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Georgia [21]
3 years ago
10

Plsss answer this plss​

Physics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

1. stay away from the people and control the habits

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Consider the train car described in the previous part. Another experiment is conducted in it: A net force of 20N is applied to a
nordsb [41]

Answer:

No

Explanation:

The supplied information about the object and train is incomplete. Acceleration is the rate at which the velocity of a body changes with time. Here the velocity and time is not given

7 0
3 years ago
RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s
Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, R=50\ K\Omega=5\times 10^4\ \Omega

Capacitance, C=2.1\ \mu F=2.1\times 10^{-6}\ F

We need to find the expected time constant for this RC circuit. It can be calculated as :

\tau=R\times C

\tau=5\times 10^4\times 2.1\times 10^{-6}

\tau=0.105\ s

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0
3 years ago
You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its _____.
olga2289 [7]

Answer:

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>

Explanation:

By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.

This is given by :

F=m\ a

Where F represents force applied on the object , m represents mass of the object and a represents the acceleration.

In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.

4 0
3 years ago
Will give 30 points!!!
AysviL [449]
The answer is C I believe
8 0
3 years ago
Read 2 more answers
When the distance between two charges is halved, the electrical force between them?
Llana [10]
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
3 0
3 years ago
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