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3 years ago

5. When an object reaches terminal velocity its acceleration is

Physics

2 answers:

Vsevolod [243]3 years ago

7 0

I think it is B. 4.9 m/s

nlexa [21]3 years ago

3 0

Acceleration is zero. ignore the above answer

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All of the following are good tips for controlling your emotions EXCEPT:

TEA [102]

Answer:

Explanation:

Building up emotions can be bad for your health and hurt you mentally all other options are good.

8 0

3 years ago

Read 2 more answers

Why do we call the microscope a compound light microscope

Ede4ka [16]

Explanation:

Types of light microscope

1. Compound , and 2. Stereo Microscope

Compound microscope has two lens system also called compound lens system. The objective lens and the eyepiece lens. The magnification provided by the objective lens is compounded by the eyepiece lens, the a higher magnification is observed.

6 0

3 years ago

Uma massa de 500 Kg desloca-se com velocidade 58 km por hora. Calcule o módulo de sua quantidade por movimento

Simora [160]

The momentum of the object is 8050 kg m/s

Explanation:

The momentum of an object is defined as

$p=mv$

where

p is the momentum

m is the mass

v is the velocity of the object

For the object in this problem, we have

m = 500 kg is its mass

v = 58 km/h is its velocity

Converting the velocity into m/s,

$v=58 \frac{km}{h}\cdot \frac{1000 m/km}{3600 s/h}=16.1 m/s$

Therefore now we can find the momentum of the object:

$p=(500)(16.1)=8050 kg m/s$

Learn more about momentum:

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3 years ago

Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di

olga nikolaevna [1]

<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

☆...hope this helps...☆

_♡_mashi_♡_

6 0

2 years ago

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

tiny-mole [99]

Explanation:

Charges, $q_1=8\ \mu C=8\times 10^{-6}\ C$

$q_2=-5\ \mu C=-5\times 10^{-6}\ C$

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N$

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0

3 years ago