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3 years ago

A. Hydrogen B. Sodium C. Neon D. Magnesium

Physics

2 answers:

solmaris [256]3 years ago

6 0

Answer:

It’s B/sodium

:)

umka2103 [35]3 years ago

4 0

The answer is B. Sodium because it has 11 proton

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The distance between two consecutive nodesof a standing wave is 20.9cm.Thehandgen-erating the pulses moves up and down throughac

irina1246 [14]

Answer:

Velocity, v = 0.239 m/s

Explanation:

Given that,

The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m

The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.

For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

$\dfrac{\lambda}{2}=0.209\ m\\\\\lambda=0.418\ m$

Frequency is number of cycles per unit time.

$f=\dfrac{2.57}{4.47}\\\\f=0.574\ Hz$

Now we can find the velocity of the wave.

Velocity = frequency × wavelength

v = 0.574 × 0.418

v = 0.239 m/s

So, the velocity of the wave is 0.239 m/s.

4 0

3 years ago

For most substances, the distance between particals is smallest when the substance

mamaluj [8]

Its smallest when the substance is a gas

7 0

3 years ago

Read 2 more answers

The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

5 0

3 years ago

PLEASE ANSWER ASAP

Mekhanik [1.2K]

Answer:

yes h h jj h v gu i kj

Explanation:

jjejejejejj

8 0

3 years ago

Calculate the work done (in J) by a 90.0 kg man who pushes a crate 4.25 m up along a ramp that makes an angle of 20.0° with the

Mrrafil [7]

Answer:

W = 3.4x0³ J.

Explanation:

The work done by the man is given by the following equation:

$W = F_{t}\cdot d$ (1)

<em>where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.</em>

We need to find first the total force Ft, which is:

$Ft = Fm + W$

<em>where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°. </em>

$F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N$

Hence, the work is:

$W = 802.0 N \cdot 4.25 m = 3.4 \cdot 10 ^{3} J$

Therefore, the work done by the man is 3.4x10³ J.

I hope it helps you!

8 0

3 years ago