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3 years ago

Sodium has an atomic mass of 23 and an atomic number of 11, so the

Physics

1 answer:

horrorfan [7]3 years ago

5 0

Answer:

12 Neutrons

Explanation:

So the mass of sodium is 22.990. You round it up to get 23(as stated in the problem). So, <em>what exactly is atomic mass?</em>

Atomic Mass is the total amount of neutrons and protons added up to form a total mass. So when you subtract 23-11 you get 12 Neutrons.

<u>Tip: </u>Don't know if you need this but-

The neutrons and protons are typically close in number (unless it's an isotope). So say that you subtract and the numbers of protons and neutrons aren't close at all. Well if that's the case, it's probably wrong.

hope this helps!!

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A car is traveling at a speed of 37 m/s.

Vika [28.1K]

1. The speed in kilometers per hour (Km/h) is 133.2 Km/h

2. Yes, the speed is exceeding the 125 Km/h limit

<h3>How to convert 37 m/s to Km/h</h3>

From the question given above, the following data were obtained:

Speed (in m/s) = 37 m/s
Speed (in Km/h) =?

We can convert 37 m/s to kilometers per hour (Km/h) by doing the following:

1 m/s = 3.6 Km/h

Therefore,

37 m/s = 37 × 3.6

37 m/s = 133.2 Km/h

Thus, 37 m/s is equivalent to 133.2 Km/h

<h3>2. How to determine if the speed is exceeding the limit</h3>

Speed of car = 133.2 Km/h
Speed limit = 125 Km/h

From the above, we can see that the speed of the car is greater than the speed limit.

Thus, we can conclude that the speed of the car is exceeding the speed limit.

Learn more about conversion:

brainly.com/question/10893215

#SPJ1

3 0

2 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i

Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

5 0

4 years ago

A high-speed drill rotating ccw at 2400 rpm comes to a halt in 2.5 s. What is the magnitude of the drill’s angular acceleration?

WARRIOR [948]

Answer:

The magnitude of the drills angular acceleration is $-32\pi s^{-2}$ .

The drill makes 50 revolutions before it stops.

Explanation:

The revolutions that the drill makes in 1 second is

$\dfrac{2400\:rev}{60s} =40\:rev/s$

And the angular velocity is

$\omega= \dfrac{40 (2\pi )}{1s}$ (<em>one revolution is </em> $2\pi$ <em> radians)</em>

$\boxed{\omega =80\pi\:s^{-1}}$

Now, the drill comes to a halt in 2.5 s, which means the magnitude of it's angular acceleration is

$a= \dfrac{\Delta \omega}{\Delta s}$

$a=\dfrac{80\pi-0 }{0-2.5s}$

$\boxed{a=-32\pi \:s^{-2}}$

In other words, the magnitude of the angular acceleration is $-32\pi\: s^{-2}.$

Now, we find the angular displacement of the drill which is given by the equation

$\theta = \omega t +\dfrac{1}{2} \alpha t^2$

putting in $\omega = 80\pi s^{-1}$ , $\alpha = -32\pi s^{-2}$ , and $t=2.5s$ , we get:

$\theta = (80\pi s^{-1} *2.5s)+\dfrac{1}{2} (-32\pi s^{-2})(2.5s)^2$

$\theta = 100\pi$

which is

$\dfrac{100\pi }{2\pi } =50\:rev$ <em>(one revolution is </em> $2\pi$ <em> radians)</em>

50 revolutions.

In other words ,the drill makes 50 revolutions before it stops.

3 0

3 years ago

A syringe containing 1.54 ml of oxygen gas is cooled from 92.8 ∘c to 0.4 ∘c. what is the final volume vf of oxygen gas? (assume

Andrews [41]

Givens
=====
Ti = 92.8 oC = 92.8 + 273 = 365.8
T2 = 0.4 oC = 0.4 + 273 = 273.4

Vi = 1.54 mL
Vf = ?????

Formula
======
Vi/Ti = Vf / Tf
1.54 / 365.8 = Vf / 273.4

Calculations
=========
1.54 * 273.4 / 355.8 = Vf
vf = 1.15 mL

Comment
=======
if your teacher is concerned about sig digs, the answer is 1 mL. I'm not sure how to show that it is 1 sig dig. You can ask about this.

5 0

3 years ago