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Ksenya-84 [330]
2 years ago
6

A truck was carrying a substance in a tank. The molecules of that substance were moving away from each other. The truck parked o

vernight in a place where energy transferred out of the substance. In the morning, the substance was a gas. How were the molecules moving in the morning? Explain why the molecules were moving that way after energy was transferred out of them.
(Make it ask long as you need)
Or (short)
Physics
1 answer:
DochEvi [55]2 years ago
5 0

Answer:

In the morning the molecules were moving away from each other with a smaller speed than when the truck was carrying the substance.

Explanation:

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A cube of mass m1=9.6 kgm1=9.6 kg is sitting on top of a second cube of the same size and mass m2=0.8 kgm2=0.8 kg while both are
GalinKa [24]

Answer:N=0

Explanation:

Given

m_1=9.6 kg

m_2=0.8 kg

both blocks experiencing free fall so net weight of block during free fall is zero thus there is no normal reaction between them.

N=0

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3 years ago
Find the magnitude of the side and the angle. Please help!
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Answer:

Explanation:

<em>Assuming the triangle is a right triangle,</em>

the magnitude of R can be found using the Pythagorean theorem,

R = sqrt(9.7^2+6^2) = 11.41 m

The angle can be found by arctangent, which is

angle = atan(9.7/6) = 58.26 degrees.

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An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt
rusak2 [61]
If the velocity is constant then the acceleration of the object is zero.
a=0 (m/s^2)
Thus when we apply the equation
\Delta X=vt+(at^2/2)
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or equivalent
t=(\Delta X/v) =7500/278 =26.98 (seconds)

7 0
3 years ago
A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha
stira [4]

Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

T=\dfrac{mv^2}{r}+mg

\dfrac{T}{m}-g=\dfrac{v^2}{r}

v=\sqrt{(\dfrac{T}{m}-g)r}

v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

8 0
3 years ago
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0
3 years ago
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