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3 years ago

Can some one help me :< its music

Physics

1 answer:

Bingel [31]3 years ago

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Which of the objects below has the greatest acceleration? *

lesya692 [45]

$\\ \bull\sf\dashrightarrow F=ma$

$\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}$

Now

$\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}$

$\\ \bull\sf\dashrightarrow a\propto F$

Lower mass=Higher acceleration
Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

8 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

An ideal gas in a piston is placed in a refrigerator, which removes 413 J of energy from the gas via cooling. The plunger on the

maria [59]

Answer:

The net change in the internal energy of the gas in the piston is -343J

Explanation:

Because heat and workdone are the only means of energy transfer between the system and the surrounding, change in internal energy is given by;

∆E = q + w

q = heat transfer

w = workdone

Because heat is lost by the system, the heat transfer is negative

q = -413J

Because work is done on the system, workdone is positive

w = +70J

∆E = -413J + 70J

∆E = -343J

5 0

3 years ago

With countercurrent flow, diffusion happened in all regions of the filter. Explain why

jeka94

Answer:

When the blood and the dialysate are flowing in the same direction, as the the dialysate and the blood move away from the region of higher concentration of the urea, to a region distant from the source, the concentration of urea in the blood stream and in the dialysis reach equilibrium and diffusion across the semipermeable membrane stops within the higher filter regions such as II, III, IV or V

However, for counter current flow, as the concentration of the urea in the blood stream becomes increasingly lesser the, it encounters increasingly unadulterated dialysate coming from the dialysate source, such that diffusion takes place in all regions of the filter

Explanation:

3 0

4 years ago

One object is thrown vertically upward with an initial velocity of 100 m/s and

gavmur [86]

We have that for the Question it can be said that The maximum height reached by the first object will be 100 times that of the other.

$(H_{max})_1=100*(H_{max})_2$

From the question we are told

One object is thrown vertically upward with an initial velocity of 100 m/s and another object with an initial velocity of 10 m/s. The maximum height reached by the first object will be

that of the other.

a. 10,000 times

b. none of these

c. 1000 times

d. 100 times

e. 10 times

Generally the equation for the velocity is mathematically given as

$v=\frac{d}{t}\\\\Where\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{(V_1)^2}{(v_2)^2}\\\\\frac{H_{max}_1}{H_{max}_2}=\frac{10000}{(10}\\\\$

$(H_{max})_1=100*(H_{max})_2$

Therefore

The maximum height reached by the first object will be 100 times that of the other.

$(H_{max})_1=100*(H_{max})_2$

For more information on this visit

brainly.com/question/23379286

8 0

3 years ago

Can some one help me :< its music​

Can some one help me :< its music