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just olya [345]
3 years ago
7

As you move from left to right across the periodic table, what happens (in terms of metals/nonmetals, etc.)?

Physics
2 answers:
Andru [333]3 years ago
8 0

Answer:

Hope it helps;)

Have a good day

stepan [7]3 years ago
8 0
As you move from left to right across the periodic table, you start to leave the metals and run into the metalloids like boron and silicon. As you keep going right, you eventually run into the non metals like carbon, nitrogen, and oxygen. Metals and non metals are separated by a diagonal line.
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A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele
Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

3 0
3 years ago
Options are:<br>a)4Cn<br>b)5Cn<br>c)6 Cn<br>d)3 Cn<br>​
nasty-shy [4]

Answer:

Option B. 5 nC

Explanation:

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

Capicitance (C) = 100 pF

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Next, we shall determine the quantity of charge. This can be obtained as follow:

Capicitance (C) = 1×10¯¹⁰ F

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Q = CV

Q = 1×10¯¹⁰ × 50

Q = 5×10¯⁹ C

Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:

1 C = 1×10⁹ nC

Therefore,

5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C

5×10¯⁹ C = 5 nC

Thus, the quantity of charge is 5 nC

3 0
2 years ago
The magnetic field produced by a long straight current-carrying wire is
alexdok [17]

Answer:

proportional to the current in the wire and inversely proportional to the distance from the wire.

Explanation:

The magnetic field produced by a long, straight current-carrying wire is given by:

B=\frac{\mu_0 I}{2 \pi r}

where

\mu_0 is the vacuum permeability

I is the current intensity in the wire

r is the distance from the wire

From the formula, we notice that:

- The magnitude of the magnetic field is directly proportional to I, the current

- The magnitude of the magnetic field is inversely proportional to the distance from the wire, r

Therefore, correct option is

proportional to the current in the wire and inversely proportional to the distance from the wire.

8 0
3 years ago
What area of science compares and describes quantities and movements of matter and energy?
qaws [65]

Answer:

the answer is physics

4 0
3 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
3 years ago
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