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3 years ago

How many protons are there in any te atom

Physics

1 answer:

Andreyy893 years ago

4 0

Answer:

Explanation:

Tellurium is the element of the periodic table with atomic number 52.

The atomic number of a chemical element represents the number of atoms contained in the nucleus of the atom: therefore, this means that an atom of tellurium contains exactly 52 protons in its nucleus.

Tellurium is classified as a metalloid, having properties in between metals and non-metals, and it appears with a silver color.

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In a weak acid solution, _____.

erastovalidia [21]

Answer:

Explanation:

8 0

3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago

What is the resistance of a circuit with a voltage of 10 V and current of 5A is?

wolverine [178]

Resistance = (voltage) / (Current)

Resistance = (10 V) / (5 A)

Resistance = 2 ohms.

5 0

4 years ago

Read 2 more answers

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experie

Harman [31]

Answer:

Explanation:

Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².

Now, the pilot is traveling in a circle of radius

r = 3340 m

And the speed is

v = 495 m/s

Then, acceleration?

The acceleration of a circular motion can be determine using centripetal acceleration

a = v² / r

a = 495² / 3340

a = 73.36 m/s².

Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully

4 0

3 years ago