Which formula is used to find fluctuation of the shape of body

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

4 0

2 years ago

What is the rotational speed of the second hand on a clock that measures the seconds?

pickupchik [31]

Explanation:

Below is an attachment containing the solution

5 0

3 years ago

two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

Tomtit [17]

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0

2 years ago

A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud

NARA [144]

<h2>Answer: 540 J</h2>

Explanation:

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (1)

In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work $W$ performed is the product of the force exerted to push the box $F=45N$ by the distance traveled $d=12m$ .