Ask question

Ask question

All categories

3 years ago

10

Signs that say "do not use near an open flame" refer to a(n) A) energy property. B) chemical property. C) physical property. D)

mechanical property.

1 answer:

sveta [45]3 years ago

8 0

Answer: B chemical

Explanation: it will make a chemical reaction

You might be interested in

How did planck find the correct curve for the specturm of light emitted by a hot obkect?

Neko [114]

Planck find the correct curve for the specturm of light emitted by a hot object by vibrational energies of the atomic resonators were quantized.

<h3>Briefing :</h3>

The energy density of a black body between λ and λ + dλ is the energy E=hc/λ of a mode times the density of states for photons, times the probability that the mode is occupied.
This is Planck's renowned equation for a black body's energy density.
According to this, electromagnetic radiation from heated bodies emits in discrete energy units or quanta, the size of which depends on a fundamental physical constant (Planck's constant). The basis of infrared imaging is the correlation between spectral emissivity, temperature, and radiant energy, which is made possible by Planck's equation.

Learn more about the Planck's constant with the help of the given link:

brainly.com/question/27389304

#SPJ4

3 0

1 year ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10

Ilya [14]

The solution is in the attachment

4 0

3 years ago

Read 2 more answers

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago

Explanation for question i & ii. Thank you.

suter [353]

Do u know Chinese ?if yes I can explain to u easily

6 0

3 years ago