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Hitman42 [59]
3 years ago
14

How would you describe the appearance of the substance after the phase change?

Physics
1 answer:
Leona [35]3 years ago
4 0

Answer:

Substances can change phase—often because of a temperature change. At low temperatures, most substances are solid; as the temperature increases, they become liquid; at higher temperatures still, they become gaseous. The process of a solid becoming a liquid is called melting.

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A racecar on a straight track, starting from rest*, steps on the
kirill115 [55]
I think the answer is C
6 0
3 years ago
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.​
docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

7 0
3 years ago
A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5
pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

v_w=v_s+v=1+1.5=2.5m/s

Distance=35 m

Time=\frac{distance}{speed}

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving=\frac{35}{v_w}=\frac{35}{2.5}=14s

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground=v_w=v_s+v=1-1.5=-0.5m/s

Time=\frac{-35}{-0.5}=70 s

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0
3 years ago
Friction is just hot air rubbing off on the inevitable force of solid<br><br>prove me wrong
algol13

Answer:

lol ur not wrong

Explanation:

8 0
3 years ago
Read 2 more answers
In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
2 years ago
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