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4 years ago

The temperature scale used in science is the _______ scale.

1 answer:

3 0

The temperature scale used in science is the Kelvin scale. The correct option among all the options given in the question is option "a". This is the standard scale used for measuring in astronomical and thermodynamics fields. The other scales can create complications as different countries use different type of scales.

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Identify two factors that determine the state of matter

vova2212 [387]

Two factors determine whether a substance is a solid, a liquid, or a gas: The kinetic energies of the particles (atoms, molecules, or ions) that make up a substance. Kinetic energy tends to keep the particles moving apart. The attractive intermolecular forces between particles that tend to draw the particles together.

4 0

3 years ago

Read 2 more answers

A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land

erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + $\frac{1}{2}$ at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = $\sqrt{\frac{2h}{g} }$ .

⇒ t = $\sqrt{\frac{2(1.3)}{9.81} }$ = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = $\frac{s}{t}$ .

so, u = $\frac{0.75}{0.51}$ = 1.45 m/s .

6 0

3 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

Pls help i give brainliest

san4es73 [151]

Answer:

Explanation:

cuz whenever the ball was traveling up the staircase it was building up energy to use (potential energy) unlike b.

5 0

3 years ago

Read 2 more answers

A spaceship whose rest length is 350m has a speed of .82c

igomit [66]

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

4 0

3 years ago