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3 years ago

What describes the particles in a liquid

Physics

1 answer:

Gnoma [55]3 years ago

8 0

Answer:

liquid a particles slides past pother

Explanation:

mark brainliest :))

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Help !!!! Estimate the number of breaths taken by a person during 44 years?

Lisa [10]

44 x 12. I got the 12 from the total of 12 months in a year.

44 > 40

x
12 > 10
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The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.

40 x 10 = 400.

Therefore, your estimate is 400.

The real answer is 520 breaths.

6 0

2 years ago

Write down at least 3 equations as you solve them

Step2247 [10]

Answer:

3+7= 7x3= 21➗7=

Explanation:

3+7= 10 7 8 9 10

7x3= 21 7 14 21

21➗7=3 21 14 7

mark me brainliest please

4 0

1 year ago

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 years ago

A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s

9966 [12]

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

$L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m$

The velocity of the wave in the string is given by :

$v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s$

So, the velocity of the waves in the string is 72.54 m/s.

3 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

2 years ago