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Tatiana [17]
3 years ago
15

Particle with the symbol ( - ) but it not negative is different mean

Physics
1 answer:
IrinaK [193]3 years ago
8 0
Neutron. and I need 20 more characters to post, so just ignore this.
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The pressure in a compressed air storage tank is 1200 kPa. What is the tank's pressure in (a) kN and m units, (b) kg, m, and s u
Sever21 [200]

Answer:

a) 1200 kN/m²

b) 1,200,000 kg/ms²

c) 1.2 × 10⁹ kg/km.s²

Explanation:

Given:

Pressure = 1200 kPa

a) 1 Pa = 1 N/m²

thus,

1000 N = 1 kN

1200 kPa = 1200 kN/m²

b)  1 Pa = 1 N/m² =  1 kg/ms²

Thus,

1200 kPa = 1200000 Pa

or

1200000 Pa = 1200000 × 1 kg/ms²

or

= 1,200,000 kg/ms²

c) 1 km = 1000 m

or

1 m = 0.001 Km

thus,

1,200,000 kg/ms² = \frac{1,200,000}{0.001}\frac{kg}{km.s^2}

or

= 1.2 × 10⁹ kg/km.s²

3 0
3 years ago
How do you define radiation (whats the definition for radiation?)​
stiv31 [10]

Answer:

<em>Radiation is energy that comes from a source and travels through space and may be able to penetrate various materials. Light, radio, and microwaves are types of radiation that are called nonionizing.</em>

Explanation:

7 0
3 years ago
When you stretch a spring, there is energy in the spring's __________ potential energy store. What word completes the sentence?
astraxan [27]

Answer:

Elastic potential energy store - which is the energy transferred to a spring as it's stretched.

5 0
3 years ago
A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0
3 years ago
NEED THIS DONE ASAP HELP
ExtremeBDS [4]

Answer: Exercise Physiology

Explanation:

4 0
3 years ago
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