All categories

3 years ago

When a liquid has vapor pressure equal to atmospheric pressure, it?

Physics

2 answers:

Olenka [21]3 years ago

5 0

When vapor pressure equals atmospheric pressure, it's called boiling point of that liquid.

galina1969 [7]3 years ago

5 0

<span>When a liquid has vapor pressure equal to atmospheric pressure, It boils. So the temperature is the boiling point.

Hope it helps!
</span>

You might be interested in

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

Mekhanik [1.2K]

Answer:

$v\approx 16.956\,\frac{m}{s}$

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

$\Sigma F = f = m\cdot \frac{v^{2}}{R}$

The maximum speed is:

$v = \sqrt{\frac{f\cdot R}{m} }$

$v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }$

$v\approx 16.956\,\frac{m}{s}$

7 0

3 years ago

I’ll give brainliest and 10 points

schepotkina [342]

I think it is D. I hope this helps

4 0

3 years ago

Dust settling out of the air<br> chemical change or physical change

mars1129 [50]

Physical changes: melting, evaporating, and condensation. This is a physical change.

7 0

3 years ago

Read 2 more answers

A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is

tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

$v_x = 4.8 m/s$

now the acceleration is given as

$\vec a = -3.2 \hat i + 4.6 \hat j$

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

$v_f = v_i + at$

$0 = 4.8 - 3.2 t$

$t = 1.5 s$

Part a)

the velocity of the particle at this moment in Y direction is given as

$v_f_y = v_i + at$

$v_f_y = 0 + 4.6(1.5)$

$v_f_y = 6.9 m/s$

Part b)

X coordinate of the particle at this time

$x = v_x t + \frac{1}{2}a_x t^2$

$x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)$

$x = 3.6 m$

Y coordinate of the particle at this time

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)$

$y = +5.175 m$

so position is given as (3.6 m, 5.175 m)

5 0

3 years ago

A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the

ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0

3 years ago