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3 years ago

It’s not what happens to you. It’s

Physics

2 answers:

Sphinxa [80]3 years ago

7 0

Answer:

B.how other people feel about it.

Explanation:

Rudiy273 years ago

3 0

Answer:

It's not what happens to you. It's A.) how you face it

Explanation:

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PLEASE HELP Due today!

BigorU [14]

So i believe is exercise:)

7 0

3 years ago

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

loris [4]

Answer:

The acceleration of the crate is $1.82\ m/s^2$ .

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

$F=ma\\\\F-f=ma$

f is frictional force, $f=\mu N=\mu mg$

$F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2$

So, the acceleration of the crate is $1.82\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

Vsevolod [243]

Answer:

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by $d sin\Theta =1.22\lambda$ --------eqn 1

The angle through which the first minima is diffracted is given by $tan\Theta =\frac{y_1}{D}$ ---------eqn 2

As $\Theta$ is very small so we can write $sin\Theta =tan\Theta$

So from eqn 1 and eqn 2 we can write

$y_1=\frac{1.22\lambda D}{d}$ --------eqn 3

Here $y_1$ is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so $y_1=\frac{14.7}{2}=7.35 \ cm$

Now putting all these value in eqn 3

$d=\frac{1.22\lambda D}{y_1}$

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

8 0

3 years ago

In a simple electric circuit, the current in resistor is measured as (2.50+_0.05)mA. the resistor is marked as having a value of

lesya692 [45]

Here is a hint for you
P=VI=IR*I=I^2R

8 0

3 years ago

Read 2 more answers

A clock battery wears out after moving 10,400 C of charge through the clock at a rate of 0.530 mA. (a) How long (in s) did the c

Sati [7]

Answer:

Explanation:

Given that:

Charge Q = 10400 C

current I = 0.530 mA

From our previous knowledge, we know that charge Q can be expressed as:

Q = I×t

where

I = current and t = time

∴

replacing our values

10400 = (0.530×10⁻³) t

t = 10400/(0.530×10⁻³)

t = 19622641.51 seconds

However, the required flow of electrons per seconds can be calculated by using the formula:

number of electrons/second = current (I)/ charge on electron

= (0.530×10⁻³)/(1.6×10⁻¹⁹)

= 3.3125×10¹⁵ electrons

7 0

3 years ago