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2 years ago

Which change of state do particles in a material become farther apart?

Physics

1 answer:

crimeas [40]2 years ago

6 0

Particles become farther apart from each other in the change of state between liquid and gas.
Hope that helped =)

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A baseball (m=140g) traveling 32m/s moves a fielders

posledela

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average force exerted by the ball on the glove. It is solved using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

F = mg

$\dfrac{1}{2}mv^2=Fh$

$F=\dfrac{mv^2}{2h}$

$F=\dfrac{0.14\times (32)^2}{2\times 0.25}$

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

7 0

2 years ago

Which planet is most likely to have acid rain? A. Mercury B. Venus C. Mars D. Uranus

timofeeve [1]

Its VENUS because the planet is basicily a hot planet

3 0

2 years ago

Graph are blank a relationship

Maru [420]

I have no clue sorry maybe try applying the answers and questions then I'll answer for u

7 0

2 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

kumpel [21]

Answer: $\frac{D_A}{v_B-v_A}$

Explanation:

Given

car A had a head start of $D_A$

and it starts at x=0 and t=0

Car B has to travel a distance of $D_A and d_a$

where $d_a$ is the distance travel by car A in time t

distance travel by car A is

$d_a=v_A\times t$

For car B with speed $v_B$

$d_B=D_A+d_a$

$v_B\times t=D_A+v_A\times t$

$t=\frac{D_A}{v_B-v_A}$

7 0

3 years ago