested by the same people who aren't ready to let their
fourth-grader choose his own clothing, dinner menu, or school.
And it sounds reasonable to a vast mass of citizens who have decided
for them selves that the jury is still out on climate change.

What I'm saying is this:

-- The Scientific Method is a METHOD of investigation that's designed
and developed to remove the effects of human prejudice from the
collection and evaluation of evidence, and to be able to tell bogus
conclusions apart from true ones. It's the most reliable way we have
of asking and answering questions about the natural world.

-- Some questions CAN'T be studied with the Scientific Method,
because experiments generally can't be constructed. These include
matters of religion and faith. Nobody can flatly state that those are
right or wrong. We have no reliable way to say, either way.
The only way to decide is . . . faith.

-- It is illegitimate to take the answer to a question of faith that can't be
derived scientifically, and a scientifically derived conclusion, set them
down next to each other on the same table, and pretend that they can be
compared.

-- When you put them next to each other, say that they're equivalent,
and tell people "go ahead and choose one or the other", the situation
is bogus, the comparison is dishonest, and people who are untrained
or uneducated or immature are not qualified to "choose".

That's why.

This is my opinion. I could be wrong.

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8 0

3 years ago

A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block tha

Anna007 [38]

Answer:

v = 2.82 m/s

Explanation:

For this exercise we can use the conservation of energy relations.

We place our reference system at the point where block 1 of m₁ = 4 kg

starting point. With the spring compressed

Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂

final point. When block 1 has descended y = - 0.400 m

Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y

as there is no friction, the energy is conserved

Em₀ = Em_f

½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y

½ k x² - m₁ g y = ½ m₂ v²

v² = $\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y$

let's calculate

v² = $\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)$

v² = 2.7 + 5.23

v = √7.927

v = 2,815 m / s

using of significant figures

v = 2.82 m/s

5 0

3 years ago

A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle

dalvyx [7]

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N\dfrac{d\phi}{dt}$

$\dfrac{d\phi}{dt}$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA\dfrac{dB}{dt}\times cos30$

$\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

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4 years ago

A person with mass mp = 75 kg stands on a spinning platform disk with a radius of R = 1.59 m and mass md = 186 kg. The disk is i

Ad libitum [116K]

Answer:

1. $I_0=424.7208\,kg.m^2$

2. $I_f=256.1808\,kg.m^2$

3. $\omega_f=3.3158\,rad.s^{-1}$

4. $\Delta KE=558.8432\,J$

5. $a_c=5.8271\,m.s^{-2}$

6. $\omega_0=2\,rad.s^{-1}$

Explanation:

Given:

mass of the person, $m_p=75\,kg$

radius of the disk, $r=1.59\,m$

mass of the disk, $m_d=186\,kg$

initial angular speed of the disk, $\omega_0=2\,rad.s^{-1}$

distance of the person from the center of the disk, $d=0.53\,m$

Initial moment of inertia of the system when the man stands at the rim of disk:

Moment of inertia of the disc:

$I_D=\frac{1}{2} .m_d.R^2$

$I_D=\frac{1}{2} \times 186\times 1.59^2$

$I_D=235.1133\,kg.m^2$

Now for the person, we treat the mass to be a point revolving around R:

$I_P=m_p.R^2$

$I_P=75\times 1.59^2$

$I_P=189.6075\,kg.m^2$

∴We have the moment of inertia of the system in this case as:

$I_0=I_D+I_P$

$I_0=235.1133+189.6075$

$I_0=424.7208\,kg.m^2$

Moment of inertia when the person stands at 0.53 m from the center of the disk:

Moment of inertia of the disk will be constant:

$I_D=235.1133\,kg.m^2$

For the person, we treat the mass to be a point revolving around radius 0.53 m:

$I_P=75\times 0.53^2$

$I_P=21.0675\,kg.m^2$

∴We have the moment of inertia of the system

$I_f=I_D+I_P$

$I_f=235.1133+21.0675$

$I_f=256.1808\,kg.m^2$

The final angular velocity of the disk:

Using the conservation of angular momentum:

$I_0.\omega_0=I_f.\omega_f$

$424.7208\times 2=256.1808\times \omega_f$

$\omega_f=3.3158\,rad.s^{-1}$

Change in Kinetic Energy:

∵ $KE=\frac{1}{2} I.\omega^2$

∴ $\Delta KE= \frac{1}{2} (I_f.\omega_f^2-I_0.\omega_0^2)$

$\Delta KE=\frac{1}{2} (256.1808\times 3.3158^2-424.7208\times 2^2)$

$\Delta KE=558.8432\,J$

Centripetal acceleration of the person when she is at R/3:

Centripetal acceleration is given as:

$a_c=r'.\omega^2$

we have ω=3.3158 radian per second at R=0.53 m

$a_c=\frac{R}{3} .\omega^2$

$a_c=0.53\times 3.3158^2$

$a_c=5.8271\,m.s^{-2}$

If the person now walks back to the rim of the disk:

Then by the law of conservation of angular momentum the initial angular speed of $\omega_0=2\,rad.s^{-1}$ will be restored.

8 0

3 years ago