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3 years ago

Witch will delsove faster 1 gram of sugar or 1 gram of salt

Physics

1 answer:

natka813 [3]3 years ago

4 0

Answer:

Sugar dissolves faster than salt

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A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

3 years ago

Given that: = 2i + 9j ; and ⃗ = -i – 4j . Find . ⃗

Marat540 [252]

Answer:

A.B = -38

Explanation:

A = 2i + 9j and B = -i - 4j.

So, A.B = (2i + 9j).(-i - 4j)

= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)

= -2i.i - 8i.j - 9j.i - 36j.j

since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have

A.B = -2(1) - 8(0) - 9(0) - 36(1)

A.B = -2 - 0 - 0 - 36

A.B = -38

5 0

3 years ago

Iwill need to use more force to stopa O Lighter mass O Heavier mass

trasher [3.6K]

Answer:

a heavier mass

Explanation:

7 0

3 years ago

As a student in a physics lab, you are asked to classify two springs. Right away you notice some differences. The first spring,

strojnjashka [21]

Answer:

C. The spring constant of spring 2 is larger than spring 1.

Explanation:

As we know that spring constant also called the stiffness constant of a spring is the value of force required for the deformation of the spring by a unit length.

Mathematically given as:

$k=\frac{F}{\delta x}$

where:

$k=$ stiffness constant

$F=$ force on the spring

$\delta x=$ change in length of the spring

It is given in the question that spring 1 is easily compressible and requires less force to compress as compared to spring 2 and hence it has lesser stiffness constant than that of spring 2.

3 0

4 years ago

Robin fired a bullet of mass 100 gm from a gun of mass 5 kg. The bullet leaves the gun with a speed of 400 m/s. After penetratin

Marina CMI [18]

(a) the recoil or backward velocity of the gun is 8 m/s.

(b) the bullet cannot completely penetrate the plank.

The given parameters include;

mass of the bullet, m₁ = 100 g = 0.1 kg

mass of the gun, m₂ = 5 kg

initial velocity of the bullet, u₁ = 400 m/s

thickness of the plank, x = 10 cm

To calculate the following:

(a) the backward velocity of the gun

let the backward velocity of the gun = u₂

Apply the principle of conservation of linear momentum.

m₁u₁ + m₂u₂ = 0

m₂u₂ = -m₁u₁

$u_2 = -\frac{m_1u_1}{m_2} \\\\u_2 = - \frac{0.1 \times 400}{5} \\\\u_2 = -8 \ m/s$

Thus, the recoil or backward velocity of the gun is 8 m/s.

(b) Can the bullet penetrate the plank of the wood completely ?

the bullet traveled 4 cm and lost ¹/₃ of u₁

the remaining distance to completely penetrate the plank = 6 cm

the final velocity of the bullet at 4 cm, v = 400 - ¹/₃ x 400 m/s = 266.67 m/s

the acceleration of the bullet is calculated as;

v² = u₁² + 2as

2as = v² - u²

$a = \frac{v^2 -u_1^2}{2s} \\\\a = \frac{(266.67)^2 -(400)^2}{2\times 0.04} = -1.111 \times 10^6 \ m/s^2$

Finally, determine the distance traveled by the bullet when it comes to a complete stop, that is the final velocity = 0

$v_f^2 = v^2 + 2ad\\\\2ad = v_f^2 - v^2\\\\d = \frac{v_f^2 - v^2}{2a} \\\\d = \frac{(0) - (266.67)^2}{2(-1.111\times 10^6)} \\\\d = 0.032 \ m$

d = 3.2 cm

The total distance traveled by the bullet inside the plank = 4 cm + 3.2 cm = 7.2 cm

Therefore, the bullet cannot completely penetrate the plank.

To learn more about linear momentum visit: brainly.com/question/15869303

6 0

3 years ago