All categories

3 years ago

Someone pls answer! i will give u brainliest

Physics

2 answers:

serious [3.7K]3 years ago

5 0

I think it’s liquid.

olya-2409 [2.1K]3 years ago

4 0

Answer:

Liquid

Explanation:

Before anything evaporates, it is in a liquid state.

You might be interested in

The probability that a battery will last 10 hr or more is 0.8, and the probability that it will last 15 hr or more is 0.11. Give

podryga [215]

We are given with the data that says the probability that a battery will last 10 hr or more is 0.8 and the probability <span>that it will last 15 hr or more is 0.11. In this case, the probability that the battery lasts at least 10 hours and even 15 hrs more is 0.11 / 0.8 or equal to 13.75 percent.</span>

5 0

2 years ago

In an open system such as a campfire matter can

eimsori [14]

In an open system such as a campfire, matter can lose particles, gain particles or exchange particles.

4 0

3 years ago

A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

$v=at+v_o$ where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

$v=at+v_o$

$(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})$

$1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t$

$\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}$

$10\text{ [s]}=t$

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is $a=\dfrac{100 [\frac{m}{s}]}{1[s]}$ , the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0

2 years ago

During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.

Allushta [10]

Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0

2 years ago

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

$d_1 = v* t$

$d_1 = 18* 0.200 = 3.6 m$

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

$v_f^2 - v_i^2 = 2 a s$

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

$0^2 - 18^2 = 2*(-3.35)(s)$

$s = 48.35 m$

So total distance after which car will stop is

$d = d_1 + d_2$

$d = 48.35 + 3.6 = 51.95 m$

So car will not stop before the intersection as it is at distance 20 m

3 0

3 years ago