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2 years ago

13

Which quantity can be calculated using Newton’s second law of motion?

1 answer:

DiKsa [7]2 years ago

3 0

Newton's 2nd law of motion:

Net Force = (mass) x (acceleration) .

The law shows the relationship among an object's mass
and acceleration, and the net force acting on it.

If you know any two of the quantities in the formula,
the law can be used to calculate the third one.

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Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo

kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

$W=F*d*cos(\theta)$

here we have the work done by Paul and the friction force, so:

$W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J$

$W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J$

Now the change of energy is:

$W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s$

4 0

3 years ago

Read 2 more answers

A target behavior is

crimeas [40]

Answer:

Target Behavior would be the behavior known to be changed, it can be defined by function or topography

3 0

2 years ago

An inductor with an inductance of .5 henrys (H) is to be connected to a 60 Hz circuit. What will the inductive reactance (X L) b

Anika [276]

Answer:

1885.2 ohms

Explanation:

Step one:

given data

L=5H

f=60Hz

Required

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Step two:

Applying the expression

XL= 2πfL

substitute

XL=2*3.142*60*5

XL=1885.2 ohms

4 0

2 years ago

Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues

Llana [10]

Let say the height of two balls from the ground is H

now we can use kinematics

$s = v_i * t + \frac{1}{2} at^2$

now we have

$H = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2H}{g}}$

now in the same time ball on the left will cover the horizontal distance between them

$v_x = \frac{d}{ t}[/tex[tex]v_x = \frac{3}{\sqrt{\frac{2H}{g}}}$

<em>so above is the horizontal speed of the left ball</em>

8 0

3 years ago

Consider a 400 nm thick film of water (n-1.33) on plastic (-1.9). Visible light is incident. on the film firom air. (25 pts) Ske

Gnoma [55]

Answer:

Explanation:

For constructive interference in thin films by reflected light , the condition is

Path difference = 2μd = odd multiple of λ / 2

Where μ is refractive index of water , d is thickness of water medium and λ

is wavelength of light

λ = 4μd or 4μd / 3 or 4μd / 5

= 4 x 1.33 x 400 nm = 2128 nm

or 2128 /3 = 709.33 nm

or 2128 /5 = 425.6 nm

For destructive interference ( dim reflection )

2μd = multiple of λ = λ or 2λ or 3λ

2 x 1.33 x 400 nm =1064

or 1064 x 2 = 2128 nm

or 1064 x 3 = 3192 nm.

7 0

3 years ago