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zvonat [6]
3 years ago
8

If a block is in equilibrium the magnitude of the block's acceleration is

Physics
2 answers:
kaheart [24]3 years ago
3 0

zero.

from newton's first law of motion

Nostrana [21]3 years ago
3 0

Answer:

zero.

Explanation:

The condition for an object to be in equilibrium is that there is no resulting force acting on it, in other words, that the force on the object is zero:

F=0

and by Newton's second law we know that

F=ma

where m is the mass of the object, and a its acceleration.

So combining the two things:

F=ma=0

⇒ma=0, This equation is fulfilled when a=0

Wich tells us that for the the block to be in equilibrium, the acceleration must be zero:

a=0

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The carrot hangs from the ceiling by two ropes (1 and 2).
Vera_Pavlovna [14]

Answer:

The Answer is D!

Explanation:

I checked it on Khan Academy.

3 0
3 years ago
The diagram below shows the velocity vectors for two cars that are moving relative to each other.
scoundrel [369]

Answer:

The answer is "5 \ \frac{m}{s} \ west"

Explanation:

\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to  \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\

velocity of car | respect to car :

\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\

          =\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west

7 0
3 years ago
A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1
kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

\mu = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s

Frequency is given by

f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz

Angular frequency is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s

Maximum velocity of a particle is given by

v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0
3 years ago
What is the velocity of a wave with a wavelength of 1.5m and a frequency of 500 Hz
Schach [20]
Wave speed  =  (wavelength) x (frequency)

                       = (1.5 m) x (500 / sec)

                       =       750 m/s .  
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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