Substance of an atom of 2 or more elements

A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant s

Monica [59]

Hey There,

Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"

Answer: C. Force Of Friction
B. Force

If This Helps May I Have Brainliest?</span>

7 0

3 years ago

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In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the

padilas [110]

The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.

4 0

3 years ago

How man significant figures does the following value gave 43.023

Varvara68 [4.7K]

Answer:

It has 5 I hope this helps you

7 0

3 years ago

A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.

Ainat [17]

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

$\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg$

We can simplify and rearrange the equation to solve for 'v'.

$\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}$

Plugging in values:

$v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}$

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

$\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg$

Rearranging for 'T":
$T = \frac{mv^2}{r} + mg\\\\$

Plugging in the appropriate values:
$T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}$

5 0

2 years ago

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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago