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puteri [66]
3 years ago
11

A force of 40N is applied to a 28 g mass, what is the acceleration? (round to the hundredths place)

Physics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

1428.6m/s²

Explanation:

Given parameters:

Force applied on the body  = 40N

Mass of the body  = 28g

                    1000g  = 1kg

                      28g will therefore be 0.028kg

Unknown:

Acceleration  = ?

Solution:

To solve this problem, we use the expression derived from Newton's second law of motion.

         Force  = mass x acceleration

Insert the parameters and solve;

            40  = 0.028 x acceleration

           Acceleration  = \frac{40}{0.028}   = 1428.6m/s²

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What is the charge on a hypothetical ion with 35 protons and 33 electrons?
kenny6666 [7]

+2 electron charges = 2x1.6x10^-19Coulombs

7 0
2 years ago
Marina is staring at an optical illusion where he sees a version of the American flag that is colored green, yellow, and black.
k0ka [10]

Explanation:

The reason why Marina sees the colour red, white and blue or the original colour of the American flag is that because of a phenomenon known as Afterimage. The retina in our eyes have mainly three receptors that are colour sensitive known as cones. These receptors can perceive the colour green, red and blue. Now when we look or stare at a particular colour for a long time, what happen is that our retina becomes tired and they ignore the colours that stared at. And now they work to form other colours at the retina just like the way when we produce other colour from the primary colour.

If the red receptor gets exhausted we will see the colour red. Likewise when we see the colour orange when we stare at the colour blue.

This explains the optical illusion of the American flag.

6 0
3 years ago
A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the
siniylev [52]

Answer:

K=58.8N/m

Explanation:

From the question we are told that:

Mass M=0.442

Drop distance d=0.150

Generally the equation for Spring Constant is mathematically given by

 K=\frac{2mg}{x}

 K=\frac{2*0.442*9.8}{1.150}

 K=58.8N/m

6 0
3 years ago
The magnetic field of a long, straight, and closely-wound solenoid, inside the solenoid at a point near the center, is 0.645 T.
Savatey [412]

Answer:

B'=1.935 T      

Explanation:

Given that

magnetic field ,B= 0.645 T

We know that magnetic filed in the solenoid is given as

B=\mu _0 n\ I

I=Current

n=Number of turn per unit length

μ0 =magnetic permeability

Now when the current increased by 3 factors

I'=3 I

Then the magnetic filed

B'=\mu _0 n\ I'

B'=\mu _0 n\ (3I)

B'=3 B

That is why

B' = 3 x 0.645 T

B'=1.935 T

Therefore the new magnetic filed will be 1.935 T.

3 0
3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
3 years ago
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